A 2.00-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 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

(a) The tension on the first book is 7.5 N and the tension of the second book is 18.15 N.

(b) The moment of inertia of the pulley about its rotation is 0.016 kgm².

The given parameters:

The acceleration of the book is calculated as follows;

The tension on the first book is calculated as follows;

The tension on the second book is calculated as follows;

The moment of inertia of the pulley about its rotation is calculated as follows;

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Given that,

Mass of textbook Mb = 2kg

Initially at rest Ub=0m/s

diameter of pulley d = 0.15m

Then, the radius is r = d/2 =0.15/2 = 0.075m

A hang book of mass Mh = 3kg

Initial at rest too Uh = 0m/s

Distance moved = 1.2m in t = 0.8s

Check attachment for free body diagram and better understanding

Analysis of the mass of textbook on table

Using Newton second law

ΣFx = m•ax

T1 = Mb•ax

T1 = 2•ax, equation 1

Now, analysis of the hanged body

Using the same newton's law

ΣFy = m•ay

T2 — Mh•g = —3•ay,

ay is negative because it is moving in the negative direction

Then, T2 = Mh•g — 3•ay

T2 = 3×9.81 —3•ay

T2 = 29.43 — 3•ay, equation 2

Now, the body moves 1.2m in t=0.8second

The initial velocity of the body is 0m/s

Using equation of motion to calculate the acceleration (a)

S=ut+½at²

1.2 = 0•t + ½ × a × 0.8²

1.2 = 0 + 0.32a

0.32a =1.2

a = 1.2/0.32

a = 3.75m/s²

Since the body are connected by an I inextensive string, their acceleration are the same i.e. ax=ay=a=3.75m/s²

So, back to equation 1

T1 = 2•ax

T1 = 7.5 N

Also, back to equation 2

T2 = 29.43 — 3•ay

T2 = 29.43 — 3 × 3.75

T2 = 29.43 — 11.25

T2 = 18.18 N

b. Torque?

Modelling the pulley as a rigid body

Then, applying equilibrium of torque

Clockwise torque = anti-clockwise torque

Σ τ = (T2-T1)r = Iα

Where α is angular acceleration

Relationship between angular acceleration and radial acceleration is given as a=αr

Therefore,

(T2-T1)r = Iα

Since α=a/r

(T2-T1) = Ia/r

Cross multiply

(T2-T1)r² = Ia

(18.18 — 7.5) × 0.075² = I(3.75)

10.68×0.075² = 3.75I

0.060075 = 3.75I

Then, I = 0.060075/3.75

I = 0.01602 kgm²

I ≈ 0.016 kgm²

Then, the moment of inertia is 0.016 kgm²

there are no contact forces mentioned on the list of choices that you've provided.