Ball rolls without slipping in an accelerating car

[astrocytosis]

Homework Statement

A bowling ball sits on the smooth floor of a subway car. If the car has a horizontal acceleration a, what is the acceleration of the ball? Assume that the ball rolls without slipping.

Homework Equations

torque = R x F = Iα
a_ball= Rα
I_sphere = (2/5) MR²

The Attempt at a Solution

This seems like a simple problem but I'm struggling with it. If the ball is rolling without slipping, then there must be some friction force ƒ causing a torque on the ball equal to Rƒ = (2/5) MR² * (a_ball/R). It's not on a ramp so the weight of the ball doesn't contribute to the torque. I guess I just can't figure out how to relate the acceleration of the car to all of this.

I had a thought that without friction, in the frame of the car, the ball would simply be accelerating at the same rate as the car, so the difference in acceleration must somehow be the result of friction. Friction here is ƒ = (2/5)Ma_ball, so the difference in net force on the ball in the cases with and without friction is ma_car-ma_ball=ƒ. But solving for the acceleration here does not get me the right answer so this reasoning must be flawed.

The correct answer according to the text is 2a/7. With my reasoning I get 5a/7.

 

[mfb]

The approach is good, you have two equations with two unknowns. Apparently something went wrong in the calculations you didn’t show.

 

[astrocytosis]

Ma_car-Ma_ball = (2/5) Ma_ball

M (mass of ball) cancels

a_car=a_ball + (2/5) a_ball = (7/5) a_ball

So a_ball = (5/7) a_car.

Am I missing something in this equation?

 

[kuruman]

You can get the answer by doing the problem correctly in the accelerated frame. In that frame, there is a fictitious force $F=m_{ball}~a_{car}$ acting in a direction opposite to the actual acceleration of the car. This force generates a torque $\tau$ about the point of contact P. Write the torque equation $\tau = I_P~\alpha_{ball}$ where $I_P$ is the moment of inertia about point P, not the center of the ball. Solve the equation and the correct answer will plop out.

 

[astrocytosis]

By the parallel axis theorem, I_P = (2/5)MR² + MR² = (7/5)MR²

Then the torque is (7/5)MR² *(a_ball/R) = R*F = RM_car

I'm still getting a_ball = 5a_car/7 when I solve this equation, though.

 

[kuruman]

Don't forget that the a_ball calculated this way is in the accelerated non-inertial frame. What is it in the inertial frame of the track?

 

[astrocytosis]

Oh I see... I just have to subtract the acceleration of the car then. Thanks!