Q.349. Pulley with mass and two blocks - Mechanics

Question 349.
Consider the system as shown in the figure with m1 = 20 kg, m2 = 12.5 kg, R = 0.2 m, and the mass of the pulley M = 5 kg. Object m2 is resting on the floor and object m1 is 4 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.
b) How would your answer change if the pulley were massless?

Answer 349.
Since pulley is not massless, tension on both sides of the string will be different.

Also, m1 > m2 will result in m1 moving downwards and rotating the pulley in anticlockwise direction
=> tension T1 on m1 side > tension T2 on m2 side

Let a = acceleration of m1 downwards and m2 upwards
=> angular acceleration of the pulley = a/R

For the motion of m1 and m2,
m1g - T1 = m1a ... ( 1 )
T2 - m2g = m2a ... ( 2 )

For the motion of the pulley,
(T1 - T2) * R = M.I. of pulley * angular acceleration
=> (T1 - T2) * R = 5 *(1/2) R^2 * a/R
=> T1 - T2 = 2.5 a ... ( 3 )

Adding eqns. ( 1 ), ( 2 ) and ( 3 ),
m1g - m2g = m1a + m2a + 2.5 a
=> a
= (m1 - m2)g / (m1 + m2 + 2.5)
= (20 - 12.5) * (9.81) / (20 + 12.5 + 2.5)
= 2.102 m/s^2

a)
Time to cover a distance of 4 m moving down with an acceleration a = 2.194 m/s^2
= √[(2*4)/(2.102)] sec
= 1.9509 sec.

b)
acceleration when pulley is massless
= (m1 - m2)g/(m1 + m2)
= (7.5 * 9.81) / (32.5)
= 2.2638 m/s^2
Time for m1 to hit the floor
= √[(2*4)/(2.2638)] sec
= 1.8798 sec
=> time reduces by
(1.9509 - 1.8798) x 100 / (1.9509) %
= 3.64 %.

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