Question

A textbook of mass 2.08 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.160 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.30 m over a time interval of 0.770 s . What is the moment of inertia of the pulley about its rotation axis?

Answer 1

we find the linear acceleration by knowing the masses moved 1.3 m in 0.770s

distance = 1/2 at^2 => a=2d/t^2
a=2*1.3m/0.5930s^2=4.385m/s/s

now apply newton's second law to the 2.08 kg book:

T1= 2a =>T1=2.08kg*4.385m/s/s=9.12N

apply newton's second law to T2:

T2-mg = -ma
T2=3.1g-3.1a=3.1(9.8-4.385)= 16.77N

now, consider the pulley

T1 exerts a force of 9.12 N in one direction, and T2 exerts a force of 16.77N in the opposite direction, the net force on the pulley of
7.65N generates a torque

the amount of torque =(T2-T1)R since this force acts a distance R from the rotation axis of the pulley

this torque produces an angular acceleration equal to

torque = I alpha where I is the moment of inertia and alpha is the angular acceleration

alpha is related to linear acceleration according to

a=R alpha or alpha =a/R, so we combine all these and get

(T2-T1)R=I(a/R)
I=(T2-T1)R^2/a =7.65N*(0.08)^2/4.385
I= 11.1653x10^-3kg-m^2