Question

A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the textbook passes over a pulley whose diameter is 0.120 m, to a hanging book with mass 3.06 kg. The system is released from rest, and the books are observed to move a distance 1.17 m over a time interval of 0.810 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?

Answer 1

A. T₁ = 7.204 N

B. T₂ = 19.07 N

C. I = 0.011979 kg * m²

Step-by-step explanation:

Given

m₁ = 2.02 kg , d = 0.120 m, m₂ = 3.06 kg, s = 1.17 m, t = 0.810s

The tension can be find using the conserved energy in each axis so:

∑Fₓ = T₁ = m₁ * a₁

∑Fy = m₂ * g - T₂ = m₂ * a₁

Now to find the acceleration can use the equation of Newton motion

s = v₁* t + ¹/₂ * a₁ * t ²

Solve to a'

a₁ = 2* s / t ² = 2 * 1.17 m / (0.810 s)²

a₁ = 3.566 m/s²

A).

T₁ = m₁ * a₁ = T₁ = 2.02 kg * 3.566 m/s²

T₁ = 7.204 N

B).

T₂ = m₂ *(g - a₁) = 3.06 kg *(9.8 - 3.566) m/s²

T₂ = 19.07 N

C).

Finally to find the moment of inertia use the equation

∑ T = (T₂ - T₁) * r = I * α

α = a₁ / r

I = (T₂ - T₁ )* r²/ a₁

I = ( 19.07 - 7.204)* (0.06 m)² / 3.566 m/s²

I = 0.011979 kg * m²