PHYSICS 206
Problem set 6 1999, solution

Problem 1
Introduce a coordinate system as in figure:

The z-axis points out of the plane of the paper. The moment of inertia about a vertical axis through the center of mass is

$$I_{zz}=\frac{M\l ^2}{12}$$

a:
The center of mass velocity is

$$\vec{V}_{cm}=\frac{P}{M}\hat{j}$$

while the angular momentum is

$$\vec\Lambda=-\frac{P\l }{2}\hat{k}$$

so that the angular velocity is

$$\vec\omega=\frac{\vec\Lambda}{I_{zz}}=-\frac{6P}{Ml}\hat{k}$$

$\vec{V}_{cm}$ and $\vec\omega$ stay constant during the subsequent motion, since no further forces or torques act on the plank.

b:
The translational velocity of the point of contact is

$$\vec{V}_P=\vec{V}_{cm}+\vec\omega\times\vec{r}_P$$

The position of the point of contact P at time t is

$$\vec{r}_P=-\hat{i}\frac{\l }{2}\cos(\omega t)+\hat{j}\frac{\l }{2}\sin(\omega t)+ \vec{V}_{cm}t$$

The velocity of $\vec{V}_P$ is thus

$$\vec{V}_P=\frac{d\vec{r}_P}{dt}=\vec{V}_{cm}+\frac{\omega\l }{2}( \hat{i}\sin(\omega t)+\hat{j}\cos(\omega t))$$

$$=\frac{P}{M}[(1+3\cos(\omega t))\hat{j}+3\cos(\omega t)\hat{i})]$$

The angular velocity $\vec\omega$ is independent of the reference point.

c:
The position of an arbitrary point on the plank which initially was located a distance x to the right of the center of mass is

$$\vec{r}_x=\hat{i}x\cos(\omega t)-\hat{j}x\sin(\omega t)+\vec{V}_{cm}t$$

The initial velocity of this point is thus

$$(V_{cm}-\omega x)\hat{j}=\frac{P\hat{j}}{M}(1-\frac{6x}{l})$$

We conclude that the initial velocity is zero a distance $x=\l /6$ to the right of the center on mass.

Problem 2
The moments of inertia of a box with sides a,2a,3a about the center of mass in a coordinate system with axes parallel to the sides of the box is

$$I=M\left(\begin{array}{ccc} \frac{b^2+c^2}{12}&0&0\\ 0&\fr... ... 0&\frac{10}{12}&0\\ 0&0&\frac{5}{12}\\ \end{array}\right)$$

We have

$$\vec\omega=\frac{\omega}{\sqrt{14}}(1,2,3)$$

The kinetic energy is thus

$$\frac{1}{2}\vec\omega\cdot I\vec\omega=\frac{Ma^2\omega^2}{28... ...13}{12}+\frac{40}{12}+\frac{45}{12})= \frac{7Ma^2\omega^2}{24}$$

Problem 3
The moment of inertia of a sphere of radius r, mass $\mu$ about its center is

$$I_{cm}=\mu r^2\left(\begin{array}{ccc} \frac{2}{5}&0&0\\ 0&\frac{2}{5}&0\\ 0&0&\frac{2}{5}\\ \end{array}\right)$$

The moment of inertia tensor of the same sphere about (0,0,-r) is

$$I_{r}=\mu r^2\left(\begin{array}{ccc} \frac{7}{5}&0&0\\ 0&\frac{7}{5}&0\\ 0&0&\frac{2}{5}\\ \end{array}\right)$$ (1)

Let M be the mass of the solid sphere in the absence of a cavity and m the mass with the cavity present. We have

$$m=\frac{7M}{8}$$

a:
The z-coordinate of the center of mass is

$$z_{cm}=\frac{1}{m}(M\cdot 0-\frac{Ma}{2\cdot 8})=-\frac{a}{14}$$

b:
Substituting $\mu=-\frac{M}{8}$ and $r=\frac{a}{2}$ into the expression for I_r gives for the moments of inertia about the center of the sphere

$$I_0=Ma^2\left(\begin{array}{ccc} \frac{2}{5}-\frac{7}{5\cdot... ...0&\frac{57}{140}&0\\ 0&0&\frac{31}{70}\\ \end{array}\right)$$

c:
We have

$$I_{cm}=ma^2\left(\begin{array}{ccc} \frac{57}{140}+\frac{1}{... ...&\frac{101}{245}&0\\ 0&0&\frac{31}{70}\\ \end{array}\right)$$

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