Question

A uniform disc 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 center of mass (CM) 'N' of the disc. The other ends of the springs are rigidly attached with vertical walls. If we shift the CM in horizontal direction by a small distance, the disc oscillates simple harmonically. Assuming a perfect rolling of the disc on the horizontal surcface, the angular frequency of oscillation is (answer upto two decimal places) $(Take{k}_1=10N/m,k_2=20N/m,m=10kg)$

Answer 1

Let us displace the CM by a small distance $x$ towards right. As a result the body rolls on the horizontal surface. Hence the spring 1 is elongated by $x_1=2R\theta$ and spring 2 is compressed by $x_2=R\theta ,$ where $\theta =$ angle of rotation of the rolling body.


The spring forces $F_1$ and $F_2$ produce anticlockwise torques $F_1\left(2R\right)$ and $F_2\left(R\right)$ respectively about $P$ . Summing up the spring torques about $P$ , we have $${\tau }_{\text{net}}=F_1\left(2R\right)+F_2\left(R\right)$$ Where $F_1=k_1{x}_1=k_1\left(2R\theta \right)$ and $F_2=k_2{x}_2=k_{2}R\theta$
This gives
${\tau }_{net}=-(4R^2{k}_1+R^2{k}_2)\theta =-(4k_1+k_2)R^2\theta$This torque produces anticlockwise acceleration $\alpha$ for clockwise displacement $\theta ,$ which can be given as
$\alpha =\frac{{\tau }_{net}}{I_P}=\frac{-(4k_1+k_2)R^2\theta }{I_P}$
Substituting $I_P=m{R}^2+I_C=m{R}^2+\frac{m{R}^2}{2}=\frac{3m{R}^2}{2}$
We have $\alpha =\frac{-2(4k_1+k_2)}{3m}\theta$
Comparing the above equation with $a=-{\omega }^2\theta$
We have $\omega =\sqrt{\frac{2(4k_1+k_2)}{3m}}=\sqrt{\frac{2(4\times 10+20)}{3\times 10}}=2$