Question

Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy (a) at mean position and (b) at extreme position. A horizontal disc is freely rotating about a transverse axis passing through its centre at the rate of $100$ revolutions per minute. A $20$ gram blob of wax falls on the disc and sticks to the disc at a distance of $5cm$ from its axis. Moment of inertia of the disc about its axis passing through its centre of mass is $2\times {10}^{-4}kg{m}^2$. Calculate the new frequency of rotation of the disc.

Answer 1

An oscillating particle possess both types of energies : Potential as well as kinetic. It possesses potential energy on account of its displacement from the equilibrium position.

Potential energy : Consider a particle of mass $m$ executing S.H.M. Let $x$ be its displacement from the equilibrium position at any instant $t$. Since in S.H.M. the force $F$ acting upon the particle is proportional to and opposite to the displacement $x$, we have

$F=-kx$

Where, the constant $k$ gives the force per unit displacement $x$. The force can also be express in terms of potential energy $U$ of the particle as :

$F=-\frac{dU}{dx}$

Thus, $\frac{dU}{dx}=kx$

On integrating, we get

$U=\frac{1}{2}k{x}^2+C$

Where $C$ is a constant. If we assume the potential energy zero in the equilibrium position, i.e.,

If $U=0$ at $x=0$, then $C=0$.

Therefore, $U=\frac{1}{2}k{x}^2$

(P.E.) $U=\frac{1}{2}m{\omega }^2{x}^2$

Hence from here it is clear that the potential energy of a particle doing S.H.M. is directly proportional to the square of the displacement (P.E. $\propto x^2$).

(a) Potential energy at mean position : At mean position, velocity of a particle executing S.H.M. is maximum and displacement is minimum i.e., $x=0$

P.E. $=\frac{1}{2}m{\omega }^2{x}^2=0$

(b) Potential energy at extreme position : At extreme position, velocity of a particle executing S.H.M. is minimum and displacement is maximum, i.e., $x=\pm a$

P.E. $=\frac{1}{2}k{a}^2=\frac{1}{2}m{\omega }^2{a}^2$

Given : $n_1=100r.p.m.$

$=\frac{100}{60}=1.66r.p.s.$

$m=20gm=2\times {10}^{-2}kg$

$h=r=5cm=0.05m$

$I_1=$ Moment of inertia of the disclabout its axis is $=2\times {10}^{-4}kg{m}^2$

$I_2=$ Moment of inertia of the disc about same axis with the wax.

To find : $n_2=$ ?

From formula,

$I_2=I_1+m{r}^2$ [from parallel axis theorem]

$I_1{\omega }_1=I_2{\omega }_2$

$I_1\left(2\pi n_1\right)=(I_1+m{r}^2)\left(2\pi n_2\right)$

$I_1{n}_1=(I_1+m{r}^2)n_2$ [$\because 2\pi$ is cancelled out]

$n_2=\frac{I_1{n}_1}{I_1+m{r}^2}$

$n_2=\frac{2\times {10}^{-4}\times 1.66}{2\times {10}^{-4}+(2\times {10}^{-2}\times 0.05\times 0.05)}$

$=\frac{2\times {10}^{-4}\times 1.66}{2\times {10}^{-4}+0.5\times {10}^{-4}}$

$=\frac{2\times {10}^{-4}\times 1.66}{2.5\times {10}^{-4}}=1.28\cong 1r.p.s.$

$n_2=1r.p.s.$