Work/Rotation of a Rigid Body

1. Oct 27, 2007

1. The problem statement, all variables and given/known data
A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m and its moment of inertia about its rotation axis is 0.800MR^2. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 3500 J. Calculate h.

Mass "M" = w/g = 392/9.8, or 40 kg.
2. Relevant equations
$$I=cMR^{2}$$

$$W_{total}=\Delta K - W_{friction}= \frac{1}{2}I \omega^{2}_{final}-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}$$

$$W_{total}=-\Delta U=Mgh_{initial}-Mgh_{final}$$

3. The attempt at a solution
I set the two equations for Work equal to get:

$$Mgh_{initial}-Mgh_{final}=\frac{1}{2}I \omega^{2}_{final}-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}$$

Since initial height is 0 and final angular velocity is 0, this simplifies to:

$$-Mgh_{final}=-\frac{1}{2}I \omega^{2}_{initial}-W_{friction}$$

Multiplying this equation by -1 to make the negatives a little more friendly and writing out "I" (moment of inertia) yields:

$$Mgh_{final}=\frac{1}{2}cMR^{2} \omega^{2}_{initial}+W_{friction}$$

Isolating to solve for "h":

$$\frac{(\frac{1}{2}cMR^{2} \omega^{2}_{initial}+W_{friction})}{Mg}=h_{final}$$

Plugging in numbers:

$$\frac{(\frac{1}{2}(0.800)(40)(0.600)^{2} (25.0)^{2}+3500)}{(40)(9.80)}=h_{final}$$

For h I get 18.1 m, but the answer according to the book is 11.7 m.

I tried several approaches and couldn't get the right answer, so I would appreciate help.

2. Oct 27, 2007

hotcommodity

You have an equation where energy is not conserved, which is good, but it's missing two major things. You take the rotational kinetic energy of the wheel into account, but you leave out the wheel's translational kinetic energy. Additionally, you're assuming that the friction does negative work on the wheel. Think about which way the friction points along the hill.

3. Oct 27, 2007

Ah, okay, I get it now. Since there is both rotational and translational motion, $$\Delta K$$ should include both types of kinetic energy. And work due to friction is positive because it acts in such a direction that makes angular velocity positive? So my new equation would be:
$$-Mgh_{final}=-\frac{1}{2}I \omega^{2}_{initial}-\frac{1}{2}mv^{2}_{initial}+W_{friction}$$.
And knowing that $$v=\omega R$$, plug in numbers to get h=11.7 m.