Question

A 1.4kg block and a 2.7kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is 27cm in diameter and has a mass of 0.78kg .When the blocks are released, what is the acceleration of the lighter block?

Answer 1

Answer:

$32.67 m/s^2$

Explanation:

27 cm = 0.27 m

As the blocks are hanging on 2 sides of the pulley, their difference in mass would generate a net force toward the heavier side:

Let g = 9.81 m/s2

F = g(2.7 - 1.4) = 9.81*1.3 = 12.753 N

This force would generate a torque at 0.27 m moment arm

T = F*r = 12.753*0.27 = 3.44 Nm

The moment of inertia of the solid disk:

$I = mR^2/2$

Where m = 0.78 kg is the disk mass and R = 0.27 m is the radius of the disk.

$I = 0.780*27^2/2 = 0.028431 kgm^2$

So the angular acceleration of the torque is

$\alpha = T/I = 3.44 / 0.028431 = 121 rad/s^2$

So the linear acceleration is its angular component times radius

$a = \alpha r = 121 * 0.27 = 32.67 m/s^2$