Question

A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.190 m, to a hanging book with mass 2.97 kg. The system is released from rest, and the books are observed to move a distance 1.15 m over a time interval of 0.820 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?

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

Answer 1

a. T₁ = 7.18 N

b. T₂ = 18.95 N

c. The moment of inertia of the pulley about its rotational axis is given by;

I is 3.104×10⁻² kg·m²

Step-by-step explanation:

Here we have

The equation of motion of the book given by

Δx = v₁ₓ·t + 1/2·a·t²

Where:

Δx= Distance moved by the book = 0.190 m

a = Acceleration of the book

t = Time of motion = 0.820 s

Since the book is initially at rest, we have

v₁ₓ = 0 m/s

∴ Δx = 1/2·a·t² → 0.190 m = 1/2·a·(0.820 s)²

a₁ = 3.42 m/s²

Force on string = Tension in string

= Mass of textbook on surface × Acceleration of textbook on surface

T₁ = m₁ × a₁ = 3.52 m/s² × 2.1 kg = 7.18 N

The tension in the part of the cord attached to the textbook = 7.18 N

b. The tension in the part of the cord attached to the book is given by

∑Fy = Mass of book × acceleration due to gravity - Tension in cord attached to the book = Mass of book × Acceleration of book

= m₂g - T₂ = m₂a₂

Since The book and the textbook are connected via a chord that cannot be extended, we have

a₁ = a₂ and

T₂ = m₂g -m₂a₁ = 2.97×9.8 - 2.97 × 3.42 m/s² = 18.95 N

c. Here the turning force is given by

(T₂ - T₁)r = Iα

Where:

α = Angular acceleration

r = Radius of pulley = 0.19/2 = 0.095 m

I = Moment of inertia of the pulley

Since α = a₁/r, we have

I = (T₂ - T₁)r/α = (T₂ - T₁)r²/a₁

I = (18.95 N - 7.18 N) (0.095 m)²/3.42 m/s² = 3.104×10⁻² kg·m²

Moment of inertia of the pulley about its rotational axis, I = 3.104×10⁻² kg·m².