PHYSICS 206
Problem set 8 2001. Solution.

Problem 1:
Perhaps the most direct method to solve this problem is to start from the expression for the angular moment for a sum of mass points dm

$$\vec{l}=\sum_\alpha m\vec{r}_\alpha\times\vec{v}_\alpha$$

Consider a mass element dm=mds/a a distance s from the center of mass of the rod. The z-axis is the axis of rotation. At some instance the mass element will be located at

$$x=s\sin\theta\cos(\omega t),\;\;y=s\sin\theta\sin(\omega t);\;\;z=s\cos\theta$$

The corresponding velocities are

$$v_x=-s\omega\sin\theta\sin(\omega t),\;\;v_y=s\omega\sin\theta\cos(\omega t),v_z=0$$

a:
The angular momentum is

$$\int_{-a/2}^{a/s}dl=\frac{m}{a}\int_{-a/2}^{a/s}ds\;\vec{r}\times\vec{v}$$

$$=\frac{ma^2\omega}{12}\{-\cos\theta\sin\theta[\hat{i}\cos(\omega t)+ \hat{j}\sin(\omega t)]+\hat{k}\sin^2\theta\}$$

b:
The torque is

$$\vec{N}=\frac{d\vec{l}}{dt}=\frac{ma^2\omega^2}{12}\cos\theta\sin\theta[ \hat{i}\sin(\omega t)-\hat{j}\cos(\omega t)]$$

Problem 2:

$${\cal T}=\frac{mv^2}{2}+\frac{I\omega^2}{2}=\frac{v^2}{2}(m+\frac{I}{a^2})$$

$${\cal V}=-msg\sin{\theta}$$

$$(1+\frac{I}{ma^2})\ddot{s}=g\sin\theta$$

a:

$$I=\frac{2}{5}ma^2$$

$$\ddot s=\frac{5}{7}g\sin\theta=0.7142857143g\sin\theta$$

b:
The volume of the ball excluding the cavity is

$$\frac{4\pi}{3}(a^3-\frac{a^3}{27})$$

If the higher density ball had been solid its mass would have been 27m/26. The moment of inertia is thus

$$\frac{2ma^2}{5}(\frac{27}{26}-\frac{1}{26\times 9})=ma^2\frac{242}{585}$$

$$\ddot{s}=\frac{g\sin\theta}{(1+\frac{I}{ma^2})}=\frac{585g\sin\theta }{827} =.7073760579g\sin\theta$$

Problem 3:




We let A be the origin of our body centered coordinate system, 1 the direction out of the plane of the paper, 2 the axis from A to the center of the wheel and 3 the vertical. For the wheel to stay on its path (without falling or flying up) $\omega_1=0$. From Euler's equations

$$N_1=\omega_2\omega_3(I_3-I_2)$$

We have

$$\omega_3=\Omega$$

For the wheel to roll without slipping

$$D\Omega_2=\omega_2R$$

The moment of inertia of a disk about an axis perpendicular to the plane of the disk is

$$I_2=\frac{mR^2}{2}$$

The moment of inertia of a disk about an axis in the plane of the disk through its center is mR²/4 giving

$$I_3=\frac{mR^2}{4}+mD^2$$

Hence the torque is

$$N_1=\frac{\Omega^2 D}{R}(mD^2-\frac{mR^2}{4})=(mg-n)D$$

where n is the normal force. We find

$$n=m[g-\Omega^2(\frac{D^2}{R}-\frac{R}{4})]$$

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