In the figure, block 1 has mass m1 = 440 g, block 2 has mass m2 = 510 g, and the

Question

In the figure, block 1 has mass m1 = 440 g, block 2 has mass m2 = 510 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.2 cm. When released from rest, block 2 falls 73 cm in 5.3 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.

Explanation / Answer

here,

mass of block, 1, m1 = 0.440 kg
mass of block, 2, m2 = 0.510 kg

radius of pulley, r = 5.2 cm = 0.052 m

distance fallen, d = 73 cm = 0.73 m
time taken to fall, t = 5.3s

From Second eqn of motion,
h = 0.5*a*t^2

solving for acceleration, a = 2*h/t^2
a = 2*0.73/(5.3)^2
a = 0.05198 m/s^2

Net force acting on block2 , from Newton law of motion Fnet = m*a

Fg - T2 = m2*a ( Fg is force due to gravity)
T2 = m2*g+m2*a (T2 is tension in string due to block 2)
T2 = m2*(g+a)
T2 = 0.510(9.81 + 0.05198)
T2 = 5.02961 N

Since String is unstretchable and massless, so
T1 = T2 = 5.02961 N
a1 = a2 = a = 0.05198 m/s^2

Since angular acceleration, alpha = a/r

also, alpha = 0.05198/0.052 = 0.99962 rad/s^2

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