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Question: A force \(F\) acting on a particle of mass \(m\) acts along the radius of the circle and is directed...

A force FF acting on a particle of mass mm acts along the radius of the circle and is directed towards the centre of the circle. Find the square root of the magnitude of the force FF . (The time period of the motion is denoted by TT ).
A) 2πTmr\dfrac{{2\pi }}{T}\sqrt {mr}

B) Tmr4π\dfrac{{Tmr}}{{4\pi }}

C) 2πTmr\dfrac{{2\pi T}}{{\sqrt {mr} }}

D) T2mr4π\dfrac{{{T^2}mr}}{{4\pi }}

Explanation

Solution

The force acting on a particle undergoing a circular motion is called centripetal force. This centripetal force directed inwards is what keeps the particle moving in its circular path without getting thrown off/losing the circular path.

Formula Used:

  1. The centripetal force acting on a particle is given by, F=mv2rF = \dfrac{{m{v^2}}}{r} where mm is the mass of the particle, vv is its linear velocity and rr is the radius of the circular path along which the particle moves.
  2. The relation between the linear velocity vv and angular velocity ω\omega of a particle undergoing circular motion is given by, v=rωv = r\omega ; rr is the radius of the circular path along which the particle moves.

Complete step by step answer:
Step 1: Write down the parameters involved in the problem at hand.
The mass of the particle performing circular motion is represented by mm .
It is given that a force represented by FF acts on the particle and is directed inwards.
Also, it is given that the radius of the circular path along which the particle moves is represented by rr .
Step 2: Express the relation of the force acting on the particle.
The force acting on the particle is the centripetal force and it is given by, F=mv2rF = \dfrac{{m{v^2}}}{r} ------ (1)
where mm is the mass of the particle, vv is its linear velocity and rr is the radius of the circular path along which the particle moves.
The linear velocity vv of the particle can be expressed in terms of its angular velocity ω\omega as v=rωv = r\omega .
When the above relation for the linear velocity is substituted in equation (1) we get, F=m(rω)2r=mrω2F = \dfrac{{m{{\left( {r\omega } \right)}^2}}}{r} = mr{\omega ^2} ------- (2)
Now, the angular velocity ω\omega can be expressed in terms of the period of the motion TT as ω=2πT\omega = \dfrac{{2\pi }}{T} .
Substituting the expression ω=2πT\omega = \dfrac{{2\pi }}{T} in equation (2) we get, F=mr(2πT)2F = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2} ------ (3)
Step 3: Find the square root of the force FF using equation (3).
Equation (3) gives us F=mr(2πT)2F = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2}
Taking the square root of equation (3) on both sides we get, F=2πTmr\sqrt F = \dfrac{{2\pi }}{T}\sqrt {mr}
\therefore The square root of the force acting on the particle is 2πTmr\dfrac{{2\pi }}{T}\sqrt {mr} . Hence, the correct option is (A)

Note: Here, the linear velocity of the particle at every point along the circular path will be directed along the tangent of the circle at each point. The centripetal force can also be described as a force that constantly pulls the particle inwards. The apparent force that is directed away from the centre is called the centrifugal force. Also one should note that the square root of the centrifugal force is what is asked and not only the centrifugal force.