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## Homework Statement

Ball of mass m and radius r rolls without sliding in a bowl of radius R.

Express it's angular velocity ω as a function of the angle θ.

What are the normal force and friction force as a function of θ.

## Homework Equations

Moment of inertia of a solid ball round it's center: ##I_c=\frac{2}{5}mr^2##

Sheteiner's theorem: ##I_A=I_c+mr^2##

Kinetic energy of a rigid body: ##E=\frac{1}{2}I\omega^2##

Centripetal force: ##F=mr\omega^2##

Torque and angular acceleration: ##M=I\alpha=I\dot\omega##

## The Attempt at a Solution

The descending in height is transferred to kinetic energy:

$$mg(R-r)\sin\theta=\frac{1}{2}\left( \frac{2}{5}mr^2+mr^2 \right)\omega^2\;\rightarrow\; \omega^2=\frac{10}{7}\frac{R-r}{r^2}g\sin\theta$$

$$\omega=\left( \frac{1}{r}\sqrt{\frac{10}{7}g(R-r)} \right) \cdot (\sin\theta)^{\frac{1}{2}}$$

$$\dot\omega=\left( \frac{1}{2r}\sqrt{\frac{10}{7}g(R-r)} \right) \frac{\cos\theta}{\sqrt{\sin\theta}}$$

The normal force is the substarction of the weight's radial component and the centripetal one:

$$F=mg\sin\theta-\frac{1}{2}I\omega^2=mg\sin\theta-\frac{10}{7}\frac{R-r}{r^2}g\sin\theta=mg\sin\theta\left( \frac{10}{7}\frac{R-r}{r}+1 \right)$$

The friction force:

$$M=f\cdot r=I_c\dot\omega\;\rightarrow\; f=\frac{2}{5}mr\cdot\left( \frac{1}{2r}\sqrt{\frac{10}{7}g(R-r)} \right) \frac{\cos\theta}{\sqrt{\sin\theta}}$$