JEE MAIN - Physics (2018 (Offline) - No. 15)
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely
proportional to the nth power of R. If the period of rotation of the particle is T, then :
T $$ \propto $$ Rn/2
T $$ \propto $$ R3/2 for any n
T $$ \propto $$ Rn/2 +1
T $$ \propto $$ R(n+1)/2
Explanation
We know, Central force in circular motion, F = $$m{\omega ^2}R$$
According to the question,
$$F \propto {1 \over {{R^n}}}$$
$$\therefore$$ $$m{\omega ^2}R$$ $$ \propto {1 \over {{R^n}}}$$
$$ \Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$$
$$ \Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$$
$$\therefore$$ $$\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$$ .......(1)
And we know, $$T = {{2\pi } \over \omega }$$
$$\therefore$$ $$T \propto {1 \over \omega }$$ ...... (2)
From (1) and (2) we can conclude that,
$$T \propto {R^{{{n + 1} \over 2}}}$$
According to the question,
$$F \propto {1 \over {{R^n}}}$$
$$\therefore$$ $$m{\omega ^2}R$$ $$ \propto {1 \over {{R^n}}}$$
$$ \Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$$
$$ \Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$$
$$\therefore$$ $$\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$$ .......(1)
And we know, $$T = {{2\pi } \over \omega }$$
$$\therefore$$ $$T \propto {1 \over \omega }$$ ...... (2)
From (1) and (2) we can conclude that,
$$T \propto {R^{{{n + 1} \over 2}}}$$
Comments (0)
