Question

Two blocks of mass m1 = 2 kg and m2 = 1 kg are suspended over a pulley of mass M with a string of neglible mass. The pulley is a solid cylinder of radius R = 4 cm. The string does not slip on the pulley, which is a solid cylinder. The magnitude of downward accleration of mass m1 is measured to be a = g/4.

A) Write down Newton's second law for m1, m2, and M using the relevant sumbols m1, m2, M, R, g, and a.

B) What is the mass of the pulley?

Answer 1

rotational inertia of the pulley = I = 0.5 MR^2

from Newtons second law

on mass m1,   let T1 be the tension and a be its acceleration

then m1g - T1 = m1 a

let T2 be the tension in the string 2,

then

T2 - m2g = m2 a

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from Rotational Equillibirum

T1 r - T2 r = Net Torque = I * alpha

solving for accleration a

a = (m1 -m2)*g/(m1 +m2 + I/R^2)

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here given than a = g/4

so

9.8 /4 = (2 -1)* 9.8/((2+1 + (0.5 M))

solving for M

M = 2 kg