Consider the system shown in the figure with m 1 = 27.00 kg, m 2 = 14.60 kg, R =

Question

Consider the system shown in the figure with m1 = 27.00 kg, m2 = 14.60 kg, R = 0.24 m, and the mass of the uniform pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.40 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) Calculate the time interval required for m1 to hit the floor if the pulley were massless?

Explanation / Answer

a) 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) = 2.5a ................(3)

adding equations (1), (2) and (3), we get

a = (m1 - m2)g/ (m1 + m2 +2.5) = [(27 - 14.6) * 9.8]/(27 + 14.6 +2.5)

a = 2.8m/s2

so time taken to cover a distance of 4.4m with an acceleration of 2.8m/s2, t = sqrt(2h/a)

so, t = sqrt [(2 * 4.4)/2.8]

therefore, t = 1.77secs

b) acceleration when the pulley is massless, a = (m1 - m2)g/(m1 + m2) = [(27 - 14.6) * 9.8]/(27 + 14.6)]

so, a = 2.92m/s2

now, the time taken, t = sqrt(2h/a)

so, t = sqrt [(2 * 4.4)/2.92]

t = 1.73secs

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