Question

The radius of the semicircular trough in the image is 5.0 m
is D = 10 cm and the moment of inertia is J = mD2 / 8, rolls by rotation and non-slip in the trough. The roll coefficient is zero, and there is no air resistance.
a) What is the speed of spinning of the ball at the lowest point of the trough?
b) How high is the ball rising on the opposite side?

Answer 1

a) Total initial energy of the ball relative to the bottom of the trough:

$$E_1=mgR.$$

Total final energy of the ball in the bottom:

$$E_2=\frac{1}{2}mv^2+\frac{1}{2}J\omega^2.$$

Equate the initial energy with final, substitute $J$ and divide both sides by $m$:

$$gR=\frac{1}{2}v^2+\frac{1}{16}D^2\omega^2.$$

Remember, with the given size of the ball there is obvious dependence between $v$ and $w$:

$$v=\omega\frac{D}{2}.$$

Therefore:

$$gR=\frac{1}{2}(\omega\frac{D}{2})^2+\frac{1}{16}D^2\omega^2,$$ $$\omega=\frac{4}{D}\sqrt{\frac{gR}{3}}=\frac{4}{0.1}\sqrt{\frac{9.8\cdot 5}{3}}=161.658\text{ s}^{-1}.$$

b) According to law of conservation of energy, to the height R=5 m.