# Homework Help: Rolling without slipping on a cylindrical surface - need clarification

1. Feb 10, 2010

### klg.amit

Hi,

The following is a general question which doesn't have to do with any particular problem. (therefore I am not including the template).
I understand that when a circular object (e.g: a hoop) rolls on a *flat* surface without slipping with angular speed $$\omega$$, its contact point, and thereby its center, moves with speed $$\omega r$$.
My difficulty is to understand the situation such as a pure roll of a hoop on a cylinder. I would like to understand what is wrong about my following analysis of the situation:

Let the hoop have radius r and the cylinder radius R, then:
1. Let the hoop rotate through an angle $$\theta$$ then its contact point moved a distance $$r\theta$$.
2. Let at the same time the angle from the center of the hoop to the center of the cylinder vary by $$\varphi$$ then the same distance is $$R\varphi$$
3. Therefore $$r\theta = R\varphi \Rightarrow r\omega=R\dot{\varphi}$$ by differentiating.

However my Professor says that since there is no slipping, the *center* moves with speed $$r\omega$$, not the contact point. But according to my analysis, the center should move somewhat faster, since it seems to me that it covers a longer arc than $$r\theta$$ which the contact point covered after a rotation through $$\theta$$.

A clarification: when I say the "contact point moved" I understand that it's changing all the time. I actually refer to the total distance covered by the "bottom" of the hoop.

Your kind help will be very much appreciated,
Amit

2. Feb 11, 2010

### ehild

See attached picture.
Assume uniform angular velocities.
The loops rolls upward on the cylinder. At start, its contact point is marked with blue. When the blue dot is contact point again, the loop has a trace of length s on the rim of the cylinder.The length of this trace is equal to the perimeter of the loop,

$$s=2r\pi$$

This arc corresponds to the central angle $$\phi=2r\pi/R$$

The loop turned by the angle

$$\theta=2\pi+\phi$$

during the time when the blue dot made its next contact with the cylinder. If the angular velocity of the loop is wL, this took the time period of

$$t=(2\pi+\phi)/\omega_L$$

The angular velocity of the centre of the loop is

$$\phi/t= \omega_L\frac{\phi}{(2\pi+\phi)}=\omega_L*\frac{2\pi r/R}{2\pi+2\pi r/R}=\omega_L*\frac{r}{R+r}$$

The linear velocity of the centre of the loop is

$$v_c=\omega_L*\frac{r}{R+r}*(R+r)=\omega_L*r$$

ehild

Last edited: Jun 29, 2010
3. Feb 11, 2010

### klg.amit

I think I understand now, the hoop rotates through an angle $$\phi$$ solely on the account of its movement on the cylinder (simply because of the geometry, on flat surfaces this contribution doesn't occur) and through $$\theta$$ due to rolling without slipping, is that a good way to see it ?

I think that this is the crucial point I failed to understand until now. Then you derived very clearly why this means that the center moves with linear speed wL*r. I thank you.

4. Feb 11, 2010

### ehild

Yes, you see it right. The case of Earth orbiting around the sun and rotating at the same time is a similar case. The day is longer than the period of rotation. Or the Moon: We see the same face all the time, but it completes a full rotation around its axis while completing its orbit around the Earth.

ehild

5. Feb 11, 2010

### klg.amit

I see, indeed these are good examples of the same thing. Thanks again!