Question

A uniform disc of mass M and radius R is hinged at its lowest point and placed in gravity free
space. It is attached with two identical springs in natural length and an unstreched string of
cross-sectional area A, length L and Young's modulus Y. The time period of angular oscillation
of small amplitude of the disc is​

Answer 1

Answer:

Time of the motion of the disc is given as

$T = \pi(\sqrt{\frac{3m}{10k}} + \sqrt{\frac{3mL}{(10kL + 8YA)}})$

Explanation:

When we displace the center of disc towards right by small displacement "x" then we have

$kx R + k(2x)(2R) + (\frac{YA}{L})(2x)(2R) = - I\alpha$

so we have

$(5kR + \frac{4YAR}{L}) (R\theta) = -\frac{3}{2}mR^2\alpha$

so we have

$\frac{(10kR + \frac{8YAR}{L})}{3mR}\theta = -\alpha$

so we have

$\omega = \sqrt{\frac{(10k + \frac{8YA}{L})}{3m}}$

so half time period of the disc to move to and fro towards right is given as

$\frac{T_1}{2} = \pi\sqrt{\frac{3mL}{(10kL + 8YA)}}$

Now when it is moving towards left the string will lose and hence torque due to string will be zero

so we have

$kxR + k(2x)(2R) = - I \alpha$

$5kR(R\theta) = - (\frac{3}{2}mR^2)\alpha$

$\alpha = - \frac{10k}{3m} \theta$

so half time period of the disc to move to and fro towards right is given as

$\frac{T_1}{2} = \pi \sqrt{\frac{3m}{10k}}$

So total time period of the motion of disc is given as

$T = \frac{T_1}{2} + \frac{T_2}{2}$

$T = \pi(\sqrt{\frac{3m}{10k}} + \sqrt{\frac{3mL}{(10kL + 8YA)}})$

#Learn

Topic : SHM

https://brainly.in/question/6897559