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Byju's Answer

Standard XII

Physics

Work Energy Theorem Application

A particle of...

Question

A particle of mass, $m$ , is tied to a light string and rotated with a speed, $v$ , along a circular path of radius, $r$ . If $T =$ tension in the string and $m g =$ gravitational force on the particle, then the actual forces acting on the particle are

m g

and

T

only.

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m g

T

and an additional force of

m v^{2} / r

directed inwards.

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m g

T

and an additional force of

m v^{2} / r

directed outwards.

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Only a force

m v^{2} / r

directed outwards.

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Solution

The correct option is A $m g$ and $T$ only.
The force $m v^{2} / r$ directed outwards, called centrifugal force, is not a real force.
At A, $m v_{1}^{2} / l = T_{1} + m g$
The resultant of $m g$ and tension ( $T$ ) will provide the centripetal force.

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A particle of mass, m, is tied to a light string and rotated with a speed, v, along a circular path of radius, r. If T= tension in the string and mg= gravitational force on the particle, then the actual forces acting on the particle are

The correct option is A mg and T only.The force mv2/r directed outwards, called centrifugal force, is not a real force.At A, mv21/l=T1+mgThe resultant of mg and tension (T) will provide the centripetal force.

A particle of mass, $m$ , is tied to a light string and rotated with a speed, $v$ , along a circular path of radius, $r$ . If $T =$ tension in the string and $m g =$ gravitational force on the particle, then the actual forces acting on the particle are

The correct option is A $m g$ and $T$ only.
The force $m v^{2} / r$ directed outwards, called centrifugal force, is not a real force.
At A, $m v_{1}^{2} / l = T_{1} + m g$
The resultant of $m g$ and tension ( $T$ ) will provide the centripetal force.