Question

The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14 cm in diameter and has a mass of 2.9 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.55 N.m. 

Part A If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Answer 1

m2*g - T2 = m2*a

T1 - m1*g = m1*a

Adding, (T1 - T2)+ (m2 - m1)*g = (m1 + m2)*a

(T2 - T1) = (m2 - m1)*g - (m1 + m2)*a

For pulley, Torque = (T2 - T1)*R - 0.55 = Iα

α = a/R and I = 1/2*m_p*R^2

So,  (T2 - T1)*R - 0.55 = (1/2*m_p*R^2)*(a/R)

(T2 - T1) - 0.55/R = 1/2*m_p*a

(m2 - m1)*g - (m1 + m2)*a - 0.55/R = 1/2*m_p*a

a = [(m2 - m1)*g - 0.55/R] / (m1 + m2 + 1/2*m_p)

a = [(4.0 - 2)*9.81 - 0.55/(0.14/2)] / (2.9+ 4.0 + 1/2*2)

a = 2.381 m/s^2

Using h = ut + 1/2*at^2 we get

1 = 0 + 1/2*2.381*t^2

t = 0.916 s