Consider the system shown in the figure below with m 1 = 21.0 kg, m 2 = 14.8 kg,

Question

Consider the system shown in the figure below with

m1 = 21.0 kg,

m2 = 14.8 kg,

R = 0.220 m,

and the mass of the pulley

M = 5.00 kg.

Object m2 is resting on the floor, and object m1 is 4.70 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 = 1 s

(b) How would your answer change if the pulley were massless?

t2 = 2 s

Explanation / Answer

Writing equations of motion to m2

T2-m2g = m2a

T2 = m2a + m2g = m2(a+g)

similarly for m1

m1g - T1 = m1a

T1 = m1(g-a)

using Net Torque is Tnet = I*alpha

I is the moment of inertia of the pulley = 0.5*M*R^2 = 0.5*5*0.22^2 = 0.121 Kg-m^2

alpha is the angular accelaration = a/R

then Tnet = 0.121*(a/R)

T2*R-T1*R = 0.121*a/R

(T2-T1)*R = 0.121*a/R

T2-T1 = 0.121*a/0.22^2

m2*a + m2*g - m1*g + m1*a = 0.121*a/0.22^2

(14.8*a)+(14.8*9.8)-(21*9.8)+(21*a) = 0.121*a/0.22^2

accelaration is a = 1.83 m/s^2

Using kinematic equation for m1

initial velocity is Vo = 0 m/sec

a = 1.83 m/s^2

S = 4.7 m

then V^2 -Vo^2 = 2*a*S

V^2 -0^2 = 2*1.83*4.7

V = 4.14 m/sec

and using S = (Vo*t)+(0.5*a*t^2)

4.7 = (0*t)+(0.5*1.83*t^2)

t = 2.26 sec

b) if the pulley were mass less ,then

tension in the two cables is same

T1 = T2 = T

m2*a+ m2*g = m1g-m1*a

accelaration is a = (m1-m2)*g/(m1+m2) = (21-14.8)*9.8/(21+14.8) = 1.69 m/s^2

then using S = (Vo*t)+(0.5*g*t^2)

4.7 = 0.5*1.69*t^2

t = 2.35 sec

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