Question

A textbook of mass 2.08 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.130 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move a distance 1.30 m over a time interval of 0.830 s.

1) What is the tension in the part of the cord attached to the textbook?

2) What is the tension in the part of the cord attached to the book?
Take the free fall acceleration to be g = 9.80 m/s2 .

3) What is the moment of inertia of the pulley about its rotation axis?
Take the free fall acceleration to be g = 9.80 m/s2 .

Answer 1

a) T1 = 7.48 N

b) T2 = 18.09N

c) I = 0.0115 kgm^2

Step-by-step explanation:

Mass of textbook= 2.08kg

Diameter of cord = 0.130m

Radius = 0.130/2 = 0.065m

Distance = 1.30m

Time interval = 0.830 sec

From equation of motion,

S = ut + 1/2at^2

U = initial velocity = 0

S = 1/2at^2

2S = at^2

a = 2S/t^2

a = 2(1.3) / 0.830^2

a = 3.77 m/s^2

T1 = m1*a ..........(1)

m2g - T2 = m2a..........(2)

T1 = (2.08)(3.77)

T1 = 7.84N

m2g - T2 = m2a

T2 = m2g - m2a

T2 = m2(g - a)

T2 = 3(9.80 - 3.77)

T2 = 3(6.03)

T2 = 18.09N

Moment of inertia (I)

I = [(T2 - T1) r^2]/a

I = [(18.09 - 7.84)(0.065)^2]/ 3.77

I = (10.25)(0.065)^2 / 3.77

I = 0.0115 kgm^2