Solving a Puzzling Problem: A Disk and a Hoop

[Geoffrey F. Miller]

I'm having some trouble with a conceptual problem. Your help or input would be greatly appreciated. Here's the problem.

A disk and a hoop, both having the same mass, are rolled down a hill. The disk rolls faster than the hoop. Why is this so?

Thank you for your time!

 

[Doc Al]

Which one has the greater rotational inertia? (And thus is harder to get rolling.)

 

[Geoffrey F. Miller]

I guess the hoop would have a greater moment of inertia, but why is it greater? How is rotational inertia calculated?

 

[Pyrrhus]

Moment of Inertia

$$I = \sum_{i=1}^{n} m_{i} r_{i}^2$$

For a a system composed of particles with defined mass the above can be calculated easily, but for a continuous mass system we will take a small mass [itex] \Delta m [/tex] and applying the moment of inertia definition we should get.

$$I = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \Delta m_{i} r_{i}^2$$

Which is the integral of an infinitessimal mass

$$I = \int r^2 dm$$

Now the moment of inertia around the z axis for a Uniform hoop is

$$I_{z} = \int r^2 dm$$

every particle with mass dm is at a constant distance R, so

$$I_{z} = R^2 \int dm$$

$$I_{z} = MR^2$$

For more detail we could calculate the moment of inertia for ring shaped cylinder will be:

$$I_{z} = \int r^2 dm$$

Using

$$\rho = \frac{dm}{dV}$$

$$I_{z} = \int r^2 \rho dV$$

Applying cylindric shells we get, where L is the height of our cylinder

$$dV = 2 \pi r dr L$$

so

$$I_{z} = 2 \pi L \rho \int^{r}_{r_{o}} r^3 dr$$

we get

$$I_{z} = \frac{1}{2} M(r^2 + r_{o}^2)$$

A hoop can be considered a ring shaped cylinder of very thin walls so we could say

$$I_{z} = MR^2$$

Now an uniform disk could be considered a solid cylinder so

$$I_{z} = \int r^2 dm$$

r being the radius of our cylinder

Using
$$\rho = \frac{dm}{dV}$$


$$I_{z} = \int r^2 \rho dV$$

Applying $dV = dA L$

$$I_{z} = \int r^2 \rho 2 r \pi L dr$$

$$I_{z} = \int r^3 \rho 2 \pi L dr$$

$$I_{z} = \rho 2 \pi L \int^{r}_{0} r^3 dr$$

$$I_{z} = \frac{1}{2} MR^2$$

Note: All the z- axis go throught the center of mass

Also:

Moment of Inertia for the disk
$$I_{z} = \frac{1}{2} MR^2$$

Moment of Inertia for the hoop
$$I_{z} = MR^2$$

From this two moment of inertia you can see which will get down before the other.

and i found this movie for this:
http://solomon.physics.sc.edu/~tedeschi/demo/demo12.html

 

[Tide]

Did the problem specify that the disk and hoop have the same radius?

 

[Doc Al]

I guess the hoop would have a greater moment of inertia, but why is it greater? How is rotational inertia calculated?

Cyclovenom gave the details, but the gist of it is that rotational inertia depends on how the mass is distributed: The more mass farther from the rotational axis, the greater the rotational inertia. It's harder to rotate something if its mass is far from the axis of rotation.

 

[Doc Al]

Did the problem specify that the disk and hoop have the same radius?

It doesn't matter.

 

[Tide]

It doesn't matter.

I thought it would be something Geoff should think about!

 

[Doc Al]

An excellent point!