In the figure, two 9.30 kg blocks are connected by a massless string over a pull

Question

In the figure, two 9.30 kg blocks are connected by a massless string over a pulley of radius 2.00 cm and rotational inertia 7.40 × 10-4 kg·m2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 0.600 rad in 154 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension T1, and (d) string tension T2? Assume free-fall acceleration to be equal to 9.81 m/s2.

Explanation / Answer

Given,

m = 9.3 kg ; I = 7.4 x 10^-2 kg-m^2

r = 2 cm ; theta = 0.6 rad in t = 154 ms

a)We know from eqn in circular motion

theta = w0t + 1/2 alpha t^2

0.6 = 0 + 1/2 x alpha (154 x 10^-3)^2

alpha = 2 x 0.6/(154 x 10^-3)^2 = 50.59 rad/s^2

Hence, alpha = 50.59 rad/s^2

b)a = R alpha

a = 0.02 x 50.59 = 1.012 m/s^2

Hence, a = 1.012 m/s^2

c)T1 = m1 (g + a)

T1 = 9.3 (9.81 - 1.012) = 81.82 N

Hence, 81.82 N

d)T2 = T1 - I alpha/R

T2 = 81.82 - 7.4 x 10^-4 x 50.59/0.02 = 79.95 N

Hence, T2 = 79.95 N

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