Question

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 10.7 kg and m2 = 46.0 kg. The pulley can be treated as a uniform, solid, cylindrical disk. The downward acceleration of the 46.0-kg block is observed to be exactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley.

Answer 1

Tension on 46 Kg side be T1 and than on 10.7 Kg side be T2
a = g/2 = 9.8/2 = 4.9 m/s^2

T1 = m1*(g -a) = 46*(9.8 - 4.9) = 225.4 N
T2 = m2*(g+a) =10.7*(9.8+4.9) = 157.3 N

For pulley:
net torque = I*alpha
(T1-T2) *r= I*alpha

Let mass if pulley be m kg and radius be r m
I = 0.5*m*r^2
alpha = a/r = 4.9 / r

putting values:
(T1-T2) *r= I*alpha
(225.4 - 157.3)* r = (0.5*m*r^2) * 4.9 / r
(225.4 - 157.3) = (0.5*m) * 4.9
m = 27.8 Kg
Answer: 27.8 Kg