Consider the system shown in the figure with m 1 = 29.00 kg, m 2 = 14.90 kg, R =
ID: 1371694 • Letter: C
Question
Consider the system shown in the figure with m1 = 29.00 kg, m2 = 14.90 kg, R = 0.30 m, and the mass of the uniform pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.90 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.
s
(b) Calculate the time interval required for m1 to hit the floor if the pulley were massless?
s
Explanation / Answer
On m1 ,
29g - T1 = 29a .........(i)
On pulley ,
torque = I x alpha
(T1 - T2)R = ( M R^2 /2 ) ( a/R)
T1 - T2 = (5a/2) = 2.50a .......(ii)
on m2,
T2 - m2g = m2a
T2 - 14.90g = 14.90a .....(iii)
form i, ii and iii,
29g - 14.90g = (29 + 2.50 + 14.90)a
a = 2.98 m/s^2
using h = ut + at^2 /2
4.90 = 0 + 2.98t^2 /2
t = 1.81 sec
b) when pulley is massless,
m1g - m2g = (m1 + m2)a
29g - 14.90g = (29 + 14.90)a
a = 3.15 m/s^2
4.90 = 0 + 3.15t^2 /2
t = 1.76 sec