QUESTION:
Can you tell me why/how this snow is holding together and
curling backwards after sliding off a metal roof? Facebook
pic mystery.
ANSWER:
This is not really so rare, at least the sliding off the
roof part. You can see, in the other picture, my son at his
home in Vermont with a similar (not curled) situation. The
roof is warmed from inside and the snow slides down. If
conditions are right, the wet snow on the bottom will
refreeze and hold it from breaking off and also refreezing
to halt the slide. I would guess that the curling would
result from repeated cycles of sliding, pausing, sagging
repeatedly. I cannot really see why the end of the slab
would be being pushed back toward the house.
QUESTION:
When you let a lead ball and an aluminium ball of the same radii to free fall from a height of 100 metres which will ground first?
ANSWER:
Assuming that the balls are not hollow, the lead ball will
have more mass. At a given speed, both balls will experience
the same drag force. But that force will have a bigger effect on
the ball with the smaller mass (Newton's second law). So the
lead ball will reach the ground first. If you are interested
in more detail, a recent answer
is very similar to yours.
QUESTION:
I am a man of some girth, 490 lbs to be exact. I also enjoy "competing" in 5K race events.
As I struggle to maintain a 3 mph pace, Cutler, skinnier people zip past me unphased and undaunted by hills, heat, etc. I have often pondered but do not have the requisite knowledge to figure the answers.
My question is how much force is required to propel a body, weighing 390 lbs, at 3.1 mph. Compare that to someone at 130 and 180...or share the formula.
Also, how can I calculate KCAL consumed at different external temps and humidity.
ANSWER:
To figure out the force is very difficult because running is
a series of accelerations (when you push off with one foot)
and decelerations (when the other foot hits the ground). So
your speed is constantly speeding up and slowing down as you
run and the forces which the ground exerts on you is
constantly changing in both magnitude and direction. It is
daunting to try to think about such a calculation. But, I
believe what you really want is energy expended over a
certain distance in a certain time. I found a handy little
calculator which you can look at
here. For example, for your current weight, a 5K race at
3.1 mph (which is about equal to 5 km/hr, so it takes one
hour for you to complete), I find "You burned 1151 calories
on your 5 kilometer run. At that pace, your calorie burn
rate is 230 calories per mile and 1151 calories per hour."
Sorry, I could not find any information on temperature and
humidity.
QUESTION:
The total amount of energy is always same in the universe. But if there were two observer who were travelling at different velocity,would their measurement of total energy differ?
ANSWER:
Kinetic energy is not invariant, different observers see
different values. Imagine a universe which is entirely at
rest and you are at rest in that universe; the universe has
no kinetic energy. Now imagine that you move through that
universe with some velocity v; you
now observe that universe to have a kinetic energy of ½Mv^{2}.
where M is the mass of the universe except for your
mass. And just a minute ago that energy was not there.
Obviously, the total energy of that universe cannot be the
same in all frames. But, if you stay in one frame and keep
track of the total energy of the universe in your frame, it
will not change.
QUESTION:
if a person dove off the back of a boat going forward at 35mph would that person actually be pulled backwards? I say no he would just de accelerate
ANSWER:
The answer would be easy if you dived from helicopter moving
35 mph—you would have a forward velocity of 35
mph and the water would be at rest, so you would experience
a drag force opposite your motion, backwards, which would
cause you to slow down (decelerate) in the horizontal
direction. But, hydrodynamics can be very complicated, and
it is quite possible that the water behind the boat is
moving with the boat; I cannot imagine, though, that it
would be moving faster than the boat (and you), so it would
not pull you toward the boat.
QUESTION:
Is the pull of gravity the same for every part of Earth? (32 feet per second?)
ANSWER:
The number you are stating is not the "pull of gravity" but
rather the acceleration due to gravity. And, the units are
ft/s/s, not ft/s. But it is an indication of how hard
gravity is pulling, so I can go ahead and answer your
question. We (physicists) usually prefer to work in SI
units, meters instead of feet; in SI units, g≈9.8
m/s^{2}. The acceleration due to gravitation does
vary at various places on the earth. The animation shows you
how it varies. The variation is small, no larger than 50
milligals; a gal is 1 cm/s^{2}=0.01 m/s^{2},
so 50 milligal=0.005 m/s^{2}. So, compared to 9.8
m/s^{2}, the variations are only on the order of 0.05%. There
are several reasons for gravitational variations, notably:

altitude (the farther you are from
the center of the earth, the weaker the gravity);

local geology (large volumes of
more dense earth will increase the gravity);

the earth
is not a perfect sphere.
If you want a
lot more detail, see the
Wikepedia article on the earth's gravity.
QUESTION:
when a stone is thrown up with a velocity and it collides with a particle in mid air(the particle being at rest during collision) , in this situation , is vertical momentum conserved?
ANSWER:
Linear momentum is conserved only if there are no external
forces. But there is an external force acting, the force of
gravity, so the vertical component will not be conserved for
a "real" collision by which I mean one which lasts for some
nonzero time. But, you have stated that the struck particle
remains at rest during the collision; the only way this can
be true is if the collision is instantaneous. So the
momentum will be conserved because the collision lasts for
zero time. The reason this is not a physically possible
collision is that it requires the struck particle to
instantaneously acquire its velocity which requires an
infinite force. Nevertheless, these kind of problems with an
external force present are often approximated as having
momentum conserved if the collision happens very quickly.
Another example is shooting a bullet into a block of wood
sitting on a horizontal surface which has friction; only if
the collision happens very quickly will momentum be
approximately conserved.
QUESTION:
Newton's third law of motion says that for every action, there is an equal and opposite reaction. If i throw a ball towards the wall with some Force it will not Bounce back to me with the same force. Why is that? I mean like if i have a bowling ball and I threw it Towards the ground it will not bounce back. This is driving me crazy
ANSWER:
You are thinking of Newton's third law incorrectly. It
applies only to forces during the collisions, it does not
tell you how fast the rebound will be. When the ball hits
the wall it exerts a force on the wall, and the wall exerts
an equal and opposite force on the ball. But knowing the
force on the ball does not tell you how fast the ball will
be going after the collision. What happens is that some
energy is lost by the ball during its collision so it comes
off the wall more slowly than it went into it. The bowling
ball is an example of the same thing except it loses all its
energy when it collides with the ground. These are called
inelastic collisions, the bowling ball is a perfectly
inelastic collision. If the ball were to leave the wall with
exactly the same speed it came in with, it would be called
an elastic collision.
QUESTION:
What the heck is spin? Quantum objects don't literally have angular momentum, do they? Does the concept of spin have any relationship to angular momentum? Or is it just a name like
"color" that isn’t descriptive of what’s happening?
ANSWER:
They do "literally have angular momentum". So, how can we
observe experimentally whether an object has an angular
momentum? One way is that if you can find a way to exert a
torque on an object which has angular momentum, the torque
will cause the angular momentum to change. For example, a
spinning top has angular momentum and if the top does not
have its symmetry axis vertical, there will be a torque
because its own weight exerts a force which makes it want to
fall down. But, because it has angular momentum, it does not
fall but instead precesses about the vertical. Now, we know
that an electron has an intrinsic angular momentum, a spin.
Presumably because of this spin the electron also has a
magnetic dipole moment—it looks like a tiny bar
magnet. So when we put a magnetic dipole in a magnetic
field, there is a torque exerted on it which tries to line
it up with the magnetic field. This torque, because the
electron has angular momentum, causes the dipole to precess
just like the top did. Without going into detail, it is
possible to infer this precession from measurements. What is
not possible, though, is to understand this intrinsic
angular momentum classically. For example, if you try to
make some classical model imagining a little sphere with
some reasonable radius and mass of an electron spinning on
an axis, you will get nonsensical results like the surface
of the sphere having a speed greater than the speed of
light.
QUESTION:
If a space vehicle left earth at 18000 mph in a direct line to the sun, how fast would vehicle accelerate to taking gravitational pull into consideration?
ANSWER:
The escape velocity from the earth's surface is about 25,000
mph, so your vehicle would fall back to earth.
QUESTION:
If you built a perfect sphere in space that was 187,000 miles wide, the inside was a perfectly polished mirror, and you fired a Laser inside for 1 second, how long would that beam of light bounce around?
Provided the sphere was dust free and a vacuum inside.
This question has bothered me for 35 years.
ANSWER:
You have chosen the distance to be about the distance which
the light would move in 1 second. I cannot imagine the
reflectance of the sphere would be greater than 99%, so if
you solve the equation 0.99^{N}=0.001, N would be the number of
seconds it would take for the intensity of the light to drop
to 0.1% of its original value. Using
Wolfram Alpha, I
find N≈687, a little more than 11 minutes.
QUESTION:
If two objects that weigh the same are attached by a string will they sink at the rate of combined mass or will they sink equally?
ANSWER:
Since you refer to "sink" I will assume that they are
falling in water. There is no simple answer to this question
because the drag force on a sinking object depends on its
geometry. For example, suppose one object is a big flat
sheet and the other is a long thin needle suspended below
the sheet which behaves like a parachute; clearly, this will
sink with a much smaller speed than if both objects were
long thin needles. The most appropriate approximation of the
magnitude of the drag force f in water is a
quadratic
function of the velocity v, f=½ρ_{w}C_{D}Av^{2}, where
ρ_{w}=10^{3}
kg/m^{3} is the density of water, A
is the area presented to the onrushing water as the
object falls, and C_{D} is a constant which depends on the
geometry of the object, called the drag coefficient. For a sphere of radius R
the drag coefficient is given by C_{D}=0.47. The upward force f is
only one force on the object as it sinks; it also has a
weight down W=mg and a buoyant force up B=Vρ_{w}g
where V=4πR^{3}/3
is the volume of the sphere. If the sphere is
solid and has a density ρ=m/V, we can
write B=mg(ρ_{w}/ρ).
Also note that we can write R as R=^{3}√[3m/(4πρ)].
Finally, we can write the total force on a sphere: F=[mg(ρ_{w}ρ)/ρ]+½ρ_{w}C_{D}Av^{2}.
Note that when a speed is reached where F=0, the
sphere will fall with a constant velocity called the
terminal velocity, v_{t}=√{[mg(ρρ_{w})/ρ]/[½ρ_{w}C_{D}A]}.
So, let's do a specific example to be able to answer your
question for a particular situation. Suppose we have two 1
kg balls of solid iron for which ρ=7.87x10^{3}
kg/m^{3}. Then the radius of each ball will be R=0.0312
m. Doing the arithmetic, F=8.56+0.72v^{2}.
So each of two balls of mass 1 kg, tied together will
accelerate until their speeds would be v_{t}=√(8.56/0.72)=3.45
m/s. On the other hand, if we have one sphere of mass 2 kg,
its radius will be 0.0393 m. So, F=17.1+1.14v^{2}.
So one ball of mass 2 kg will accelerate until
v_{t}=√(17.1/1.14)=3.87 m/s.
You could do more examples to come to the conclusion that
the answer to your question is that, in general, two spheres
tied together will not sink at the same rate as one sphere
with their combined mass. I am sure, by carefully tailoring
the shapes of the objects, you could force the rates be the
same, but that is not you would find in general.
QUESTION:
If we go inside a mine and drop a 10 kg iron ball and 1kg aluminium ball from the top of a high platform. what will happen?
ANSWER:
I do not understand the point of going inside a mine. I will
assume that the atmospheric pressure and acceleration due to
gravity are the same as at sea level. In that case the air
drag f can be approximated as f≈¼Av^{2}
where A is the area presented to the onrushing air
and v is the speed. (You will note that Av^{2}
is not the same dimensionally as a force; this approximation
only works if you work in SI units, A in m^{2},
v in m/s and f in N.) So the net force
F on each ball (which I assume to be solid) is F=¼Av^{2}mg
where m is its mass; and, by Newtons second
law, the acceleration a is a=(¼Av^{2}/m)g;
and there is a velocity U, called the terminal
velocity, at which the drag equals the weight so the
acceleration is zero and U=2√(mg/A).
Iron and aluminum have different densities, but it is pretty
straightforward to compute the radii of each
ball and from that compute A for each. I find A_{Fe}=0.014
m^{2} and A_{Al}=0.0064 m^{2}.
The accelerations are a_{Fe}=(9.83.5x10^{4}v^{2})
and a_{Al}=(9.81.6x10^{3}v^{2}).
At any given speed the magnitude of the acceleration for the
iron ball is larger than for the aluminum ball. The terminal
velocities would be U_{Fe}=170 m/s and
U_{Al}=78 m/s. The iron ball always has a
larger speed than the aluminum ball and will hit the ground
first.
It is also enlightening to find the distance traveled as a
function of time, s=(U^{2}/g)ln(cosh(gt/U)).
This may beyond your math ability (ln is a natural logarithm
and cosh is a hyperbolic cosine), but the graph shows the
behavior of each ball and clearly the iron ball is falling
faster. Note that for the first few seconds (6 or 7 s) the
balls pretty much keep pace with each other and after about
15 s both have nearly reached their respecitve terminal
velocities because the curves are nearly linear.
QUESTION:
Would the weight of any object become less as it approached the center of the earth? Neglecting the obvious environmental difficulties of survival could i fall thru a tunnel bored completely thru the axis of the earth and slowly decelerate to the center of the earth. Possibly oscillating back and forth near the center and finally come to rest in the weightless middle?
ANSWER:
Do you want the Physics 101 answer or the answer based on
our best understanding of the density distribution of the
earth? I will give you both.

When this kind of problem is
presented as an exercise in elementary physics, it is
usually assumed that the density of the earth is uniform
throughout—a cubic meter of earth from the surface
has the same mass as a cubic meter at the center. In
this case, it is easy to show that the acceleration due
to gravity, g(r), is linear: g(r)=9.8(r/R)
m/s^{2} where r is the distance from
the center and R is the radius of the earth.
Since the acceleration is always toward the center, you
speed up until you reach the center (not decelerate as
you suggest) and then slow down until you reach the
other end, neglecting any frictional forces. More
realistically there will be at least air drag which will
get bigger as you speed up until you eventually reach a
constant speed; then at the center, you start slowing
down until you stop somewhere before the other end. Then
you speed up back to the center getting there with a
smaller speed than the last time you were at the center,
etc. eventually stopping at the center as
you
suggest. One thing you might note is that, for the
frictionless case, your motion is exactly as if you were
attached to a spring with spring constant k=9.8/R
N/m.

But, the earth is not a uniform
sphere. The core is much denser than the mantle and
therefore the acceleration is not a nice linear
function. The figure shows that the acceleration is
almost constant until about one third of the way to the
center. The above qualitative description of your motion
remains the same except for the similarity to the simple
spring.
ADDED THOUGHT:
I never answered your first question, regarding the weight
of the object. In the uniform density earth the weight (mg)
decreases linearly to zero at the center. In the realistic
model, the weight remains constant for a while, then
increases slightly, then falls almost linearly to zero at
the center.
QUESTION:
I am a professional tennis coach and would like to explain the relationship between mass and spin of a tennis ball. So in my words, I experience hitting the ball with allot of spin makes the ball heavier. When a fast flat ball is hit, I feel it is faster but on contact lighter than the spin ball. Does this mean the increase in rpm, will increase the mass exerted on the inside of the ball, thus making it heavier? And is there a formula for this?
ANSWER:
I am surprised, if you are a professional coach, that you do
not tell me what kind of spin the ball has. The two most
common spins are top spin and back spin (the reversed
direction of top spin); there is also apparently something
called slice spin.
First, though, let me
address your reference to the ball being "heavier" or
"lighter" depending on the spin on the ball. This is totally
incorrect as the ball always has the same weight, a force
which points vertically downward. The way that the ball
moves results from the forces which the ball experiences
after it leaves your racquet; there is the weight straight
down, the force the ground exerts on it when it hits, and
the forces which the air exerts on it. The figure above
shows a ball with top spin; the top of the ball is spinning
in the same direction as the ball is moving to the right.
(If you are confused, the diagram is shown in the frame of
the ball, so the air is shown like a wind blowing in the
opposite direction as the ball is moving.) Here the ball is
experiencing the force of its weight straight down (not
shown), a drag force to the left slowing it down (not
shown), and an aerodynamic force called the magnus force
F straight down. If the ball were
not spinning, there would be no magnus force; if the ball
had back spin, the magnus force would be straight up; if the
ball had slice spin, the magnus force would act
horizontally. (These are all also important for a pitched
baseball.) I am guessing that the ball you judged as
"heavier" is the top spin case because, due to F,
the ball will drop faster than a nonspinning ball. I am
guessing you call the nonspinning ball "flat" and, indeed,
it will not fall as fast as the ball with top spin. A ball
with back spin would fall even less fast than the
nonspinning ball, so you would call it even lighter. (In
fact, I believe that the magnus force on a ping pong
ball with sufficient backspin, can actually be larger than
the weight of the ball.) But in all cases the ball has
exactly the same weight. (I also note that an actually
heavier ball, in the total absense of air, would move
identically to a lighter ball because the acceleration due
to gravity is the same for all masses.)
For
completeness I would like to talk about what happens when
the ball hits the ground because these are strategically
important in tennis. Shown in the second figure are the
forces (in red) which the ground exerts on the ball in top
and back spin cases. For clarity, I have not shown the
forces due to the weight or the air. In each case there is a
normal force N
upwards which is what lifts the ball back in the air; but
there is also a frictional force
f which slows down the spin
in both cases. In the the top spin case, the friction both
slows down the spin and speeds up the ball so it comes up
faster and lower than a nonspinning ball would; in the the
back spin case, the friction both slows down the spin and
slows down the ball so it comes up slower and higher than a
nonspinning ball would. If the spin were large enough, the
back spin ball could actually bounce straight up or even
backwards!
QUESTION:
When an electron absorbs photon does its mass increase?
ANSWER:
I guess you are thinking about the photoelectric effect. A
photon strikes an atom and disappears. An electron is
ejected from the atom. You are thinking that the photon is
totally absorbed somehow by the electron. But what really
happens is that the energy, hf, which the photon
had is given to the electron; the electron ends up with a
kinetic energy equal to hf minus the
binding energy it had in the atom. The rest mass of the
electron remains unchanged.
QUESTION:
"If a ladder you are standing on begins to fall...hold on... it will reach the ground slower than a falling body."
Is the above advise true?
ANSWER:
One could work out an analytic solution to this problem, but
that is not necessary if you just want to be convinced that
the advice is good. All the forces on you are your own
weight, straight down, and the force the ladder exerts on
you. The vertical force which the ladder exerts is upward,
so the net force in the vertical direction is smaller than
if you were just falling.
QUESTION:
Is there anything that can block the effects or decrease the effect of gravity? Any experiments that tested it? Magnetism can be blocked and why not gravity?
ANSWER:
Both magnetic and electric fields can be manipulated to be
zero, but that is much more difficult for gravitational
fields. The main reason is that there are two kinds of
electric charge but only one kind of mass. That is not to
say it is impossible. For example, there is a point between
the earth and the moon where the field is zero, where the
attraction to the earth is equal and opposite the attraction
to the moon. But, in order to have any effect on the field
near the earth, you need to manipulate giangantic masses. As
far as we know, there is no such thing as "antigravity".
QUESTION:
When I carry my cell phone (iPhone 7) in my left pocket and it vibrates, I feel the vibe on my right hand, while my left hand feels nothing. It happens quite often and I even check to see inside my right pocket (or in my tote bag on my right shoulder) I fear of having misplaced my cell there. What does the trick?
ANSWER:
This is not really so surprising. The brain is divided into
two halves and there is some crossing of nerve impulses
sometimes. I used to hear and feel a thumping in my right
ear when I tapped my left big toe. It seems to not do that
anymore.
QUESTION:
If you take out the spaces between all the atoms of the nuclei in your body, would you end up half the size of a flea but still weigh the same?
ANSWER:
Technically, there are no spaces between atoms in your body. I think what you are asking is that if we compressed all the matter in your body (as happens in a neutron star)
to nuclear density, how big would it be? Suppose your body
has a mass of 100 kg. The density of nuclear matter is about
2x10^{17} kg/m^{3}, so your volume would be
about 10^{2}/2x10^{17}=5x10^{16} m^{3}.
So, if you are a cube, your dimensions would be ^{3}√5x10^{16}=8x10^{6}
m≈10 microns. A flea has a size of about 1
mm=1000 microns, one hundred times bigger than the
compressed you. And, yes, you would still have a mass of 100
kg.
QUESTION:
If we are looking back in time through telescopes now at stars and galaxies, would it be possible to calculate the distance required away from earth in order to look back and see a moment in history here on earth, theoretically ?
ANSWER:
Of course, but you could never get there in time to see it.
An alien who was 5000 light years from earth would see, if
he had a good enough telescope, the earth as it was 5000
years before he was looking.
QUESTION:
Since even light cannot escape the gravitational pull of a black hole, does this mean that the theoretical particle, known as the graviton, have the potential to move faster than the photon?
ANSWER:
Sorry this took so long, but I had to consult with an
astrophysicist. The best answer he could come up with is at
this
link.
QUESTION:
Attached is a picture culled from the internet on an object rolling down a plane. I understand everything except how to explain and draw the forces to my AP physics students. Should this look like a box sliding down the plane? with mgsintheta down the plane and mgcostheta as the normal force and mg down and friction up? Or are some of these properly represented as torques?
Could you please draw how this should be properly represented?
ANSWER:
I have not included your picture because there is too much
stuff on it (the velocity, the R, the Rω,
etc. all distract from the crux of the problem, in
my opinion). I have redrawn the problem showing some axially
symmetric object of radius R, mass m rolling down
an incline of angle θ without
slipping. There are only three forces acting on the object,
the frictional force f and the
normal force N, both of which act
at the point or line of contact, and the weight
mg which acts at the center of mass of the
object. There are three unknowns, N,
f, and a, the
acceleration of the center of mass. To solve for these
unknowns, we must apply Newton's laws, both translational
and rotational (ΣF=ma
and Στ=Iα);
here I is the moment of inertial about the axis
about which torque is calculated and α
is the angular acceleration about that axis.
You are a teacher, so I will skip a lot of the details which
you can fill in easily. From ΣF=ma
I find N=mgcosθ and ma=mgsinθ
f. The first equation nails N, one of our
unknowns, the second is one equation for the two remaining
unknowns, so we are not done. So the only place to get a
second equation is the rotational form of Newton's second
law; I think this is the trickiest part of this problem. I find many students
have an inclination to choose the symmetry axis to sum
torques; they have been told, when they learned statics of
non point objects, that you can choose any axis you like,
but this is not a statics problem and there is a very
important constraint—Newton's laws are not valid
in an accelerating system, and the center of the
rolling object is accelerating. (See correction below!) So Στ=Iα
yields Iα=Ia/R=mgRsinθ,
or a=(mR^{2}/I)gsinθ;
knowing a, then f=m(gsinθa)
But, what is I? It isn't what you look up in a book
because usually what is listed is the moment of inertia
about an axis passing through the center of mass (I_{cm}).
Here the axis is a distance R from the center of
mass, so using the parallel axis theorem, I=I_{cm}+mR^{2}.
I will do one example, a solid sphere where I_{cm}=2mR^{2}/5,
so I=7mR^{2}/5, so a=5gsinθ/7
and f=2mgsinθ/7. If you sum torques
about the center of mass, you get the wrong answer.
CORRECTION:
OK, I slipped up here, albeit
in a fairly minor way. While it is true that Newton's laws
are not correct in an accelerating frame of reference, there
is one exception: the sum of the torques about an axis
passing through the center of mass is indeed I_{cm}α.
A little detail I had forgotten about classical mechanics.
Let me work that out for my example: I_{cm}α=2maR/5=fR,
so, using the result above, f=2ma/5=m(gsinθa)
or a=5gsinθ/7. Then it
easily follows that f=2mgsinθ/7.
But you should try to sum torques about an axis halfway
between the point of contact and the center of mass—you
will not get the right answer.
Nevertheless, it is important that your students be careful
in choosing the axis. I have edited the original answer
somewhat.
QUESTION:
How massive is the largest black hole ever discovered?
ANSWER:
See the Wikepedia article List of
Most Massive Black Holes. This is the most
uptodate information I could find.
QUESTION:
Does the core of the earth spin? If is does how does it spin faster than the mantel with the viscosity of magma and the 4+ billions years old the earth is? I just don't understand what I am being told.
ANSWER:
The core of the earth is composed mainly of iron and is
comprised of an inner core which is solid and an outer core
which liquid. Apparently, from what I understand from an
article I found, recent research indicates that the
outer core rotates with the same speed as the mantle (once a
day) but the inner core's rotational speed is slightly
faster by about 0.30.5 degrees per year.
QUESTION:
If I dropped a circular, triangular, and square parachute from the same height which will land first assuming the parachuted object's weight is the same and the surface area is the same for all three parachutes?
ANSWER:
I am assuming that your parachutes are flat, horizontal
plates and that drag forces due to the load and lines
attaching the load to the parachute are negligible or at
least the same. The drag coefficients C_{D}
for the three shapes, determined experimentally, are shown
in the figure. The drag force is given by F_{D}=½ρv^{2}C_{D}A
where ρ is the density of the air, v
is the speed of descent in still air, and A is the
area. Therefore the drag force is least on the circular
parachute so it reaches the ground first. (A proviso is that
the values of the drag coefficient depend on a quantity
called the Reynolds number which depends on the velocity.
However, the values given are for low Reynolds numbers which
are suitable for parachutes.)
QUESTION:
How did Einstein get the idea that the speed of light must be constant for all observer?
ANSWER:
Einstein's original
paper on special relativity, On the
Electrodynamics of Moving Bodies, states two main
problems in electromagnetic theory at the end of the 19th
century:

The first
paragraph states, essentially, that Maxwell's equations
will not have the same form in a frame of reference
moving with constant velocity relative to a frame where
they are the laws governing electromagnetism if you
transform them using Galilean transformations.*

The
beginning of the second paragraph acknowledges "…the unsuccessful attempts to discover
any motion of the earth relatively to the 'light medium' [luminiferous æther,
e.g. the
MichaelsonMorley experiment]…"
Einstein had a
stronglyheld philosophical belief that, for something to be
a true law of physics, it must be the same in all reference
frames—regardless of where you are or how you are
moving, the laws of physics for you are the same as for
anybody else in the universe; this is the principle of
relativity. Since he believed Maxwell's equations to be
laws of physics, he concluded that Galilean relativity was
not correct and this led to his postulate that the speed of
light was independent of the motion of the source or the
observer. To my own mind, I believe that the constancy of
the speed of light is not a necessary postulate, rather a
consequence
of the principle of relativity applied to Maxwell's
equations.
*A Galilean transformation is, for
example, observing a car moving 60 mph east relative to the
ground to be moving 100 mph east if you are in a car moving
40 mph west relative to the ground.
QUESTION:
What type of impact will cause the metal in the car to "splash" (eliminating all hope of a simple conservation of momentum calculation) should the New Horizons interplanetary space probe hit your car?
ANSWER:
The only thing which prevents linear momentum from being
conserved is an external force. If there are no forces other
than those exerted by the car and the probe (or pieces
thereof), linear momentum is conserved. For example, if the
collision happens in empty space with the probe of mass
M moving with velocity V in a
frame where the car is at rest, the total momentum of all
the pieces will be MV after the
collision or any time during the collision. What will not be
conserved is the kinetic energy ½MV^{2}.
QUESTION:
In a recent answer, you mention the Cosmic Microwave Background and say that it "provides a frame of reference for the whole universe".
It was my understanding that one of the underpinnings of the basic theory of relativity is that there CANNOT be such a universal frame of reference (UFOR). I assume you're not claiming Relativity has been disproven (well, shown to be a special case of something else), so can you explain the difference between the frame of reference meant for relativity and the CMB as you described it?
ANSWER:
Good question! Special relativity rests on the assumption
that the laws of physics must be the same in all inertial
frames (the principle of relativity); no frames are ruled
out as long as they are inertial. In other words, no one
frame is better than all the rest. But suppose that we find
a frame of reference which contains a "gas of photons" which
are all moving in random directions (average velocity of all
these photons is zero) and this "gas" pervades the whole
universe. Then, if our frame happens to be moving with some
velocity v through this gas, we
will see that gas having an average velocity v
which would be determined by looking at the doppler shift of
the photons; this is analogous to feeling a wind on your
face when you run in still air. Having found such a frame
allows us to specify an "absolute velocity" which, before
now was not possible. What makes this frame special is the
fact that, because it is a remnant of the big bang, it does
pervade the whole universe.
QUESTION:
I was curiouse at which height would a 6kg turkey have to fall for it to be perfectly cooked when it hits the ground?
ANSWER:
It is very difficult to do a quantitative calculation
because the air drag (which would cause heating) changes
with altitude as the air gets less and less dense. I doubt
that there is any height where this could work because,
giving it a high enough speed to compensate for the short
time (keep in mind that it takes several hours to roast a
turkey at 350^{0}F) of fall would result in the
outside being burnt and the inside being raw. It might be
fun to do a couple of very rough calculations just to get an
order of magnitude feeling for what you are interested in.
I will approximate the turkey to have the properties of
water, in particular that its specific heat is C=4.186
J/gram⋅^{0}C; so the energy which will raise
the temperature of a M=6000 gm turkey is E=MCΔT=6000x4.186x56=1.41x10^{6}
J. (70^{0}F→170^{0}F is 21^{0}C→77^{0}C,
hence ΔT=56^{0}C.)
The change in potential energy in falling from a height
h is V=Mgh=6x9.8h=59h, so if the turkey falls
with constant velocity the potential energy lost goes into
heat. If the heat is 1.41x10^{6} J, the height is
h=1.41x10^{6}/59=24,000 m. The atmospheric
pressure at this altitude is only about 3% what it is at sea
level; so it is clear that, if your turkey is given any
velocity at this altitude, it will not continue moving with
constant velocity. (Whereas, if the pressure were
atmospheric and you gave the turkey a speed equal to the
terminal velocity it would continue moving with that speed
as it fell.) If you dropped it from very far away
(essentially infinitely far), it would arrive at the upper
atmosphere going nearly the escape velocity, 11,200 m/s. At
this speed, which is faster than the reentry speed of the
shuttle which needed special ceramic tiles to shield against
burning up, the turkey would be burned to a crisp, probably
would not reach ground before totally burnt. Although there
would be some speed of entry to the atmosphere where the
turkey would absorb 1.41x10^{6}
J, it would almost certainly not be "perfectly cooked".
QUESTION:
In a nuclear fusion reaction hydrogen atoms combine to create a helium atom. But in a hydrogen atom there are no neutrons, then where does the neutrons in helium nucleus come from ?
ANSWER:
There is more than one answer to this question. We can look
first about how hydrogen (with no neutrons, just one proton)
becomes helium (two protons, two neutrons) in a star like
our sun. As shown by the figure, it begins with pairs of
protons which interact and fuse in a reaction which results
in one deuteron (heavy hydrogen, one proton and one
neutron), one positron (a positively charged electron), and
one neutrino. The deuterons thus created now fuse with
another proton to create a light isotope of helium, ^{3}He
(one neutron, two protons), and also
a gamma ray. Two of these ^{3}He now fuse, throwing
off two protons in the process and one alpha particle (^{4}He,
two protons and two neutrons). The two thrownoff protons
are now free to continue the process, so this chain, called
the protonproton cycle, is selfsustaining until the star
runs out of hydrogen which ends the life of the star.
For more earthbound hydrogen fusion, like fusion reactors
or hydrogen bombs, singlestep reactions are sought. The
most frequently used is the deuteriumtritium reaction,
involving the fusion of one deuteron and one triton (^{3}H,
one proton and two neutrons) shown in the other figure.
QUESTION:
If time ticks at 1 second per second in a 1g gravity at the earths surface. Then my question is would a clock tick twice as fast 2 second per second if the earth surface was 0.5g.
ANSWER:
Most certainly not! The expression for the time t
elapsed in the gravitational field of a nonrotating
spherically symmetric mass M is t=t_{0}/√(1δ)
where δ=2GM/(Rc^{2})=2aR/c^{2};
here, R is the distance from the center, t_{0}
is the time elapsed when R=∞, G
is the universal gravitational constant, M is the
mass of the sphere, and c is the speed of light,
and a is the local acceleration due to gravity. If a=g
and R=6.4x10^{6} m (earth radius), it
is easy to calculate that δ=1.39x10^{9}.
Now, t_{g}/t_{½g}=[√(1½δ)]/[√(1δ)];
using the binomial expansion
(1+δ)^{x}≈1+xδ+…
if δ<<1, we finally get t_{½g}≈t_{g}(1+¼δ)=t_{g}(1+3.5x10^{10}),
quite a bit different from t_{½g}=2t_{g}!
QUESTION:
Why does steam rise from the cooking pot when I turn off the heat after the water starts boiling?
ANSWER:
Water evaporates from the surface and the higher the
temperature, the faster it evaporates. The higher the rate
of evaporation, the more likely it is that individual
evaporated molecules will coalesce with others to form
droplets of water large enough to be seen. That is what
steam is and what clouds are. But steam can be observed even
at low temperatures under the right conditions, for example
the morning mist on the surface of a lake or fog. The water
in the cooking pot, if you observe it carefully, will begin
to steam before it starts boiling. So it should be no
surprise to find that it still steams after you turn off the
heat.
QUESTION:
what makes my radiometer occasionally accelerate wildly when bumped? I can't duplicate the action but have seen it a few times. It has happened around a desk lamp with an LED bulb. Could it be part of an electrical emission around the radiometer? I was allays taught that the photons are what push the vanes
ANSWER:
Well, as it turns out, photon pressure is not what makes the
radiometer turn. You can learn about radiation pressure by
reading an
old
answer and the links it contains. I do not know what
"accelerate wildly" means in this context but, since you
mentioned "bumped", mechanical bumping probably caused
unexpected behavior of the radiometer.
QUESTION:
I wanted to ask a question regarding airfoils. When looking at videos online about them, the instructors say that airfoils work because the top of the wing turns air down due to the coanda effect and the downturned air also turns the air under the wing down to create lift. I think I understand the law applied here to be with every action there is an equal and opposite reaction however, with the explanation they gave, it seemed like they were saying that the air on the top that was turned down was the only thing that is turning the air under the wing down. I that true? I would assume that, even though that does happen, the lift generated from the bottom of the actual wing deflecting the air down was more and more important than the coanda effect on the top. The video also said that disrupting the flow on top of the wing through too high of angle of attack would cause it to stall, but wouldn't the actual reason being that the bottom of the wing converts too much lift into drag by changing the opposing force that is lifting it from mostly up to mostly back? There was another source that said that there was a vortex at the back of the wing that pushed the air at the bottom of the wing so that the net force would cancel out a bit, slowing the air down and making a pressure difference to lift the wing. What is that about?
ANSWER:
This question violates the site
groundrule for "single, concise, wellfocused questions". I can tell you that what is learned in elementary physics courses that airplanes get lift from the Bernouli effect of lower pressure due to higher velocity of air on the top of a wing is a relatively small part of how an airplane is able to fly. As you seem to refer to in your question, the main thing is that air leaves the wing with a downward component
of its velocity; this means that the wing exerted a force down on that air which means, via Newton's third law, that the air exerted a force up on the airplane. To quote Wolfgang Langewiesche, author of
Stick and Rudder,
the private pilot's bible on the art of flying, "The main fact of all heavierthanair flight is this:
the wing keeps the airplane up by pushing the air down."
QUESTION:
How many lightyears or parsecs away from 2018 is 1961? The solar system, the earth and the galaxy would have been in a different part of the universe. Do we know the direction of where our galaxy, solar system used to be before we arrived here in the present moment? How many miles has the Earth traveled in space in fives minutes? Including the movement of the solar system and galaxy moving through the vacuum? Millions of parsecs? Billions?
ANSWER:
It never makes sense to talk about velocity of something
without specifying the frame of reference relative to which
the velocity is measured. And when you are talking about the
earth, the motion is very complicated because it rotates
about its center (about 0.5 km/s), revolves around the sun
(about 30 km/s), and the solar system revolves around the
center of the Milky Way galaxy (about 220 km/s). What about
the motion of the center of the galaxy? Until recently, you
could only measure that speed relative to another object
outside our galaxy, e.g., another galaxy or the
center of our local cluster of galaxies. However, the
discovery of the cosmic microwave background (CMB) has now
given us a reference frame which seems to define the
reference frame of the whole universe. Very careful
measurements and data analyses from the satellite COBE have
shown that our galaxy moves with a speed of about 580 km/s
relative to the CMB.
So, what can we do to answer your question which is,
essentially, how far does the earth move in 57 yr=1.8x10^{9}
s? (Your 5 minute question is essentially impossible to
answer since it depends on the time of year. Over years the
orbital motion averages out to zero.) Relative to the center
of the galaxy, the distance is D=220x1.8x10^{9}
km= 4x10^{11} km=0.013 pc. The center of the galaxy
in that time will move 580x1.8x10^{9} km=10^{12}
km=0.032 pc. But I do not know the relative directions of
these two velocities, the sum of which would be the average
velocity of the solar system over these 57 years; this speed
would be anything between 360 and 800 km/s. However, in my
researching this question I found that the speed of sun
relative to the CMB is about 370 km/s, so the final answer
is that the earth moved about 370x1.8x10^{11}=6.7x10^{11}
km=0.022 pc. It is interesting to me that, at this time, the
sun and the center of the galaxy are moving in nearly
opposite directions (370 km/s compared to 360 km/s); since
the time for the sun to go all the way around the galaxy is
about 225 million years, in about 110 million years our
speed relative to the CMB will be about 800 km/s.
Keep in mind that I am not an
astronomer/astrophysicist/cosmologist! I have gathered these
data as best I can.
QUESTION:
I have a potentially stupid question about pressure in a vacuum. So I just found out that in a vacuum chamber, microns can be used for measuring vacuum pressure because it's representative of a DISTANCE the distance being the displacement of columns of Mercury. In understanding Leak Up Rate tests in a vacuum furnace, the Leak Up Rate is a function of Difference of Vacuum Levels/Elapsed time. My question is, does the displacement of Mercury happen faster with more time? IE, does the displacement of Mercury gain momentum with extended periods of time under vacuum? Or, IE, is it exponential instead of linear?
ANSWER:
Very likely there is not a column of mercury being used to
measure pressure in your chamber. There is probably some
mechanical device using strain gauges which is calibrated in
mm or μm of Hg. So, you do not have to worry about what a
column of mercury is going to do as the pressure changes. 1 μm
(micron of Hg)=0.133322 Pa (N/m^{2})=1.93368x10^{5}
PSI=1.31579x10^{6} atm.
QUESTION:
After a math class yesterday (homeschool) , my son and I were discussing an article I read a few days earlier regarding a exploded star that went super nova about 150 million years ago. The event bathed the earth in a stream of high energy cosmic rays for a couple of months and possibly caused an extinction level event that effected ocean life of the day.
As the conversation progressed, we considered the implications and began to wonder if it was possible to shield against cosmic radiation? To the best of my knowledge, I explained that no substance I know of can block cosmic radiation and that the Earth's magnetic field would have little effect either. The only effects I know of are when they impact atoms in the atmosphere or the earth. If I understand this, if it doesn't hit something fairly dense, it just keeps going, understandable given the space between atoms.
I have read about the gravitational lensing effects on light by massive objects like neutron stars and galaxies that bend light around them. I have also read about the possibility of warping space as a method space travel by changing the shape of space in front of and behind a craft so that it "falls" in a particular direction.
While we were discussing these effects, he looked at me and said something to the effect, "Maybe we could bend space and make them go around?". To me, this was a pretty profound leap in my book. So I decided to ask physicist if he could be right.
Please forgive the length of this post before asking our question. I wanted to provide sufficient context before asking. So, here it is. Is it possible to make cosmic rays go around something, either by gravitational or microwave warping of space?
B.T.W. I really really enjoy your web site. It is awesome!
ANSWER:
I am not a paleontologist, so I had to do a little research
on mass extinctions. I discovered that there is no known
mass extinction 150 million years ago. In addition, no known
extinction is associated with a nearby supernova. (The
timeline here is clickable for an enlarged picture.) I would
conclude that the article which you read is a "fringe
theory" of mass extinctions, not a mainstream fact.
But, I can speak to your main question and comment on some
of your statements.
Cosmic
radiation refers to a broad range of radiations but
mostly protons and light nuclei; but also it could refer to
neutrinos, electrons, positrons, antiprotons, gamma rays,
xrays, and others. What is not the case, though, is that "no
substance…block cosmic radiation"; just about any
cosmic radiation can be blocked by a dense solid or liquid.
Neutrinos are the exception and most pass all the way
through the earth without any interactions. You may know
many neutrino experiments are performed in deep abandoned
mines to block out all other radiation. The atmosphere also
is very effective since a single energetic proton will
strike an atom and many other particles, all with less
energy than the original proton, will form a "shower" of
particles. And magnetic fields are not very effective
deflecting the very highest energy charged particles but are
known to be an effective shield for lower energy particles.
Finally, let's address
the question of warping of space as a way to deflect the
rays. The only way to warp space that I know of is to have
an enormous mass nearby. The earth already warps the space
around it which would cause a particle passing close by, but
not on a collision course with earth, to be deflected a tiny
bit (viz. gravity). But putting a star sized object near
enough the earth to "shield" us would obviously be a
catastrophic thing to do! The whole solar system would be
disrupted. I have no idea what "microwave warping of space"
means.
I commend your son for his hypothesizing and you for your
tutoring and encouragement of his curiosity.
QUESTION:
I understand that LED light won't refract a full spectrum of colors through a prism. Is this true? I want to combine a grouping of different wave length diodes to deliver a white light that will separate from indigo to red. Can this be done?
ANSWER:
Although a simple LED will not be monochromatic like a
laser, it will tend to have a fairly narrow range of
wavelengths emitted, not a full spectrum. The most common
way to
make a white LED is to take a blue or nearUV LED and coat
it with a material called a phosphor. The phosphor will
absorb some of the light from the LED and reemit that energy
in a broader spectrum creating white light. The details of
the spectrum depend on the phosphor used; varying the
phosphor used, for example, can create the whites described
as cool white or warm white or bright white; two of these
are shown in the figure. I do not understand what "separate
from indigo to red" means.
QUESTION:
As I understand it, electricity and magnetism are the same force; this is where "electromagnetism" comes from. I also understand that radio waves, light, and nonparticle radiation like xrays are all the same thing; waves in the magnetic field. However, they're carried by different particles: electrons for electricity, photons for magnetism (radio, light, etc). Furthermore, there is a significant difference between electrons and photons; electrons have mass, photons do not.
Wouldn't this mean that electricity and magnetism are actually separate forces which are so tightly coupled (have a very strong influence on eachother), that it's just easier or simpler to think of them as the same force most of the time? One example of this coupling in my hypothesis is how electrons gain energy when a photon collides with and is absorbed by an electron, and the reverse, when an electron emits a photon and looses energy.
ANSWER:
You have some serious misconceptions here. First, the waves
you refer to are not "waves in the magnetic field", they are
waves which are composed of magnetic
and electric fields—they
are electromagnetic waves. Second, both electric and
magnetic forces are "carried by" photons—electrons
never play that role. Indeed, there is only one field, the
electromagnetic field; for example, if you have a charged
particle at rest you will only see an electric field, but if
you give the charge some velocity you will now also see a
magnetic field and you will find that the electric field is
somewhat smaller which you could interpret as having
"morphed" into the magnetic field.
QUESTION:
I want to reach 90%
c in 30 days. How many Gs would I have to accelerate at to reach that velocity in the given time? (in a vacuum)
ANSWER:
You should read an
earlier answer
to understand the answer
here. The graph here is from that earlier answer and shows
(in red) v/c as a function of a_{0}t/c
where a_{0 }is the acceleration measured in
the frame of the object which is accelerating and t
is the time since the object was at rest. Reading off the
graph, you can see that v/c =0.9 when
a_{0}t/c=2. Now, t=30
days=2.6x10^{6} s, so a_{0}t/c=a_{0}(2.6x10^{6}/3x10^{8})=8.64x10^{3}a_{0}=2;
solving, a_{0}=232 m/s^{2}=23.6 Gs. I would not
suggest having anybody aboard this object!
QUESTION:
Everything being equal,including speed and mass, Like a man walking up a slope, does the amount of energy expended on say a 10% inclined slope double with a 20% inclined slope?, if traveling at the same speed and everything else the same?
In other words, is energy expended related arithmetically and constant according to slope of incline. and resistance, or is it proportional to the slope by some other factor, geometric increase.?
ANSWER:
In the figure the man of mass m is walking up a
hill with incline θ with a constant
speed of v. His potential energy is E=mgy
where y is his height above the bottom of the hill
and g=9.8 m/s^{2} is the acceleration due
to gravity. The rate at which his energy is changing is dE/dt=P=mg(dy/dt)=mgv_{y}=mgvsinθ;
so this rate (called power) depends only on the vertical,
not the horizontal component of his velocity. Now, you have
made things difficult for me because you have not specified
the angle, but rather the inclination which is actually the
tangent of the angle (rise/run) times 100%. I will call this
inclination the grade G=100tanθ. For
example, let's take a man with mass 100 kg walking with a
speed of 2 m/s up the 10% and 20% inclines you specify. The
angles would be θ_{10%}=tan^{1}(10/100)=5.71^{0}
and θ_{20%}=tan^{1}(20/100)=11.31^{0}.
So the power values are P_{10%}=100x9.8x2xsin(5.71)=195.0
Watts and P_{20%}=100x9.8x2xsin(11.31)=384.4
Watts. The ratio of these two is 1.97, close to, but not
equal to, 2.0; for small angles the sine and tangent are
approximately linear functions but as the angle (hence
inclination or grade) increases this is no longer a good
approximation. The graph shows the power for grades up to
200% (about 63.4^{0}); as you can see, the power
increases approximately linearly only up to about a grade of
about 30% which is about 17^{0}.
QUESTION:
A vehicle , unknown speed of travel, strikes a 14lb object and displaces it
29ft. Could you please tell me what the approximate speed of travel is and
how you come to that conclusion. I ask because a vehicle deliberately veered
to the opposite side of the street in order to run over my cat. The driver
was successful in killing my sweet lad and I am trying to collect as many
facts as I can before I request help in prosecuting the individual under
Arizona's Animal cruelty laws.
ANSWER:
I am afraid that you cannot deduce the speed of the vehicle from the
information you gave me. You need to know how elastic the collision was and
what the trajectory of the cat was.
FOLLOWUP QUESTION:
Thank you for responding to my question but I am at a loss as to what is meant by "elastic". As far as trajectory my cat was launched a distance of 29 ft from point of impact with the next point where he died on the street...… no evidence, ie; of blood in between with there being blood evidence only at the point of impact and final resting location. If it would help I can shoot a video of the street and the location of where Liam was struck and where he landed in relation to the surroundings.
ANSWER:
Elastic refers to the energy lost in the collision which
gives you information on how fast the cat was moving
immediately after the collision. The extremes are a
perfectly inelastic collision where the cat would be stuck
to the vehicle after the collision and perfectly elastic
which would have the cat moving with approximately twice the
speed of the car after the collision. Clearly, neither of
these would describe a car hitting a cat and I cannot
imagine how you could approximate it.
When I say "trajectory"
I refer to the path from collision point to the ground. The
horizontal distance gone from the point of collision to the
point of landing depends on the angle at which the cat
started; if launched at 45^{0} relative to the
horizontal he will go much farther than if launched at 20^{0}.
Again, unless you witnessed the accident, there is no way to
know the angle.
QUESTION:
If you are traveling on a spaceship close to the speed of light (pick a number e.g. 0.95c), With or without constant acceleration whichever works for the question.
Time will slow down relative to earth and you will be able to travel a greater distance than you would be able to without relativistic effects. You would also be able to return to earth with significant time passed relative to your experience.
My question is, what would it feel like on the spaceship...I know the standard answer is you wouldn't feel anything, but logically, if you were traveling through galaxy after galaxy in a relativistic time of human life then intuition says everything would look like it whizzing past.
Basically, what is the human experience of being able to travel interstellar distances?
ANSWER:
Of course
everything is whizzing past compared with what it would look
like if you were at rest relative to the earth. What you
would also see is that the distance to an object you were
moving toward is shorter by a factor of √(10.95^{2})=0.31;
so if you went to a star 1 light year from earth, it would
only take you 0.31/0.95=0.33 years to get there but the
earthbound clock would say it took you 1/0.95=1.05 years.
Also where the stars appear to be relative to where they
would be if you were not moving would be different; see an
earlier
answer. In the case where v/c=0.95, the graph shows
how the angle at which you see something depends on where it
would be if you were at rest; for example, a star actually
at θ=120^{0} (30^{0}
behind you) would be seen as being about θ'≈30^{0}
in front of you!
QUESTION:
how much times earth to be compressed to become black hole?
ANSWER:
A black hole is a singularity, of zero size, so compression
is infinite. If the earth were a black hole, its
Schwartzchild radius, R=2GM/c^{2},
would be about 1 cm. R is the radius within which
nothing can escape. So, if you could get the earth down to
that size, it would then collapse the rest of the way to a
black hole, I presume.
QUESTION:
If momentum is applied not on the center of mass on an object, what would its momentum be at the center of mass and is there an angular momentum at the point of impact?
ANSWER:
Linear momentum is not something which is applied, force is
what is applied. And, what happens also depends on the
details of how the force is applied. I will work out one
simple example which will illustrate the principles involved
when the force is applied in a simple collision. A mass
m approaches the end of a thin uniform rod of mass
M and length 2L with velocity v
perpendicular to the rod; when it collides with the end of
the rod it sticks. The center of mass of the rod is at its
geometrical center a distance L from where m
exerts the force. Linear momentum and angular momentum are
both conserved but, as we shall see, energy is not
conserved. After the collision the center of mass has a
velocity u; conserving linear momentum, (M+m)u=mv
or u=v[m/(M+m)]. Before the
collision m brings in an angular momentum mvL.
After the collision, the rod plus mass has angular momentum
(I+mL^{2})ω
where I=M(2L)^{2}/12=ML^{2}/3
is the moment of inertia of a thin rod about its center of
mass. Conserving angular momentum I find ω[m+(M/3)]L^{2}=mvL
or
ω=mv/{[m+(M/3)]L}.
Just for fun I put in some numbers to illustrate what would
happen in a specific situation: m=1 kg, M=3
kg, L=1 m, v=4 m/s. With these I find:
u=1 m/s, ω=2 radians/s=19.1 rpm. You can
also calculate the energy before and after the collision:
E_{before}=½mv^{2}=8
J, E_{after}=½(M+m)u^{2}+½([m+(M/3)]L^{2}ω^{2}=6
J. So, energy was not conserved in this collision, 2 Joules
were lost.
CORRECTION:
I realized that an error had been made in my analysis. After
the collision the center of mass is no longer at the center
of the rod but displaced a distance x=mL/(M+m)
toward where m is stuck. It is this center
of mass which moves forward with speed u, and u
is still v[m/(M+m)]. The rotation
is about this center of mass but now the moment of
inertia is different; the net (rod plus mass) moment of
inertia is ML^{2}/3+Mx^{2}+m(Lx)^{2},
so the corrected value of the angular velocity is
ω=mvL/[ML^{2}/3+Mx^{2}+m(Lx)^{2}].
In the example I did, x=1/4 m, so ω=16/7 radians/s=21.8 rpm and E_{after}=46/7=6.57
J (1.43 J lost).
QUESTION:
I have been trying to figure this out, but having some trouble doing it: In general, let's say I had a vehicle that could accelerate me (according to my perception) at 1G for 1 year (again according to my perception),
then turn around and decelerate (by my perception again) for another year to return home. How far would I have travelled and how old would my twin at home be when I got back?
ANSWER:
The reason you are having trouble is that relativistic
kinematics involving acceleration, even if the acceleration
in one frame is uniform, is not easy; the main reason is
that, unlike Newtonian mechanics, acceleration in one frame
is not the same as acceleration in a different frame. As a
prelude to reading my answer, you should read an
earlier answer
and the earlier answers referred to in that answer. The
graph here shows the results derived or cited in that
earlier answer. The three graphs all show what is observed
of your motion as seen by the twin on earth. The red curve
is your velocity v relative to c as seen
by the earthbound twin, v/c=(gt/c)/√[1+(gt/c)^{2}];
the blue curve is your acceleration a
relative to g, a/g=[1+(gt/c)^{2}]^{3/2};
the black curve is your position expressed (approximately)
in light years, gx/c^{2}=√[1+(gt/c)^{2}]1.
Now, since I have set this up as seen from earth, let me
specify the time you travel as the time of your trip out as
observed from earth: earthbound twin observes you to arrive
at you destination in two years. Then, as you can read off
the graph, gx/c^{2}=√[1+(2)^{2}]1=1.24
ly; the return trip will be the same length of time, so the
earthbound twin's clock will have advanced 4 years. Light
would take 2.48 yr to travel that distance, so your clock
will have advanced a time somewhere between 2.48 and 4
years. Your maximum speed would have been about v/c=(2)/√[1+(2)^{2}]=0.89.
We can get a rough estimate of your clock reading by
approximating your average relative speed as 0.89/2=0.445;
so you would have seen, on average, the distance contracted
by √[1(0.445)^{2}]=0.9 and your time would
have been 0.9x4=3.6 years.
QUESTION:
What happend to accelaration due to gravity if the earth stops rotating?
ANSWER:
Because of the centrifugal force, g appears to be
somewhat smaller than it really is, but the amount is very
tiny. This effect depends on your latitude and is zero at
the poles and greatest at the equator. Also because of this
effect, the acceleration is not directly toward the center
of the earth except at the poles and equator.
QUESTION:
Hello, is the mass of a person on a spaceship (to provide a suitable upward force) important in designing a spaceship? My teacher says otherwise but upon researching I found out that additional fuel is required for every kg of payload that will be sent out into space.
ANSWER:
If the ship and all the fuel are much bigger than the mass
of the person, then you could neglect the mass of the person
in any calculations you know; in other words, the rocket
launch would be, as close as you could hope to be measure,
identical whether or not an astronaut was on board. However
you are also right that the amount of fuel you need depends
on the payload you want to deliver, and that would include
everything which is not fuel. You might find the an
earlier answer
illuminating.
QUESTION:
How exactly do electrons in electric discharges generate radio waves?
Whenever someone uses an AM radio while there are electric sparks like lightning strikes nearby the radio waves they emit can be detected.
ANSWER:
Any time an electric charge is accelerated it emits
electromagnetic radiation. There are many electrons in the
spark which accelerate under the influence of the electric
fields they experience.
QUESTION:
I just read about the kilogram being defined by measuring the current of an electormagnet to balance a scale. How can various electromagnets have exactly the same force to current ratio? Would variations in the purity of the conductor, variations in the exact coil shape, etc. cause variations in this ratio? What am I missing about this new designation?
ANSWER:
You cannot use any old electromagnet. There is an
exquisitely accurate balance called the
Kibble balance which is used. Also see
Wikipedia for more detail
QUESTION:
Is the following statement correct?
Net radiated energy is made up of emitted and absorbed energy.
I stated that it was incorrect based on the fact that
"made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that “made up of” means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?
ANSWER:
I do not believe that there is a hard and fast definition of
what "net radiated energy" is. It is a matter of semantics.
However, radiated energy usually means energy emitted and
absorbed energy means energy absorbed; therefore, I would
say that this is an incorrect statement because net radiated
energy would mean the sum of all radiated energy, absorbed
energy having nothing to do with it. Your reason is quite
shaky because "made up of" is even more semantically
ambiguous than "net radiated"; it would simply imply that
"net" means total energy flux.
QUESTION:
How is momentum conserved for single or double slit diffraction? in other words:I have a photon gun that I can dial the intensity down to fire one photon at a time. I fire my photon gun, the gun recoils in the X direction, the photon flies off in the +X direction. The photon then passes thru two closely spaced slits (double slit experiment)instead of appearing on the screen directly behind the slits, the photon ends up to the side (after many, many photons the interference pattern appears) however, for the single event of a single photon, how is momentum conserved? dos the single event photon exchange momentum with the walls of the slit? Since the photon ended up to the side, it was no longer travelling in the purely X direction, therefore shouldn't the photon have exchanged momentum with the slit somehow? If the photon interacts with the slit, that counts as “which way” information, and the interference pattern vanishes. If the interference pattern remains, then how do we account for the photons change in momentum? This is for a single event, one photon, not expectation value of many, many photons.
ANSWER:
You are forgetting that the photon goes through both slits,
is being forced (if you like) to assume its wavelike
identity. It behaves like a wave until something makes it
act like a particle, so it is a wave spreading out in both
directions, so its net momentum in the direction parallel to
the screen is zero. When the screen is encountered it is a
"measurement" which collapses the wave function at some
point.
QUESTION:
I can imagine 2 similar objects rotating around a sphere at the same velocity but at different orbits. One object would rotate around the sphere faster than the other, right? What is puzzling to me is that if I think of both objects going now in a straight line one could not say that one object was going faster than the other. Can someone help me through this? Is there a subtlety here?
ANSWER:
You want to say same speed, not same velocity, but I get the
idea. Say one orbit is twice the radius as the other. Then
the smaller orbit has a circumference half that of the
larger. So each time the object in the larger orbit goes
around once, the one in the smaller orbit goes around twice.
But they still have the same speed. But the angular
speed of the object in the smaller orbit is twice that
of the object in the larger orbit even though their linear
speeds are the same. You cannot do this with planets or
moons, though, because the speed of a satellite in a
circular orbit depends on the radius.
QUESTION:
If space is curved, doesn't it need another dimension to curve into? If a 2d object is curved it curves into the 3rd dimension. So if 3d space is also curved it must curve into 4d.
ANSWER:
This
is really mathematics, not physics; I am not a
mathematician. But the definition of an Ndimensional space
is, I believe, that N quantities are required to specify the
location of a point in that space. Curvature of a space is
defined as a space in which the geometry is not Euclidean.
The physical existence of an Ndimensional space does not
imply the physical existence of an N+1 dimensional space.
You cannot make a generalization that curvature is
necessarily "curving into" a higher dimension. For
example, a planar sheet can be distorted without leaving the
2D. You also might "curve into" a higher dimension than N+1;
for example, a coil spring is a one dimensional space which
exists in a 3D space. You are trying to make generalizations
based on the only spatial dimensions which we know to exist;
but higher dimensions are defined by certain rules which in
no way prove that they exist physically.
While
researching this question I came upon a very interesting
little
essay about general relativity, curvature of
spacetime, gravity. Many nonscientists (and scientists as
well) are familiar with and accept as gospel that general
relativity demonstrates that the distortion of spacetime by
the presence of mass is what gravity is; the "cartoon" which
illustrates this is the "bowling ball on a trampoline"
example in which a mass causes the distortion of the
twodimensional space defined by the trampoline surface. The
implication is that gravity is just geometry. On the other
hand, many theorists are struggling to find a successful
theory of quantum gravity and to do that one must treat
gravity as a force. The takeaway from this essay, though,
is that curved spacetime is not a unique representation of
what the mathematics which are the framework of the very
successful theory of general relativity mean. The author,
Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT
[quantum field theory], exists in threedimensional space and evolves in time according to the field equations."
He also goes so far as to say "…shame on those who try to foist and force the fourdimensional concept onto the public as essential to the understanding of relativity theory."
QUESTION:
A rear wheel drive car accelerates quickly from rest. The driver observes that the car noses up. Why does it do that? Would a front wheel drive car behave differently?
ANSWER:
The car with rearwheel drive has an acceleration
a and a mass M. The forces on
the car are its own weight W (W=Mg)
acting at the center of mass a distance d from the
real axle; the normal and frictional forces on the rear
wheels, N_{1} and
f_{1}; and the normal and
frictional forces on the front wheels, N_{2}
and f_{2}; the radius of
the wheels is R and the distance between front and
rear axels is D. I will ignore the rotational
motion of the wheels, that is, I will approximate their
masses as zero. If the front wheels are not lifting up from
the ground, the sum of torques about the rear axel is (f_{1}f_{2})R+N_{2}DMgd=0;
Newton's first law for the horizontal and vertical
directions are (f_{1}f_{2})=Ma
and N_{1}+N_{2}Mg=0.
From these equations it is easy to show that N_{2}=M(gdaR)/D.
This is interesting because it shows that N_{2}=0
if a=gd/R; for larger accelerations, the
front wheels will lift off the ground. (Note that as N_{2}
approaches zero, f_{2} also approaches
zero.) I have ignored the suspension of a real car, so this
lift will not be abrupt; it will happen gradually as the
front springs decompress and the rear springs compress; so
overall the nose goes up but the rear end goes down.
If
it is a frontwheel drive, the directions of the frictional
forces reverse. The corresponding torque equation (this time
about the front axle) is
(f_{2}f_{1})RN_{1}D+Mg(Dd)=0;
Newton's first law for the horizontal and vertical
directions are (f_{2}f_{1})=Ma
and N_{1}+N_{2}Mg=0.
From these equations it is easy to show that N_{1}=M(aR+g(Dd))/D.
As a increases in magnitude, N_{1}
gets bigger and bigger until N_{1}=Mg
which necessarily means that N_{2} becomes
zero. It is easy to show that this happens when a=gd/R,
exactly the same result as for the rear wheel drive.
Finally,
it might appear that there is no difference between front
and rearwheel drive. That is not the case in the real world
at all. Keep in mind that the frictional forces are static
friction and the larger the corresponding normal force is,
the more friction you can get. Since in the rearwheel
situation the normal force gets bigger, the amount of
friction you can get without the wheels slipping can keep
increasing so that eventually (with a powerful enough
engine) you could lift the whole front end far off the
ground. However, when you have frontwheel drive, the normal
force on the front wheels gets smaller as you increase a,
so most likely the wheels will slip well before N_{2}=0
when f_{2} must be zero. For very large
accelerations, like required for drag racers, rearwheel
drive is imperative.
QUESTION:
Gravity is what makes objects orbit around other objects,and gravity is a reflection of an object's mass.So why doesn' the mass of the objects appear in kepler's third law??
ANSWER:
For the same reason why all objects have the same
acceleration in the gravitational field, i.e. all
objects near the surface of the earth have an acceleration
of 9.8 m/s^{2}. The acceleration (which is what
describes the orbit) is F/m but the force
is MmG/r^{2}, so the acceleration
is MG/r^{2}.
QUESTION:
If kinematics can give you data on a dropped object, would it be valid to use kinematics to calculate the acceleration experienced by the object upon impact with the ground? (the time from when it hits the ground to when it actually stops)
ANSWER:
When the object hits the ground until its velocity reaches
zero is some time, call it Δt. Its
momentum drops from mv (where m is the
mass and v is the speed when it hits) to zero
because an impulse I was delivered to it by the
ground. The impulse may be written as I=F_{avg}Δt.
where F_{avg} is the average
force. Therefore you can find the average acceleration a_{avg}
during the collision, but not the acceleration as a function
of time: a_{avg}=F_{avg}/m=I/(mΔt)=v/Δt.
So you cannot get detailed kinematics about the
collision without measuring how the speed decreases with
time.
QUESTION:
We know that air density decreases as you ascend into space, but what about looking out to the horizon? Is there a atmospheric lensing that occurs? Is there a formula for its calculation?
ANSWER:
Certainly there is refraction due to this density profile of
the air. There is no simple formula because the actual
density at any point is dependent on the pressure and
temperature of the air as well as altitude, and these vary
with time. One could calculate based on average density
profiles, but I would suppose it is a numerical calculation,
not an analytic formula. There is a nice description of this
phenomenon
here.
QUESTION:
I've been interested in physics for years but have never managed to get my head around the concept of spacetime. I understand time and i understand space but somehow i could never get this. However today i think it finally clicked when i wasn't even trying. Can you tell me if my understanding of it is correct or not?
I thought of spacetime as like if i were on a rollercoaster or train tracks. When i am on the long flats, i travel slower, and when i am on the declines, i travel faster. Is spacetime like this analogy? In that, when you are on Earth for example, you experience time moving faster than if you were floating in space millions of miles away. This is because the presence of Earth "warps" space time, much like a bowling ball on a trampoline. This "dent" in spacetime causes a gravitational pull, pulling time into it, much like if i were rolling down a hill.
I know that might sound very convoluted but it's the idea i have in my head and it has bothered me for YEARS, but i think i am on to an understanding here.
ANSWER:
Sometimes we make things more difficult than they need to
be. The main takeaway from the theory of relativity is that
time and space are not separate and universal things. In
Newtonian times, it was assumed that any clock in the
universe would run at exactly the same rate; now we know
that if a clock moves with respect to us that it runs more
slowly than ours—doesn't appear to run slower,
it runs slower. Similarly, space depends on the observer; a
meter stick moving by you is actually shorter than 1 m. What
these facts about space and time tell us is that time and
they are not different things, they are really entangled
with each other. You cannot talk about one without talking
about the other. They are two pieces of the same thing. So
the term spacetime is simply a way of acknowledging that. I
do not find your roller coaster analogy very helpful. The
bowling ball on a trampoline is a classic but keep in mind
that it is really a cartoon and not rigorous. Generally
trying to picture something in four dimensions futile!
QUESTION:
What does a wind speed (a gust is okay)in miles per hour, need to be to launch a baseball at a speed of 80 miles per hour into an object like a window.
Given:
Ball is 60 feet 6 inches from the window that it hits traveling at 80 mph Baseball is 9.25 inches in circumference and 2.94 inches in diameter and weighs 5.25 ounces
My name is Karin and I'm testing to see if I use a baseball pitching machine from 60 feet 6 inches from a window protected with a protective panel, and set the machine to pitch at 80 mph, what would the hurricane wind speed need to be to get that baseball to fly into the window at 80 mph. I hope this is clear enough.
ANSWER:
If I understand your question, you are essentially asking
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 80 mph in a distance of 60.5
ft due to the force of the wind on the ball. This is a
problem which is very tedious to do in full analytic detail
and not very illuminating. Also, all problems involving air
drag are approximate, so it would be appropriate for me to
find an approximate way to solve this problem. I am
going to work in SI units so I will rephrase your question:
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 35.8 m/s in a distance of 18.4
m due to the force of the wind on the ball? In SI units you
may approximate the force at sea level which the wind exerts
on the ball as F=¼A(uv)^{2 }
where u is the speed of the wind, v is the
speed of the ball, and A is the cross sectional
area of the ball. Applying Newton's second law, ¼A(uv)^{2}=ma=mdv/dt
. This differential equation may be shown to have the
solution t=cv/[u(uv)] where
c=4m/A and t is the time it takes the
ball to reach the speed v if the wind speed is
u. I find that c=137 s^{2}/m. Note
that after a very long time the speed will be constant at
the wind speed; we are only interested in the time it takes
to get to 35.8 m/s (80 mph). I have plotted t for
three values of u, 70, 80, and 90 m/s.
Now comes the hard part. We are interested in finding the
value of u for which v will be 35.8 m/s when the ball has
gone a distance x=18.4 m. This means that we need
to write the equation for t as
t=c{dx/dt}/[u(u{dx/dt})]
and solve that differential equation for t as a
function of x. That is actually doable, but the
solution is so messy that I looked for a simpler way to do it.
This took some trial and error, but I got a pretty good
estimate, I believe. Look at the black line in the graph; it
is not a straight line, but it is pretty close, so I am going
to approximate the acceleration as being constant over this
time interval of 1 s, a≈(35.8 m/s)/(1 s)=35.8
m/s^{2}. Now, an object with constant acceleration
starting from rest goes a distance d=½at^{2}
in a time t, so if u=90 m/s=201 mph the ball will have a
speed of 35.8 m/s when d=35.8/2=17.9 m=59 ft. As
far as I am concerned, I have adequately solved this problem—it
takes a wind of approximately 200 mph to accelerate a
baseball from rest to the speed of 80 mph in a distance of
about 60 ft.
QUESTION:
Why is light speed the fastest speed and why would people see a train going backwards if it is traveling at a velocity greater than light speed?
ANSWER:
I will not answer the second question because the site
groundrules state
that I will not answer questions of this sort. Your first
question is equivalent to the question "why can nothing with
mass travel as fast or faster than light?" which is included
in the FAQ page.