Question

# Consider the system shown in the figure below with m1 = 23.0 kg, m2 = 12.8...

Consider the system shown in the figure below with m1 = 23.0 kg, m2 = 12.8 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 4.90 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. ?t1 = s (b) How would your answer change if the pulley were massless? ?t2 = s

(a)

Let T1 = tension in the cord on m1 side

and T2 = tension in the cord on m2 side

For motion of m1,

m1g - T1 = m1a ... ( 1 )

For motion of m2,

T2 - m2g = m2a ... ( 2 )

Adding ( 1 ) and ( 2 ),

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

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

For rotational motion of the pulley,

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

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

From ( 3 ) and ( 4 ),

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

=> a

= [(m1 - m2) g] / [(1/2) M + m1 + m2]

= [(23 – 12.8) * 9.8] / [5/2 + 23 + 12.8]

= 2.60992 m/s^2

d = (1/2) at^2

=> t^2 = 2d/a = 2 * 4.9 / (2.60992)

=> t = 1.938 sec.

(b)

If the pulley were massless, taking M = 0,

a = [(m1 - m2) g] / [m1 + m2]

= [(23 – 12.8) * 9.8] / [23 + 12.8]

= 2.7922 m/s^2

d = (1/2) at^2

=> t^2 = 2d/a = 2 * 4.9 / (2.7922)

=> t = 1.873 sec.