Question

Consider the system shown in the figure below with m1 = 21.0 kg, m2 = 10.7 kg, R = 0.130 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.60 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley. (a) Calculate the time interval required for m1 to hit the floor. Δt1 = s (b) How would your answer change if the pulley were massless? Δt2 = s

Answer 1

Let T1 and T2 be tension on cord of mass m1 and m2 respectively

Thus

for mass m1 we have

m1g - T1 = m1a -------- (1)

for m2:

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

Adding and rearranging equation 1 and 2 we get

T1 - T2 = (m1 - m2)g - (m1 + m2)a ------- (3)

For given pulley we have

(T1 - T2) R = (1/2) MR^2 * a/R

T1 - T2 = (1/2) Ma ------- (4)

equating 3 and 4 we have

(m1 - m2)g - (m1 + m2)a = (1/2) Ma

therefore, a = [(m1 - m2) g] / [(1/2) M + m1 + m2] = [(21 - 10.7) * 9.81] / [5/2 + 21 + 10.7] =2.5944 m/s^2

now distance, d = 0.5at^2 (by second law of motion)

Thus t^2 = 2d/a = 2*4.6/ (2.5944)

t = 1.8831 s

part b)

for Massless pulley i.e. M = 0 we have

a = [(m1 - m2) g] / [m1 + m2] = [(21 - 10.7) * 9.81] / [21 + 10.7] = 3.187 m/s^2

d = (1/2) at^2

t^2 = 2d/a = (2 * 4.6 )/ (3.187)
t = 1.699 sec.