Question

12.
A rod of mass 'M' and length 'L' is hinged about
one end and a particle of mass 'M' is attached to
its other end. If it is kept horizontal with the help
of a spring of spring constant 'K' as shown in fig.
find the extension in spring :-​

Answer 1

Answer:

3Mg/K

Explanation:

See the attachment

Answer 2

The extension in spring is (1) 3Mg/K

Explanation:

The force at the center of the mass is:

L = ((M × L/2) + (M × L))/2M

∴ L = 3L/4

The torque is:

τ₁ = 2Mg × (3L/4)

∴ τ₁ = (3MgL)/2

The torque is:

τ₂ = τ₁

F × L/2 = (3MgL)/2

∴ F = 3Mg

Now, the force on the spring is given as:

F = kx

3Mg = kx

∴ x = (3Mg)/k