A textbook of mass 1.93kg rests on a frictionless, horizontal surface. A cord at

Question

A textbook of mass 1.93kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.140m , to a hanging book with mass 2.91kg . The system is released from rest, and the books are observed to move a distance 1.13m over a time interval of 0.850s .

Part A

What is the tension in the part of the cord attached to the textbook?

Part B

What is the tension in the part of the cord attached to the book?

Take the free fall acceleration to be g = 9.80m/s2 .

Part C

What is the moment of inertia of the pulley about its rotation axis?

Take the free fall acceleration to be g = 9.80m/s2 .

Explanation / Answer

we find the linear acceleration by knowing the masses moved 1.29 m in 0.760s

distance = 1/2 at^2 => a=2d/t^2
a=2*1.13m/0.85s^2= 3.13 m/s/s

now apply newton's second law to the 1.93kg book:

T1= 2a =>T1=1.93kg*3.13m/s/s=6.04N

apply newton's second law to T2:

T2-mg = -ma
T2=2.91g-2.91a=2.91(9.8m/s/s-3.13m/s/s)=19.44N

now, consider the pulley

T1 exerts a force of 6.04N in one direction, and T2 exerts a force of 19.44N in the opposite direction, the net force on the pulley of
13.4N 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 =13.4N*(0.07m)^2/3.13m/s/s
I=20.98x10^(-3)kgm^2

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