Question

Particles are released from rest at A and slide down the smooth surface of height h to a conveyor B. The correct angular velocity (0 of the conveyor pulley of radius r to prevent any sliding on the belt as the particles transfer to the conveyor is

Answer 1

Answer on Question #66182, Physics / Mechanics | Relativity

Particles are released from rest at A and slide down the smooth surface of height h to a conveyor B. The correct angular velocity (0 of the conveyor pulley of radius r to prevent any sliding on the belt as the particles transfer to the conveyor is

Solution:

We write the equations of motion using polar coordinates

$$m g \cos \theta = m R \dot {\theta}$$ $$- N + m g \sin \theta = - m R \dot {\theta} ^ {2}$$

Integrate the first equation

$$\int_ {0} ^ {\dot {\theta}} \dot {\theta} d \dot {\theta} = \frac {g}{R} \int_ {0} ^ {\theta} \cos \theta d \theta$$ $$\frac {\dot {\theta} ^ {2}}{2} = \frac {g}{R} \sin \theta$$ $$\dot {\theta} = \sqrt {\frac {2 g}{R} \sin \theta}$$

The conveyor pulley must turn at the rate

$$R \dot {\theta} = r \omega$$

For $\theta = \pi /2$ , so that

$$\omega = \sqrt {\frac {2 g R}{r}}$$

Answer: $\omega = \sqrt{\frac{2gR}{r}}$

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