Question:
A particle is moving in a circle of radius r under the action of a force \(F=\alpha r^2\) which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r = 0) :
A particle is moving in a circle of radius r under the action of a force \(F=\alpha r^2\) which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r = 0) :
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Get K from the centripetal condition and U from integrating the attractive force from 0 to r; add them.
Updated On: Nov 11, 2026
- \(\alpha r^3\)
- \(\frac{1}{2}\,\alpha r^3\)
- \(\frac{4}{3}\,\alpha r^3\)
- \(\frac{5}{6}\,\alpha r^3\)
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The Correct Option is D
Solution and Explanation
Centripetal condition: \(\frac{mv^2}{r}=\alpha r^2 \Rightarrow mv^2=\alpha r^3\).
Kinetic energy \(K=\frac12 mv^2=\frac{\alpha r^3}{2}\).
The force is attractive (towards the centre), so \(F=-\frac{dU}{dr}=-\alpha r^2\), giving \(U=\int_0^r \alpha r^2\,dr=\frac{\alpha r^3}{3}\) (with \(U(0)=0\)).
Total energy \(E=K+U=\frac{\alpha r^3}{2}+\frac{\alpha r^3}{3}=\frac{5}{6}\alpha r^3\).
Answer: (4) \(\frac{5}{6}\,\alpha r^3\)
Kinetic energy \(K=\frac12 mv^2=\frac{\alpha r^3}{2}\).
The force is attractive (towards the centre), so \(F=-\frac{dU}{dr}=-\alpha r^2\), giving \(U=\int_0^r \alpha r^2\,dr=\frac{\alpha r^3}{3}\) (with \(U(0)=0\)).
Total energy \(E=K+U=\frac{\alpha r^3}{2}+\frac{\alpha r^3}{3}=\frac{5}{6}\alpha r^3\).
Answer: (4) \(\frac{5}{6}\,\alpha r^3\)
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