Question

A ball of radius R, mass M, and moment of inertia 2/
5MR^2 is placed on a plane inclined at an angle θ. The plane is accelerated upwards (along its direction) with acceleration a, as shown. For what value of a is the center of mass of the ball stationary? Assume sufficient friction so that the ball does not slip.

Could I possibly ask for some assistance with this one?

This is what I have thus far.

http://imgur.com/Wgrlvcj

Answer 1

Hi. I haven't done much with physics regarding inclined planes yet, but I think if you find the acceleration of the ball if you were to let it go at the top, then that should be your answer.

Answer 2

Although I could be wrong.

Answer 3

You should find: \(a=\large \frac 52 \normalsize g \sin \theta\)

Answer 4

first of all take the components of the acceleration provided to the incline then apply the condition that the vertical component of the acceleration should be equal to g as the com has no acceleration in the vertical direction you need not worry about the component of acceleratiion in the horizontal direction as it can be the result of the force that can be taken as an inetrnal force and internal forces cannot provide acceleration to centre of mass. I'll stick to this problem as i havent started rolling in rotational motion and will definately try to get it done.

Answer 5

1step: find angular acceleration wrt a, keeping in mind that the centre of mass of the ball must remain stationary.
2 step: equate net torque and change in angular momentum
3 step: write down that net force = 0
4 step: eliminate friction force and solve for a

Answer 6

sorry for the ignorant question but is angular accleration simply R theta?

Answer 7

No problem ;-)

Angular acceleration is \(\alpha = \dot \omega = \ddot \theta\)

Angular momentum about centre of mass C is \(J \omega \)

Net torque about C is \(\tau\)

Theorem of angular momentum states that: \(\tau = J\alpha\)

Answer 8

Perfect, all good now.
Thanks again.