g 1.90 kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.10 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.900 s . Part A What is the tension in the part of the cord attached to the textbook? Express your answer with the appropriate units. T1 = nothing nothing Request Answer Part B What is the tension in the part of the cord attached to the hanging book? Express your answer with the appropriate units. T2 = nothing nothing Request Answer Part C What is the moment of inertia of the pulley about its rotation axis? Express your answer with the appropriate units. I = nothing nothing Request Answer Provide Feedback

Explanation:

First of all we shall calculate the acceleration of the system . Let it be a .

s = ut + 1/2 a t²

1.2 = 1/2 x a x .9²

a = 2.963

A ) Applying 2nd law of motion on textbook

T₁  = ma , m is mass of the textbook a is its acceleration

T₁ = 1.9 x 2.963

= 5.63 N .

B ) Applying 2nd law of motion on hanging book

mg - T₂ = ma , m is mass of hanging book , a is its acceleration

T₂ = mg - ma

= m ( g - a )

= 3.1 x ( 9.8 - 2.963 )

= 21.2 N

C )

Torque on the pulley

= ( T₂ - T₁ )  x R

R is radius of the pulley

= (21.2 - 5.63 ) x .075

= 1.16775  Nm

angular acceleration of pulley

α = linear acceleration /  radius

= 2.963 /  .075

= 39.5 rad / s²

torque on pulley = moment of inertia x angular acceleration

1.16775 = moment of inertia x 39.5

moment of inertia = 29.56 x 10⁻³ kg m² .

this would be about 43%