Question

Consider the system shown in the figure. A rope goes over a pulley. A mass, m, is hanging from the rope. A spring of stiffness, k, is attached at one end of the rope. Assume rope is inextensible, massless and there is no slip between pulley and rope.



The pulley radius is r and its mass moment of inertia is J. Assume that the mass is vibrating harmonically about its static equilibrium position.
The natural frequency of the system is

• A. $\sqrt{\frac{k{r}^2}{J+m{r}^2}}$
• B. $\sqrt{\frac{k{r}^2}{J}}$
• C. $\sqrt{\frac{k{r}^2}{J-m{r}^2}}$
• D. $\sqrt{\frac{k}{m}}$

Answer 1

The correct option is A $\sqrt{\frac{k{r}^2}{J+m{r}^2}}$


$E=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2+\frac{1}{2}I{\omega }^2$

$=\frac{1}{2}m{r}^2{\dot{\theta }}^2+\frac{1}{2}k{x}^2{\theta }^2+\frac{1}{2}J{\theta }^2$

$\frac{1}{2}\left[\right(J+m{r}^2){\dot{\theta }}^2+k{r}^2{\theta }^2]=0$

$\frac{dE}{dt}=0$

$(J+m{r}^2)\times 2\dot{\theta }\ddot{\theta }+k{r}^{2}2\theta \dot{\theta }=0$

$\left(\ddot{\theta }\frac{k{r}^2}{J+m{r}^2}\right)\theta =0$

${\omega }_n=\sqrt{\frac{k{r}^2}{J+m{r}^2}}$