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प्रश्न
A stone of mass m tied to a spring of length l is rotating along a circular path with constant speed v. The torque on the stone is ______.
पर्याय
mv2l
zero
mlv
`(mv^2)/l`
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उत्तर
A stone of mass m tied to a spring of length l is rotating along a circular path with constant speed v. The torque on the stone is zero.
Explanation:
Torque depends on the force and its perpendicular distance from the axis.
Here, the stone is moving in a circle, so the force acting on it is the centripetal force provided by the spring:
F = `(mv^2)/l`
This force always acts along the radius (towards the centre), i.e., along the same direction as the position vector `vecr`.
`vecτ = vecr xx vecF`
Since `vecr and vecF`are in the same direction, the angle between them is 0°.
τ = r F sin 0°
= 0
So, even though the stone is rotating, the torque about the centre is zero because the force is purely radial.
