Question

A rigid massless rod of length L is connected to a disc (pulley) of mass m and radius r = L/4 through a friction-less revolute joint. The other end of theat rod is attached to a wall through a frictionless hinge, A spring of stiffness 2k is attached to the rod at its mid-span. An inextensible rope passes over half the disc periphery and is securely tied to a spring of stiffness k and point C as shown in the figure. There is no slip between the rope and the pulley. The system is in static equilibrium in the configuration shown in the figure and the rope is always taut.


Neglecting the influence of gravity, the natural frequency of the system for small amplitude vibration is

• A. $\sqrt{\frac{k}{m}}$
• B. $\frac{3}{\sqrt{2}}\sqrt{\frac{k}{m}}$
• C. $\sqrt{3}\sqrt{\frac{k}{m}}$
• D. $\sqrt{\frac{3}{2}}\sqrt{\frac{k}{m}}$

Answer 1

The correct option is C $\sqrt{3}\sqrt{\frac{k}{m}}$


MI about point 'O':
$I=\frac{m{r}^2}{2}+m{l}^2$

$=\frac{m}{2}{\left(\frac{l}{4}\right)}^2+m{l}^2=\frac{33m{l}^2}{32}$

Mi about disc centre:
$I_{disc}=\frac{m{r}^2}{2}$

When the rod rotates by $\beta$

$\beta .r=l.\theta$

$\beta =\frac{l.\theta }{r}$

(If the disc is also rotating about its own centre due to static friction).

Energy Method:

$E=\frac{1}{2}I(\dot{\theta }{)}^2+\frac{1}{2}{I}_{disc}(\dot{\beta }{)}^2+\frac{1}{2}\left(2K\right)\times {(\frac{l}{2}.\theta )}^2+\frac{1}{2}K(2l.\theta {)}^2$

$=\frac{1}{2}.\frac{33m{l}^2}{32}{\dot{\theta }}^2+\frac{1}{2}.\frac{m{r}^2}{2}\left(\frac{l^2}{r^2}\right){\dot{\theta }}^2+\frac{1}{2}.2k.\frac{l^2}{4}.{\theta }^2+\frac{1}{2}K.4l^2{\theta }^2$

$E=\frac{1}{2}(\frac{33m{l}^2}{32}+\frac{m{l}^2}{2}){\dot{\theta }}^2+\frac{1}{2}(\frac{k{l}^2}{2}+4K{l}^2){\theta }^2$

$\frac{1}{2}\left(\frac{49m{l}^2}{32}\right)\dot{\theta }+\frac{1}{2}\left(\frac{9K{l}^2}{2}\right){\theta }^2$

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

$\frac{1}{2}[\frac{49m{l}^2}{32}.2\dot{\theta }\ddot{\theta }+\frac{9K{l}^2}{2}2.\theta .\dot{\theta }]=0$

$\frac{49m{l}^2}{16}\ddot{\theta }+9K{l}^2\theta =0$

$\frac{49m{l}^2}{16}\ddot{\theta }+9K.\theta =0$

$\ddot{\theta }+\frac{49m{l}^2}{16}.\theta =0$

${\omega }_n=\frac{12}{7}\sqrt{\frac{k}{m}}$

This is exact solution.

But this solution is very close to

$\sqrt{3}\sqrt{\frac{k}{m}}$

$\text{Because,}\frac{12}{7}\sqrt{\frac{k}{m}}=\left(1.714\right)\sqrt{\frac{k}{m}}$

$\sqrt{3}\sqrt{\frac{k}{m}}=\left(1.732\right)\sqrt{\frac{k}{m}}$

$\text{Therefore,}{\omega }_n=\frac{12}{7}\sqrt{\frac{k}{m}}=\sqrt{3}\sqrt{\frac{k}{m}}$

Note : If we take an approximation in moment of inertia about hinge axis.

$\text{That,}I=\frac{m{r}^2}{2}+m{l}^2\text{(r=l/4)}$

$=(\frac{m{l}^2}{32}+m{l}^2)$

Then we get, ${\omega }_n=\sqrt{3}\sqrt{\frac{k}{m}}$

But if we take exact inertia,

$I=(\frac{m{l}^2}{32}+m{l}^2)=\frac{33m{l}^2}{32}$

Then the exact answer is

${\omega }_n=\frac{12}{7}\sqrt{\frac{k}{m}}$