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:

(A) 6.2 N and 20.52 N

(B) I = 0.035 kg m^{2}

Explanation:

mass of first book (M1) = 2.1

mass of second book (M2) = 3 kg

diameter of pulley = 0.17 m

distance (s) = 1.2 m

time = 0.9 secs

(A) what is the tension in each part of the cord

    we first have to get the acceleration of the first  book

 s = ut + 0.5 at^{2}

 since the books are initially at rest u = 0

s = 0.5 at^{2}

1.2 = 0.5 x a x 0.9^{2}

a = 2.96 m/s^[2}

• Tension at T1 = M1 x A1

       T1 = 2.1 x 2.96 = 6.2 N

• Tension at T2 = m x ( g - a )

          (g-a) is the net acceleration of the first book

           T2 = 3 x ( 9.8 - 2.96 ) = 20.52 N

(B) What is the moment of inertia of the pulley?

     taking clockwise rotation of the pulley to be negative while

     anticlockwise to be positive, we can see from the diagram that T1      

     causes a clockwise rotation while T2 produces an anticlockwise rotation

       ( T2 - T1 )r = I∝

         where ∝ is the angular acceleration of the pulley relative to its radial  

         acceleration, ∝ = \frac{a}{r}

          ( T2 - T1 )r = I\frac{a}{r}

          I = \frac{(T2 - T1)r^{2}}{a}

          I = \frac{(20.52 - 6.2)0.085^{2}}{2.96}

         I = 0.035 kg m^{2}