Bowling ball goes from sliding to rolling

[MuIotaTau]

Homework Statement

A bowling ball is thrown down the alley with speed $v_0$. Initially it slides without rolling, but due to friction it begins to roll. Show that its speed when it rolls without sliding is $\frac{5}{7} v_0$.

Homework Equations

$K= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2$

The Attempt at a Solution

Because friction does no work in this case, energy is conserved. Equating the energy before and after rolling begins, I get $$\frac{1}{2}Mv_0^2 = \frac{1}{2}Mv^2 + (\frac{1}{2})(\frac{2}{5})MR^2\omega^2$$

Using the fact that $v = R\omega$, I get $$Mv_0^2 = Mv^2 + \frac{2}{5}Mv^2$$

Solving for $v$, I get $$v = \sqrt{\frac{5}{7}}v_0$$ What did I do wrong to make me off by a factor of $\sqrt{\frac{5}{7}}$?

 

[Tanya Sharma]

Energy is not conserved until the ball starts to roll without slipping .Initially the ball rolls with sliding ,that means friction does some work .

Energy conservation is not the correct way to approach the problem .Think differently .

 

[MuIotaTau]

Energy is not conserved until the ball starts to roll without slipping .Initially the ball rolls with sliding ,that means friction does some work .

Energy conservation is not the correct way to approach the problem .Think differently .

Ohhh, I see. Hmmm, well neither angular nor linear momentum are conserved, correct? But angular momentum while sliding is zero, so I could set the torque equal to angular momentum of the ball after slipping takes place, correct? Would I have to do something similar with linear momentum as well? Otherwise I'm not sure how to include $v_0$.

 

[Saitama]

Ohhh, I see. Hmmm, well neither angular nor linear momentum are conserved, correct?

Angular momentum is conserved. Can you find a point about which the torque on system is always zero?

 

[MuIotaTau]

Oh! The point of contact between the ball and the ground, right? And so using the parallel axis theorem, the moment of inertia about that point is $\frac{7}{5}MR^2$. And then I can say that $I\omega = MRv_0$, which gives me the correct result. And the reason I can do that is because torque is zero about that point, which means that the angular momentum is time independent, correct? In other words, if the torque weren't zero, then I would have a time dependence in $L$?

 

[Tanya Sharma]

Oh! The point of contact between the ball and the ground, right? And so using the parallel axis theorem, the moment of inertia about that point is $\frac{7}{5}MR^2$. And then I can say that $I\omega = MRv_0$, which gives me the correct result. And the reason I can do that is because torque is zero about that point, which means that the angular momentum is time independent, correct? In other words, if the torque weren't zero, then I would have a time dependence in $L$?

Do not consider point of contact between the ball and the ground ,even if it may give correct result.That point is continuously changing .

Instead consider any fixed point on the ground, say P .The torque due to friction about that point P will be zero .Hence Angular momentum will be conserved about P.

The angular momentum of the ball about the fixed point is the sum of angular momentum of the CM(mvR) about P + angular momentum of the ball around the CM(Iω) .

 

[rcgldr]

Angular momentum is conserved.

Only if you include the angular momentum of the Earth as part of a closed system.

Back to the problem, assume some unknown kinetic sliding friction force F, and a given initial velocity V0. You should be able to generate equations related to linear deceleration and angular acceleration to determe the velocity at which the ball transitions from sliding to rolling in terms of V0.

Not clearly specified in the problem statement is that you're to assume the ball's initial angular velocity is zero.