# Acceleration of a tire rolling

## Homework Statement

So a tire rolls with radius r. The CM of the wheel travels at a velocity of vcom.

What is the acceleration of the top of the wheel and the bottom of the wheel relative to a passenger in the car and to a street sign on the street?

a = v^2/r

## The Attempt at a Solution

Relative to the passenger, the top and bottom of the wheel are traveling at a velocity of vcom and -vcom. So the magnitudes of both accelerations are vcom^2/r.

Relative to the street sign, the top of the wheel is traveling at 2vcom. The bottom of the wheel is traveling at 0. I know that these are the wrong velocities to plug into the equation a = v^2/r since the answer for the accelerations are the same as the accelerations in the previous situation.

A friend told me that centripetal acceleration is always relative to the axle. I thought that perhaps since the translational acceleration of the top and bottom of the wheel are 0, and since the acceleration of the wheel when it is rotating is vcom^2/r for the top and bottom, the magnitudes of the accelerations of both the top and bottom could be Arotation + Atranslational = ARolling,
which would be vcom^2/r.

Are any of these explanations correct?

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ehild
Homework Helper
Relative to the sign post, the CM travels with constant velocity. The tire rotates around an axis through the CM and travels also with the same velocity as the CM. It keeps this component of velocity, so any change of the velocity of the tire comes from the rotation. Therefore the only acceleration is the centripetal one, which magnitude is related to the speed of the tyre with respect to the axis. So its magnitude is vcom^2/r, the direction points inward, towards the CM.

ehild

Okay, the centripetal acceleration is relative to the axis. Is centrip. acceleration always relative to the axis or is it because in this case centrip. acceleration is the only acceleration acting on the wheel?

tiny-tim
Homework Helper
hi joej24! (try using the X2 icon just above the Reply box )
A friend told me that centripetal acceleration is always relative to the axle.
rubbish!

acceleration is acceleration!

it isn't relative to any point

(it is of course relative to velocity, but not to any point at that velocity)

when dr/dt = 0, the acceleration perpendicular to the velocity is always speed2/r, where r is the radius of curvature of that piece of the wheel

(perhaps your friend meant that v2/r is always relative to the centre of curvature?)

in this case, relative to a stationary observer, the speed (for the top of the wheel) is 2v, but the r is unknown to you (unless you're familiar with the geometry of a cycloid ) …

that's why it's much easier to choose a frame in which you know the radius of curvature without having to do complicated geometry! ok, now here's a fun test for you …

what is the radius of curvature of the path (relative to the ground) of the piece of metal presently (a) at the top of the wheel (b) at the bottom of the wheel? Is the radius of curvature the same as the distance from a point to the center, or point of rotation? In that case
(a) r
(b) r

tiny-tim
Homework Helper
no

a point on the rim of a wheel traces out a cycloid

see the pretty picture at the top of http://en.wikipedia.org/wiki/Cycloid" [Broken] …

you need the radius of curvature at the top and bottom of that cycloid Last edited by a moderator:
2r and 0, which concur with the wheel's velocity relative to the ground, 2v for the top and 0 for the bottom. But it doesn't make sense that the acceleration of the top is
(2v)2 / 2r

and

02 / 0

for the bottom

SammyS
Staff Emeritus
Homework Helper
Gold Member
2r and 0, which concur with the wheel's velocity relative to the ground, 2v for the top and 0 for the bottom. But it doesn't make sense that the acceleration of the top is
(2v)2 / 2r

and

02 / 0

for the bottom
The radius of curvature of the cycloid is 4r.

Although, the point of contact at the bottom of the tire is is stationary for an instant, immediately before contact the point moves vertically downward, immediately after contact it moves vertically upward.

ehild
Homework Helper
v2/r is the normal component of acceleration (an) of a selected point on the tire along the cycloid, and r is the radius of the curvature of the cycloid at the selected point. But the acceleration at that point has also tangential component at which is the time derivative of the speed. The magnitude of the overall acceleration was asked, it is √(at2+an2).
The calculation is quite tiresome but you will have the result that the magnitude of acceleration is the same in both systems with respect to the car and with respect to the ground.

It is the consequence of Newton laws: all inertial frames of reference are equivalent, the forces do not change if you choose an other frame of reference. The same holds for the acceleration.
One frame of reference ( the car) moves with velocity V with respect to the other one (ground). An object (one piece of the tire) moves with velocity v' in the moving frame of reference. The velocity of the object seen from the frame of reference in rest is v=V+v'. The acceleration is the time derivative of the velocity. As V is constant, dv/dt = dv'/dt, the acceleration is the same in both inertial frames of reference, car and ground.

ehild

Last edited:
tiny-tim
Homework Helper
hi joej24! (just got up :zzz: …)
2r and 0, which concur with the wheel's velocity relative to the ground, 2v for the top and 0 for the bottom. But it doesn't make sense that the acceleration of the top is
(2v)2 / 2r
no, you're using "r" to mean two different things

since i was the one who set the problem, i think it's ok if i give the answer we know from the moving frame that (at the top of the wheel) a = v2/r

we know that a in the stationary frame must be the same

and we know that the radius of curvature in the stationary frame must be (speed)2/a …

which is (2v)2/(v2/r),

= 4r and

02 / 0

for the bottom
same reasoning … we know that the radius of curvature (at the bottom of the wheel) in the stationary frame must be (speed)2/a …

which is 0/(v2/r),

= 0 … the acceleration at that point has also tangential component at which is the time derivative of the speed …

indeed, which is why i chickened out and specified dr/dt = 0 when dr/dt = 0, the acceleration perpendicular to the velocity is always speed2/r, where r is the radius of curvature of that piece of the wheel

Okay. So the radius of curvature for the top is 4r (which can be found out from doing geometry). A then is (2v)^2 / 4r = v^2 / r

The radius of curvature for the bottom is 0. And the condition for this is that

when dr/dt = 0, the acceleration perpendicular to the velocity is always speed2/r, where r is the radius of curvature of that piece of the wheel
since the radius of curvature for the bottom is always 0 in this case.

I think that doing this
any change of the velocity of the tire comes from the rotation. Therefore the only acceleration is the centripetal one, which magnitude is related to the speed of the tyre with respect to the axis. So its magnitude is vcom^2/r, the direction points inward, towards the CM.
ehild
is easier.

However if there was acceleration in the translational motion, then i would just use
sqrt { at 2 + ar 2 } to find the magnitude of total acceleration

tiny-tim
Homework Helper
hi joej24! I think that doing this

is easier.
yes of course … that's why i wrote …
in this case, relative to a stationary observer, the speed (for the top of the wheel) is 2v, but the r is unknown to you (unless you're familiar with the geometry of a cycloid ) …

that's why it's much easier to choose a frame in which you know the radius of curvature without having to do complicated geometry! However if there was acceleration in the translational motion, then i would just use
sqrt { at 2 + ar 2 } to find the magnitude of total acceleration
yes ## Homework Statement

So a tire rolls with radius r. The CM of the wheel travels at a velocity of vcom.

What is the acceleration of the top of the wheel and the bottom of the wheel relative to a passenger in the car and to a street sign on the street?

a = v^2/r

## The Attempt at a Solution

Relative to the passenger, the top and bottom of the wheel are traveling at a velocity of vcom and -vcom. So the magnitudes of both accelerations are vcom^2/r.

Relative to the street sign, the top of the wheel is traveling at 2vcom. The bottom of the wheel is traveling at 0. I know that these are the wrong velocities to plug into the equation a = v^2/r since the answer for the accelerations are the same as the accelerations in the previous situation.

A friend told me that centripetal acceleration is always relative to the axle. I thought that perhaps since the translational acceleration of the top and bottom of the wheel are 0, and since the acceleration of the wheel when it is rotating is vcom^2/r for the top and bottom, the magnitudes of the accelerations of both the top and bottom could be Arotation + Atranslational = ARolling,
which would be vcom^2/r.

Are any of these explanations correct?

the velocity at the highest point is 2vcom and hence acceleration is (2v)^2
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