Question

The real force $'F'$ acting on a particle of mass $'m'$ performing circular motion acts along the radius of circle $'r'$ and is directed towards the centre of circle. The square root of magnitudeof such force is ($T =$ periodic time)

• A. $\frac{2\pi}{T}\sqrt{mr}$
• B. $\frac{Tmr}{4\pi}$
• C. $\frac{2\pi T}{\sqrt{mr}}$
• D. $\frac{T^{2}mr}{4\pi}$

Answer 1

Answer: (A)

Force $F$ acting on a body of mass $m$ performing circular motion of radius $r$,
$F=\frac{m v^{2}}{r}$ (centripetal force) ...(i)
where, $v=$ velocity of the particle The time period of one complete cycle,
$T =\frac{\text { perimeter of a circular path }}{\text { velocity of body }} =\frac{2 \pi r}{v} $
$\Rightarrow v =\frac{2 \pi r}{T}$...(ii)
From Eqs. (i) and (ii), we get
$F=\frac{m}{r}\left(\frac{2 \pi r}{T}\right)^{2}=m r\left(\frac{2 \pi}{T}\right)^{2}$
or $\sqrt{F}=\frac{2 \pi}{T} \sqrt{m r}$
So, option (a) is correct.