A textbook of mass 2.06 kg rests on a frictionless, horizontal surface. A cord a

Question

A textbook of mass 2.06 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 2.99 kg . The system is released from rest, and the books are observed to move a distance 1.24 m over a time interval of 0.750 s.

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

b) What is the tension in the part of the cord attached to the book? (Take the free fall acceleration to be g = 9.80 m/s2.)

c) What is the moment of inertia of the pulley about its rotation axis? (Take the free fall acceleration to be g = 9.80 m/s2.)

Explanation / Answer

here,

distance , x = 1.24 m

t = 0.75 s

distance = 1/2 at^2

a = 2 * x/t^2

a = 2*1.24 / 0.75s

a = 3.31 m/s^2

(a)

now apply newton's second law to the 2.06 kg book:

T1= 2a

T1 = 2 * 3.31

T1 = 6.62 N

(b)

apply newton's second law to T2:

T2 - mg = - ma

T2 = 2.99 * ( 9.8 - 3.31)

T2 = 19.41 N

(c)

now, consider the pulley

T1 exerts a force of 6.62 N in one direction, and T2 exerts a force of 19.41 N in the opposite direction,

the net force on the pulley of
12.79 N 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

I = 12.79 * (0.05 )^2 / 3.31

I = 9.66 * 10^-3 kgm^2

the moment of inertia of the pulley about its rotation axis is 9.66 * 10^-3 kgm^2

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