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A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will
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- remain constant
- increase by a factor of 2
- increase by a factor of 4
- decrease by a factor of 2
Correct Option: C
K.E. = L²/2I
The angular momentum L remains conserved about the centre.
That is, L = constant.
I = mr²
∴ K.E. = L²/2mr²
In 2nd case,
K.E. = L²/2(mr'²)
But r′ = r/2
| ∴ K.E' = | = | ⇒ K.E.′ = 4 K.E. | |||
| 2 | 2m.(r²/4) | 2mr² |
∴ K.E. is increased by a factor of 4.
