Question

A uniform thin disk of mass 1 kg and radius 0.1 m is kept on a surface as shown in the figure. A spring of stiffness $K_1=400$N/m is connected to the disk center A and another spring of stiffness $k_2=100$ N/m is connected at point B just above point A on the circumference of the disk. Initially, both the springs are unstretched. Assume pure rolling of the disk. For small disturbance from the equilibrium, the natural frequency of vibration of the system is rad/s (round off to one decimal place)

Answer 1

As per given date
Disc mass,

m = 1 kg
Radius, r = 0.1 m
torque equation about point 'o'
$I\ddot{\theta }+(k_1\times r\times \theta )r+(k_2\times 2r\times \theta )2r=0$
I about 'o' $=\frac{m{r}^2}{2}+m{r}^2$
$I=\frac{3}{2}m{r}^2$
$\frac{3}{2}m{r}^2\ddot{\theta }+(k_1{r}^2+k_2(4r^2\left)\right)\theta =0$
$\ddot{\theta }\left(\frac{k_1{r}^2+4k_2{r}^2}{\frac{3}{2}m{r}^2}\right)\theta =0$
${\omega }_n=\sqrt{\frac{k_1{r}^2+4k_2{r}^2}{\frac{3}{2}m{r}^2}}$
$=\sqrt{\frac{400(0.2{)}^2+4\times 100\times {0.2}^2}{1.5\times 1\times {0.02}^2}}$
${\omega }_n=23.094$ rad/s