Question

A block (mass = 74.0 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, where M = 7.9 is the mass of the pulley and R=1.4 is its radius ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of r=R=1.4 m during the block's descent, and the distance of the block at initial position to the floor is 7.0 m. What is the angular velocity in round per minute when the block drops to the floor? Use g = 10 m/s2.

Answer 1

Answer:

51.2m/s

Explanation:

Gravity force acting on the block:

F = mg = 74 * 10 = 740 N

Torque acting on the pulley:

T = FR = 740 * 1.4 = 1036 Nm

Moments of inertia of the solid pulley

$I = \frac{1}{2}MR^2 = 0.5*7.9*1.4^2 = 7.742 kgm^2$

The angular acceleration of the pulley after release:

$\alpha = \frac{T}{I} = \frac{1036}{7.742} = 133.8 rad/s^2$

The linear acceleration or the rope

$a = \alpha R = 133.8 * 1.4 = 187.34 m/s^2$

The time it takes for the block to reach the floor:

$s = \frac{1}{2}at^2$

$t^2 = \frac{2s}{a} = \frac{2 * 7}{187.34} = 0.0747$

$t = \sqrt{0.0747} = 0.273 s$

The final velocity at the floor would be

$v = at = 187.34 * 0.273 = 51.2 m/s$