A uniform dise of mass `m` and radius `R` is connected with two light springs `1` and `2`. The springs are connected at the highest point `M` and the `CM` 'N' of the dise. The other ends of the springs are rigidly attached with vertical walls. If we shift the `CM` in horizontalby a small distance , the discoscillates simple harmonically. Assuming a perfect rolling of the dise on the horizontal surface , find the angular frequency of oscillation.

Answer

`sqrt((2(4 k_(1) + k_(2)))/(3m))`

Explanation

Let us displace the CM by asmall distance x towards right. As a result the body rolls on the horizontalsurface. Hence spring 1 is elonged by `x_(1) = 2 R theta` and spring 2 is compressed by `x_(1) = 2 R theta`, where `theta` is the angle of rotation of the rolling body.

The spring forces `F_(1) and F_(2)` produce anticlockwise torques `F_(1) (2 R) and F_(2) (R)`, respectively about P. Summing up the spring tarque about P, we have
`tau_(net) = F_(1)(2 R) and F_(2) (R)`
where `F_(1) = k_(1) x_(1) = k_(1) (2 R theta)`
and `F_(2) = k_(2) x_(2) = k_(2) (R theta)`
This gives `tau_("net") = - (4 R^(2) k_(1) + R^(2) k_(2)) theta = - (4 k_(1) + k_(2)) R^(2) theta`
This tarque produces anticlokwise acceleration `alpha` for clocwise displacement `theta` , which can be given as
`alpha = (tau_(net))/(l_(P)) = (-(4 k_(1) + k_(2)) R^(2) theta)/(l_(P))`
Substituting `l_(P) = mR^(2) + l_( C) = mR^(2) + (mR^(2))/(2) = (3 mR^(2))/(2)`
We have `alpha = (-2(4 k_(1) + k_(2)))/(3m) theta`
Comparing the above equation with `alpha = - omega^(2) theta`, we have
`omega = sqrt((2(4 k_(1) + k_(2)))/(3m))`