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Lets say I have a hoop of mass $M$ and radius $R$. It is rolling down a hill without slipping, so I don't need to worry about friction doing work on it. Lets say the angular speed is $\omega = \frac{v_{CM}}{R}$. It's center of mass of the hoop and the tangential speed is $v$.

I want to know how to go about solving for the wheels final speed at ground level with an initial speed $v_i$ and initial heigh $h$.

I try to go an choose a system: the hoop and earth.

Then I have $$\Delta E_{sys} = W + Q + T_{light}$$ With my defined system, there is no external work, (also we assume no heat), and it doesn't given off light. so $\Delta E_{sys} = 0$, and $\Delta E_{sys} = \Delta K_{transitional} + \Delta K_{rotational} + \Delta U + \Delta E_{internal}$ Crossing out change in internal leaves us with

$$\Delta K_{transitional} + \Delta K_{rotational} + \Delta U = 0$$ $$ K_{transitional_f} - K_{transitional_i} + K_{rotational_f} - K_{rotational_i} + U_f - U_i = 0$$

I am having trouble trying to figure out whether we have $.5mv^2$ for transitional kinetic energy and $.5m(\omega R)^2$ for rotational, but is not $\omega R = v_{CM}$ so basically rotational and transitional energy have the same magnitude for final and initial?

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are you sure about $\omega = \frac{v_{CM}}{R}$? I guess it should be $\frac{v}{R}$.. – Vineet Menon Nov 3 '11 at 6:36
also, are you assuming initially the hoop to be at rest??? both translationally as well as rotationally... – Vineet Menon Nov 3 '11 at 7:06
well $v_{CM} = v$ since it is a hoop rolling without sliding. Also there is initial velocity as stated, $v_i$. – Salazar Nov 3 '11 at 7:48
I'm not sure but I think the energy should be divided up equally into each degree of freedom... – Nic Nov 3 '11 at 10:50
up vote 2 down vote accepted

You are right; the magnitudes of the translational and rotational kinetic energies are the same.

In general, the kinetic energy of a mass distribution can be written

$$K = K_{cm} + K_{rel}$$

$K$ is the total kinetic energy. $K_{cm}$ is the kinetic energy of a point particle whose mass is equal to the total mass of the distribution, if the point particle is moving along with the center of mass of the distribution. $K_{rel}$ is the kinetic energy of the mass distribution as viewed in the center-of-mass reference frame. This is a basic theorem you can prove by writing down the integral of $\rho v^2$ over all space ($\rho$ being the mass density).

Intuitively, here is a way to see what this says. Suppose you are pushing a car, and your goal is to get the car up to $2 \mathrm{m/s}\,$ when you push it over a flat stretch of parking lot, so there is a certain amount of work you are trying to do on the car.

Now imagine the car has a flywheel inside. This is a heavy disk that may or may not be rotating at high speed, storing a great deal of kinetic energy. The theorem says that whether or not the flywheel is going does not matter to you. The work you do on the car is the same regardless of the state of the flywheel as long as the flywheel does not change its rotational speed while you're pushing the car. Only the total mass of the flywheel matters, and then only because it is being accelerated along with the rest of the car. The exact shape, orientation, etc of the flywheel don't affect the amount of work you need to do.

The same is not true for the wheels. The rotation state of the wheel changes as you accelerate the car, so the moment of inertia of the wheels is relevant, in addition to their mass.

For the case of the hoop, the kinetic energy is

$$K = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$$

where $I$ is the moment of inertia for rotations about the center of the hoop and $\omega$ is the angular velocity.

If the mass is all located at the rim of the hoop, then $I = mR^2$, and because $v_{cm} = \omega R$ you can write

$$K = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}(mR^2)\left(\frac{v_{cm}}{R}\right)^2$$

Canceling the $R^2$ in the numerator and denominator of the second term, we see that the translation and rotational kinetic energies are the same in this case. The total kinetic energy is $mv_{cm}^2$

You can find the final speed of the hoop after it rolls down a hill by conservation of energy using this formula for the kinetic energy.

If you have a mass in the center of the hoop, you can adjust the moment of inertia depending on the relative masses of the hoop and the thing at the center and continue to use the general equation for the kinetic energy of a mass distribution. You will find that the kinetic energy is still proportional to $mv^2$, but the constant of proportionality is less than one. Otherwise you may account for the mass at the center of the hoop separately in the expression for the energy.

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