Large PHYSICS 206
Problem set 7 2001. Solution.

Problem 1:
a:
The velocity of the point of contact is zero. Therefore, the force of friction does no work and energy is conserved. Let h be the vertical drop, v the velocity of the center of mass, and $\omega$ the angular velocity, and

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

the moment of inertia. Energy conservation gives

$$mgh=\frac{ma^2\omega^2}{4}+\frac{mv^2}{2}=\frac{3mv^2}{4}$$

where we have used $v=a\omega$. If s is the distance travelled we obtain

$$v=\frac{ds}{dt}=\sqrt{\frac{4gh}{3}}=\sqrt{\frac{4gs\sin\theta}{3}}$$

which can be integrated to give

$$\sqrt{\frac{3 s\sin\theta}{g}}=t\;$$

or

$$s=\frac{gt^2\sin\theta}{3}$$

$$v=\frac{2gt\sin\theta}{3};\;\;\omega=\frac{2gt\sin\theta}{3a}$$

b:
We have for the torque with f the force of friction

$$fa=I\dot{\omega};\;\;f=\frac{mg\sin\theta}{3}$$

c:
The normal force is $mg\cos\theta$, so the the maximum angle is given by

$$\tan\theta_{max}=3\mu$$

Problem 2:
a:
The moment of inertia tensor about the origin is

$$I=m\left(\begin{array}{ccc} 2a^2&0&0\\ 0&2a^2&0\\ 0&0&2a^2\\ \end{array}\right)$$

b: The center of mass is at

$$\vec{r}_{cm}=\frac{a}{3}(\hat{i}+\hat{j}+\hat{k})$$

c:
We have from the parallel axis thorem

$$I=I_{cm}+3ma^2\left(\begin{array}{rrr} \frac{2}{9}&-\frac{1}... ... -\frac{1}{9}&-\frac{1}{9}&\frac{2}{9}\\ \end{array}\right)$$

giving

$$I_{cm}=ma^2\left(\begin{array}{lcr} \frac{4}{3}&\frac{1}{3}&... ...\\ \frac{1}{3}&\frac{1}{3}&\frac{4}{3}\\ \end{array}\right)$$

For reasons of symmetry one of the pricipal axis must extend in the direction from the origin to the center of mass

$$\hat{e}_3=\frac{1}{\sqrt{3}}[1,1,1]$$

The corresponding moment of inertia is

I₃=2ma²

Any axis perpendicular to this axis is also a principal axis (the object is a symmetric top)

I₂=I₁=ma²

A possible choice of directions is

$$\hat{e}_1=\frac{1}{\sqrt{2}}[1,-1,0],\;\;\hat{e}_2=\frac{1}{\sqrt{6}}[1,1,-2]$$

Problem 3:
a:
The kinetic energy is

$$\frac{mv^2_{cm}}{2}+\frac{I\omega^2}{2}$$

$$v_{cm}=(b-a)\dot\theta$$

The angular velocity is determined by the condition that the point of contact has zero velocity

$$a\omega=v_{cm}$$

The moment of inertia of a sphere is I=2ma²/5 so that

$${\cal L}=\frac{m}{2}(b-a)^2(1+\frac{2}{5})\dot\theta+mg(b-a)\cos\theta$$

$$\frac{7}{5}(b-a)\ddot\theta+g\sin\theta=0$$

b:

$$T=2\pi\sqrt{\frac{7(b-a)}{5g}}$$

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