Question

A block (mass = 3.0 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.2 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.033 m during the block's descent. Find (a) the angular acceleration of the pulley and (b) the tension in the cord.

Answer 1

Given that I = 1.2*10^-3 Kg-m^2

net force acting on the block during descending is

Let F be the tension in the string

Fnet = m*a

mg-F = m*a

3*9.8 - F= 3*a

F = 29.4-(3*a)

and Torque T = r*F

I*alpha = r*F

but angular accelaration alpha = a/r

I*a/r = r*F

1.2*10^-3*a = r^2*F

1.2*10^-3*a = 0.033^2*F

a= 0.033^2*F/(1.2*10^-3) = 0.9075*F

then F = 29.4-(3*a)

F = 29.4 - (3*0.9075*F)

tension is F = 7.89 N...answer for (b)

angular accelaration is alpha = a/r

a = 0.9075*F =0.9075*7.89 = 7.16 m/s^2

alpha = a/r = 7.16/0.033 = 217 rad/s^2....answer for (a)