Question

A textbook of mass 2.00kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100m , to a hanging book with mass 2.98kg . The system is released from rest, and the books are observed to move a distance 1.15m over a time interval of 0.850s

Answer 1

we find the linear acceleration by knowing the masses moved 1.3 m in 0.8s

distance = 1/2 at^2 => a=2d/t^2
a=2*1.3m/0.64s^2=4.06m/s/s

now apply newton's second law to the 2kg book:

T1= 2a =>T1=2kg*4.06m/s/s=8.12N

apply newton's second law to T2:

T2-mg = -ma
T2=3g-3a=3(9.8m/s/s-4.06m/s/s)=17.2N

now, consider the pulley

T1 exerts a force of 8.12 N in one direction, and T2 exerts a force of 17.2N in the opposite direction, the net force on the pulley of
9.08N 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 =9.08N*(0.06m)^2/4.06m/s/s
I=8.1x10^(-3)kgm^2