Question

A 2.10 kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.170 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 0.900 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?

Answer 1

Answer:

Explanation:

Answer:

Explanation:

mass of book, m = 2.10 kg

diameter of pulley, = 0.170 m

radius of pulley, R = 0.085 m

mass of hanging book, m' = 3 kg

initial velocity, u = 0 m/s

distance, s = 1.2 m

time, t = 0.9 s

Let a be the acceleration of the system and T and T' is the tension in the string which is horizontal and vertical respectively.

Use second equation of motion

s = ut + 0.5 at²

1.2 = 0 + 0.5 x a x 0.9 x 0.9

a = 2.96 m/s²

(a) Use second equation of motion

T = ma

T = 2.10 x 2.96

T = 6.216 N

m'g - T' = m'a

3 x 9.8 - T' = 3 x 2.96

T' = 20.52 N

(b) Let the moment of inertia of the pulley is I.

So, (T' - T)R = I x α

(20.52 - 6.216) x 0.085 = I x 2.96 / 0.085

I = 0.035 Kgm²