Question

A 375-g stone hangs from a thin light string that is wrapped around the circumference of a pulley with a moment of inertia of 0.0125 kg ∙ m2 and a radius of 26 cm. When the stone is released, the stone accelerates downward and the pulley rotates about its axis as the string unwinds. What is the magnitude of the acceleration of the stone in m/s2 ?

Answer 1

Answer:

The magnitude of the acceleration of the stone is 19.87 m/s²

Explanation:

Given;

mass of stone, m = 375 g = 0.375 kg

moment of inertia, I = 0.0125 kg.m²

radius of the pulley, r = 26 cm = 0.26 m

Torque generated by the pulley on the stone is given as;

τ = F x r = Iα

where;

F is applied force on the stone due to its weight

r is the radius of the pulley

I is moment of inertia

α is angular acceleration (rad/s²)

Force, F = mg = 0.375 x 9.8 = 3.675 N

Torque, τ = F x r

τ = 3.675 x 0.26

τ = 0.9555 N.m

τ = Iα

Angular acceleration, α = τ / I

α = 0.9555 / 0.0125

α = 76.44 rad/s²

Finally, determine linear acceleration, a,  in m/s²

a = αr

a = 76.44 x 0.26

a = 19.87 m/s²

Therefore, the magnitude of the acceleration of the stone is 19.87 m/s²