# [Classical Mechanics] Just need some conceptual help (rolling and friction)

• nonequilibrium

## Homework Statement

"A ball with radius R and mass m turns around a horizontal axis through his center with an angular speed $$\omega_0$$. In that condition the ball, without an initial velocity in the center, is put on top of a table. The friction coëfficient between the ball and the table is µ. Calculate the distance that the ball travels until it doesn't slip anymore.
($$I_{ball} = \frac{2}{5} m R^2$$)"

## Homework Equations

$$I_z \frac{\mathrm d \omega_z}{\mathrm d t} = \tau_z$$

## The Attempt at a Solution

Hello,

I think I solved it ($$x = \frac{2R^2\omega_0^2}{49\mu g}$$),
but there were no answers with the exercises, so I just want to check if I did the right conceptual steps:

So first of all, no matter how slow the ball was spinning, there will always be slipping at first, because v = 0 (when there's no subscript, I'm talking about the CM) and friction thus needs to deliver positive work (at the cost of kinetic energy, also producing heat, hence making an energy approach futile).

Now I thought, the necessary and sufficient condition for rolling without slipping is $$\omega R = v$$, so basically I just need to calculate how long it will take for v to RISE and omega to DROP until they reach this relation.
You can calculate v with the kinematic equation $$v^2 = 2ax$$, assuming $$F = \mu m g$$. One gets an expression of v in function of the distance x.
We can also calculate the time derivative of omega (see "Relevant Equations") and then we use two other kinematic equations: one to find the time duration of slipping (in function of x) and then using that we use $$\omega = \omega_0 + \frac{\mathrm d \omega_z}{\mathrm d t} \Delta t$$, giving us omega in function of x.
We then plug both expressions into v = wR and solve for x.

One more concrete question:
This implies that the change in applied friction is abrupt, correct? Before the contact W = 0, then as soon as there is contact F = µmg, until v is reached at which point it goes abruptly from µmg to 0 again. There are no transient zones, correct?

Nice work! It all looks correct to me. I got most of it independently but your clever use of v² = 2ax eluded me.

I suppose in reality there would be a very brief transition time when the full weight of the ball was not on the table but the assumption of an instantaneous jump to the full normal force seems very reasonable.

You did it well, your result is correct. The usual method to solve such problems is that we assume the effect of friction abrupt. In reality, there is some transition time: while you hold the ball it presses the ground less than its weight, and even after releasing the ball, the elastic and adhesion forces between the ground and the ball (which cause the force of friction) need some time to reach their stationary value. But this time is very short and has little influence to the final distance.
Usually we always ignore a lot of effects when solving physical problems, but we have to know the limits of our approximations.

ehild

Edit: Delphi beat me, but we have the same opinion :)

Your replies are much appreciated :)

I am getting x = R2ω2o/7µg

What is the angular velocity you got once the ball starts motion without slipping?

ω = ωo + αt = ωo - 2.5μgt 
v = at = μgt 
The condition v = Rω then gives t = 2Rωo/(7μg) 
So ω = ωo - 5R*ωo/7 and v = 2R*ωo/7
Easier to use the v to get the x.

I got the same value for 'v'.
To find x, I used the Work-Energy Theorem-

0.5mv2 + 0.5Iω2 - 0.5Iωo2 = (mu)mgR*x

If you solve this then you get x = R2ω2o/7µg

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I got the same value for 'v'.
To find x, I used the Work-Energy Theorem-

0.5mv2 + 0.5Iω2 - 0.5Iωo2 = (mu)mgx

If you solve this then you get x = R2ω2o/7µg

When you use the Work-Energy Theorem and rotation is involved you need to include the work of torque which changes the rotational energy.

ehild

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errr I forgot the expression for work done due to a torque. To find that we have to use equations of rotational motion.