The real force F acting on particle of mass m performing circular motion acts along the radius of circle r and is directed forwards the centre of the circle. The magnitude of square root of such a for

Question

The real force F acting on particle of mass m performing circular motion acts along the radius of circle r and is directed forwards the centre of the circle. The magnitude of square root of such a force is (T = periodic time)

TestHub · Accepted Answer

The centripetal force acting on a particle of mass $m$ in circular motion of radius $r$ is given by $F = m ω^{2} r$ .

Since the angular velocity $ω = \frac{2 π}{T}$ , where $T$ is the time period, the force becomes:

$F = m (\frac{2 π}{T})^{2} r = \frac{4 π ^{2} m r}{T ^{2}}$

The magnitude of the square root of this force is:

$F = \frac{4 π ^{2} m r}{T ^{2}} = \frac{2 π m r}{T}$

Answer: $\frac{2 π m r}{T}$

Physics - Circular Motion Question with Solution | TestHub

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