15-96

Like all problems that change the basic setup, we need to go back to FBD's before we can apply our spring results. The forces acting are gravity, spring, normal, and friction (rolls without slipping!). So, taking our center of torque to be the center of the wheel, we get

$$\begin{eqnarray*} \Sigma F_x &=& N-mg = 0\\ \Sigma F_y &=& -kx-F_f = ma\\ \Sigma \tau &=& -F_fR=I\alpha \end{eqnarray*}$$

where $I$ is the moment of inertia and $R$ the radius. Note that acceleration to the right means clockwise rolling, or negative angular acceleration, so $a=-\alpha R$. Eliminating $F_f$ from the second two equations yields

$$\begin{eqnarray*} -kx+\frac{I\alpha}{R} &=& ma\\ -kx-\frac{Ia}{R^2} &=& ma, \end{eqnarray*}$$

or, after collection of terms and rearranging

$$a\left(m+\frac{I}{R^2}\right) = -kx$$

Thus we have an equation of the form

$$a=\omega^2 x,$$

but with

$$\omega^2=\frac{k}{m+I/R^2},$$

that is, we still have simple harmonic motion, but the rolling inertia makes our `effective mass' larger. In this case $I=mR^2/2$, so

$$\omega^2=\frac{k}{3m/2}=\frac{2}{3}\frac{k}{m}.$$

It is straightforward to show the period is as advertised at this point.

The total energy any point is the sum of three terms:

$$E=K_{rot}+K_{trans}+U_s.$$

Notice that $\omega=v/R$, so that

$$K_{rot}=\frac{1}{2}I\omega^2=\frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{v}{R}\right)^2=K_{trans}/2.$$

Now,

$$E=\frac{3}{4}mv^2+\frac{1}{2}kx^2,$$

and the time derivative of $E$ is

$$\begin{eqnarray*} \frac{dE}{dt}&=&\frac{3}{2}mv\frac{dv}{dt}+kx\frac{dx}{dt}\\ &=&\frac{3}{2}mva+kxv\\ &=& v\left( \frac{3}{2}ma+kx\right)=0, \end{eqnarray*}$$

where the last follows since $E$ is constant. This equation is identical to the one derived above.