Question

In figure, a rod of mass m = 1 kg is held
vertically by an unknown mass M hanging
over a pulley as shown. Assuming h = 50
cm and H 100 cm, find the minimum
value of M (in kg) that makes the
equilibrium stable for small displacement.
​

Answer 1

Answer:5kg

Explanation:

F*d=F*d

0.1*M=0.05*10

M=5kg

Answer 2

Answer:

Required minimum value of M is m/2.

Explanation:

When the rod is slightly disturbed from the position shown, that is from equilibrium position by a small angle θ, then the situation is as shown in the figure given below,

Refer to the attached figure

When θ is very small ,the string on the right side is almost vertical and hence the restoring torque due to tension is given by-

Torque due to tension about point O,

             =T.h.Sinθ

             =T.h.θ(Since θ is very small)

Torque due to weight of the rod about O,

=0.5mghSinθ

=0.5m×g×h×θ

Now for stable equilibrium,

Restoring torque ≥ Overturning torque

T×h×θ≥ 0.5×m×g×h×θ

T≥0.5×mg

Mg≥0.5mg

=M≥ m/2

minimum value of M=m/2.

Hence the required minimum value of M is m/2.