Question

A particle of mass 'm' moves in x-y plane so that its position vector r = x_i + y_j with a parametric equation x= a cos wt, y= b sin wt, where a&b are constants, a>b and w is a positive constant. 1. Show that the force field is conservative. 2. Find the potential energy at point a&b. 3. Find the work done by the force in moving the particle from point a-b. 4. Find the total energy of the particle.

Answer 1

Given,

Mass of the particle = m

Position vector of the particle $(r) = x\hat{i} + y\hat{j}$

$x=a\cos(\omega t)$

$y=b\sin(\omega t)$

$a>b$ and $\omega>0$

Now, we can write the position vector as,

$\overrightarrow{r}=a\cos(\omega t)\hat{i}+b\sin (\omega t)\hat{j}$

Now, $\overrightarrow{v}=\frac{dr}{dt}=-a\omega \sin(\omega t)\hat{i}+b\omega\cos(\omega t)\hat{j}$

Acceleration of the particle, $(\overrightarrow{a}) = \frac{dv}{dt}$

$\overrightarrow{a}=-a\omega^2 \cos(\omega)\hat{i}-b\omega^2\sin(\omega t)\hat{j}$

i) $F= m\overrightarrow{a}$

$F=-m\omega^2(acos(\omega t)\hat{i}+b\sin(\omega t)\hat{j})$

$=-m\omega^2(\overrightarrow{r})$

So, the force is depending on the angle.

ii) Velocity will be maximum for the case a, when $\omega t=0$

In that case $V=a$

Velocity will be maximum, when $\omega t = \pi/2$

$V=b$

Hence, the required potential energy $PE = \frac{1}{2}ma^2\omega^2$ and $PE=\frac{1}{2}mb^2\omega^2$

iii)