# Free cylinder on an accelerating platform

1. Oct 4, 2013

### ash1262

Suppose a cylinder is resting on a horizontal platform on plane x-y with its axis parallel to the y axis, and the platform accelerates in the x direction. Assume that the axis remains parallel to the y axis and the surface is rough.
What is the motion (acceleration:linear as well as rotational) of the cylinder?

2. Oct 4, 2013

### tiny-tim

hi ash1262!

tell us what you think, and why, and then we'll comment!

3. Oct 4, 2013

### ash1262

The solution says that the linear acceleration of the cylinder is 3/2 times acceleration of the platform, which I can't visualize. I feel it should be same as that of the platform as there is no slipping (the surface is rough). Again, the cylinder rolls in the opposite direction of the platform's acceleration, so how is the motion visualized.

4. Oct 4, 2013

### tiny-tim

call the friction force on the cylinder "F", the mass "m" and the radius "r"

what is the equation showing the effect of F on the linear motion?

what is the equation showing the effect of F on the rotational motion?

5. Oct 4, 2013

### BruceW

a bit more advice, since this is a tricky question:
That is good, you are thinking along the right lines. The 'no-slip' condition is the starting place for this problem. And as you suggest, this means the total acceleration of the bit of the cylinder next to the surface must be equal to the acceleration of the surface. But, the total acceleration of the bit of the cylinder near the surface is not equal to the linear acceleration of the cylinder. What other kind of acceleration contributes?

6. Oct 4, 2013

### dauto

There is no slipping but there is rolling

7. Oct 6, 2013

### ash1262

Thanks Bruce, your reply was very useful. The solution I had used a pseudo force to derive the acceleration, and I could not visualize what was really happening. Now I have derived the acceleration without using the pseudo force, and can visualize the motion!

8. Oct 6, 2013

### BruceW

hooray! nice work. yeah, pseudo forces can make things more confusing. I pretty much always use an inertial reference frame. Even when using polar coordinates, in my head, I think of
$\vec{v}= \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} = \dot{x} \hat{x} + \dot{y} \hat{y}$
So I'm thinking "don't worry, underneath it all is a nice inertial reference system, with Cartesian coordinates". haha. p.s. the 'hooray' is not meant to be sarcastic. Often people think I'm being sarcastic when I'm not.

9. Oct 6, 2013

### ash1262

Thanks for the encouragement.