Question

a rod of mass m and lenght l is hinged at one end and particle of mass m​

Answer 1

Answer:

$\boxed{\mathfrak{(1) \ \frac{3Mg}{K}}}$

Given:

Mass of rod hinged at one end = M

Length of rod = L

Mass of particle attached to rod at other end = M

Spring constant = K

To Find:

Extension in spring (x)

Explanation:

Centre of mass of the system:

$\boxed{ \bold{x_{cm} = \frac{\sum\limits_{i=1}^N m_ix_i}{M}}}$

$\sf \implies x_{cm} = \frac{M \times \frac{L}{2} + M \times L}{2M}$

$\sf \implies x_{cm} = \frac{ \frac{ML}{2} + M L}{2M}$

$\sf \implies x_{cm} = \frac{ \frac{3 \cancel{M}L}{2} }{2 \cancel{M}}$

$\sf \implies x_{cm} = \frac{ 3L }{2 \times 2}$

$\sf \implies x_{cm} = \frac{ 3L }{4}$

By rotational equilibrium we can say that torque due to centre of mass will be equal to torque due to spring.

$\sf \implies \tau_{cm} = \tau_{spring}$

$\sf \implies 2Mg \times \frac{3 \cancel{L}}{4} = Kx \times \frac{ \cancel{L}}{2}$

$\sf \implies \cancel{2}Mg \times \frac{3}{2 \times \cancel{2}} = \frac{Kx}{2}$

$\sf \implies Kx = \cancel{2}Mg \times \frac{3}{ \cancel{2}}$

$\sf \implies Kx = 3Mg$

$\sf \implies x = \frac{3Mg}{K}$

$\therefore$

$\sf Extension \: in \: spring \: (x) = \frac{3Mg}{K}$

Answer 2

Answer:

Torque on center of mass is equal to torque due to spring.

So we will get:

x = 3Mg/K