A wheel (mass M, radius R) rolls with speed v and angular velocity w (v=ωR) and collides elastically with a wall. Immediately after the collision it has v away from the wall, but the same ω as before – so it’s skidding, not rolling.
Friction at the floor will eventually re-establish rolling motion, by doing negative work on the system and applying a torque that will change ω.
What is the new speed v’ away from the wall when rolling motion is re-established?
Answer by RickB
That’s an interesting question!
First, let’s set up direction conventions. Let’s say:
* The wall is on the left, so the wheel bounces off to the right. Let’s call “rightward” the positive linear direction.
* Let’s call “clockwise” the positive angular direction. (That means the wheel’s rotation is negative immediately after the bounce.)
During skidding, the force of friction is more or less constant (given by: (normal force)×(coef. of sliding friction)), so we can say that “Ff” is the (constant) force of friction during the skidding phase.
The force Ff produces a torque on the wheel, in the amount of: τ = (Ff)R. (This is a positive torque, since Ff points to the left and therefore tends to drive the wheel clockwise.) We know that the angular acceleration is given by:
α = τ/I = (Ff)R/I
where “I” is the wheel’s moment of inertia. I’m not sure what kind of wheel this is, but its moment of inertial can be represented as: I = βmR², for some “β” that depends on how the wheel’s mass is distributed. (It’s some constant between 0 and 1.) So we can write:
α = (Ff)/(βmR)
At the same time, the ball’s linear acceleration is:
a = -Ff/m
(Here we use negative Ff, since the force points toward the left)
With these two values, we can write the ball’s angular and linear velocities as a function of time:
ω(t) = ω0 + αt = ω0 + (Ff)t/(βmR)
(where ω0 is the ball’s angular velocity right after collision)
v(t) = v0 + at = v0 + (−Ff/m)t
(where v0 is the ball’s linear velocity right after collision)
We are also given that ω0 = −v0/R, so we can rewrite the ω(t) equation as:
ω(t) = −v0/R + (Ff)t/(βmR)
Now, at some time t′, rolling is re-established. That means:
v(t′) = ω(t′)R
v0 + (-Ff/m)t′ = −v0 + (Ff)t′/(βm)
t′ = 2(v0)mβ/[Ff(1+β)]
Now plug that value back into the “v(t)” equation:
v′ = v(t′) = v0 + (-Ff/m)2(v0)mβ/[Ff(1+β)]
= v0 – 2(v0)β/(1+β)
= v0(1 – 2β/(1+β))
So for example, if you assume that the wheel has a uniform density, then β=½, in which case v′ = v0/3.