In the figure, block 1 has mass m 1 = 480 g, block 2 has mass m 2 = 580 g, and t
ID: 1459694 • Letter: I
Question
In the figure, block 1 has mass m1 = 480 g, block 2 has mass m2 = 580 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.7 cm. When released from rest, block 2 falls 78 cm in 5.0 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.
i1M ImExplanation / Answer
m1 = 480 grams = 0.48 kg
m2 = 580 grams = 0.58 kg
Let a is the acceleration of blocks.
a) Apply, d = 0.5*a*t^2
a = 2*d/t^2
= 2*0.78/5^2
= 0.0624 m/s^2
b) net force acting on m2, Fnet2 = m2*g - T2
m2*a = m2*g - T2
T2 = m2*(g - a)
= 0.58*(9.81 - 0.0624)
= 5.654 N
c) net force acting on m1, Fnet1 = T1 - m1*g
m1*a = T1 - m1*g
T1 = m1*(g + a)
= 0.48*(9.81 + 0.0624)
= 4.74 N
d) angular acceretaion of pulley, alfa = a/R
= 0.0624/0.047
= 1.327 rad/s^2
e) Net torque on pulley = I*alfa
T2*R - T1*R = I*alfa
==> I = R*(T2 - T1)/alfa
= 0.047*(5.654 - 4.74)/1.327
= 0.0323 kg.m^2