Question

A solid disc with radius a is connected to a spring at a point d above the center of the disc. The other end of spring is fixed to the vertical wall. The disc is free to roll without slipping on the ground. The mass of the disc is M and the spring constant is k. The polar moment of inertia for the disc about its center is $J=\frac{M{a}^2}{2}$


The natural frequency of this system is rad/s is given by

• A. $\sqrt{\frac{2k(a+d{)}^2}{3M{a}^2}}$
• B. $\sqrt{\frac{2k}{3m}}$
• C. $\sqrt{\frac{2k(a+d{)}^2}{M{a}^2}}$
• D. $\sqrt{\frac{k(a+d{)}^2}{M{a}^2}}$

Answer 1

The correct option is A $\sqrt{\frac{2k(a+d{)}^2}{3M{a}^2}}$




Total energy of system

$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}k{x}^2+\frac{1}{2}I{\omega }^2$

$\Rightarrow \frac{1}{2}M\times {(a\times \frac{d\theta }{dt})}^2+\frac{1}{2}k\times {\left\{\right(a+d\left)\theta \right\}}^2+\frac{1}{2}\times \frac{M{a}^2}{2}\times {\left(\frac{d\theta }{dt}\right)}^2$

$\text{Let}\frac{d\theta }{dt}=\dot{\theta }$

$E=\frac{1}{2}M{a}^2{\dot{\theta }}^2+\frac{1}{2}(a+d{)}^2.{\theta }^2+\frac{M{a}^2}{4}{\dot{\theta }}^2$

Since the energy of the system remains concerned with respect to time.

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

$\frac{dE}{dt}=\frac{M{a}^2}{2}2.\dot{\theta }.\ddot{\theta }+\frac{k}{2}(a+d{)}^2.2\theta .\dot{\theta }+\frac{M{a}^2}{4}.2\dot{\theta }\ddot{\theta }$

$0=M{a}^2\dot{\theta }\ddot{\theta }+\frac{M{a}^2}{2}.\dot{\theta }\ddot{\theta }+k(a+d{)}^2\theta .\dot{\theta }$

Divide both sides by $\dot{\theta }$

$0=M{a}^2+\frac{M{a}^2}{2}.\ddot{\theta }+k(a+d{)}^2.\theta$

$0=(M{a}^2+\frac{M{a}^2}{2})\ddot{\theta }+k(a+d{)}^2\theta$

$0=\frac{3}{2}M{a}^2.\ddot{\theta }+k(a+d{)}^2\theta$
Divide both sides by $\frac{3}{2}m{a}^2$

$0=\ddot{\theta }+\frac{k(a+d{)}^2}{\frac{3M{a}^2}{2}}\theta$

$\ddot{\theta }+{\omega }_n^2\theta =0$

Comparing the above two equations:

${\omega }_n^2=\frac{2k(a+d{)}^2}{3M{a}^2};{\omega }_n=\sqrt{\frac{2k(a+d{)}^2}{3M{a}^2}}$