Question 308: Work, Energy and Power & Circular Motion

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 magnitude of such force is ( $T=$ periodic time)

Accepted Answer

In circular motion, the force acting towards the center of the circle is known as the centripetal force ($F$). The formula for centripetal force is given by:
$$ F = \frac{m v^2}{r} $$
where:
$ m $ is the mass of the particle,
$ v $ is the velocity of the particle,
$ r $ is the radius of the circular path.
The velocity $ v $ of an object in circular motion with a period $ T $ is given by:
$$ v = \frac{2 \pi r}{T} $$
Substitute the expression for $ v $ into the centripetal force formula:
$ F = \frac{m \left(\frac{2 \pi r}{T}\right)^2}{r} = \frac{m \cdot 4 \pi^2 r^2}{T^2 \cdot r} = \frac{4 \pi^2 m r}{T^2} $
We are interested in finding the square root of the magnitude of this force:
$ \sqrt{F} = \sqrt{\frac{4 \pi^2 m r}{T^2}} $
This can be simplified as follows:
$ \sqrt{F} = \frac{2 \pi}{T} \sqrt{m r} $
Thus, the correct option is
Option A
:
$$ \frac{2 \pi}{T} \sqrt{m r} $$

Answer

$\frac{2 \pi}{T} \sqrt{m r}$

Answer

$\frac{T m r}{4 \pi}$

Answer

$\frac{2 \pi T}{\sqrt{m r}}$

Answer

$\frac{T^2 m r}{4 \pi}$

Work, Energy and Power & Circular Motion

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 magnitude of such force is ( T= periodic time)

Choose the correct answer: