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JEE Advanced 2018 · System of Particles and Rotational Motion

The potential energy of a particle of mass mm at a distance rr from a fixed point OO is given by V(r)=kr22V(r) = \frac{k r^{2}}{2}, where kk is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius RR about the point OO. If vv is the speed of the particle and LL is the magnitude of its angular momentum about OO, which of the following statements is (are) true?

Answers: B. v=kmRv = \sqrt{\frac{k}{m}}R · C. L=mkR2L = \sqrt{mk}R^{2}

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Written solution

Step 1: Find the force acting on the particle.

The potential energy of the particle at a distance rr is given.

V(r)=12kr2V(r) = \frac{1}{2}kr^2

The central force F(r)F(r) is the negative gradient of the potential energy.

F(r)=−dVdrF(r) = -\frac{dV}{dr} F(r)=−krF(r) = -kr

The magnitude of this force is:

∣F(r)∣=kr|F(r)| = kr

Step 2: Determine the speed of the particle.

For a circular orbit of radius RR, the centripetal force is provided by the central force.

mv2R=kR\frac{mv^2}{R} = kR v2=kmR2v^2 = \frac{k}{m}R^2 v=kmRv = \sqrt{\frac{k}{m}}R

Step 3: Determine the angular momentum of the particle.

The magnitude of the angular momentum LL about the center OO is defined for a circular orbit.

L=mvRL = mvR

Substitute the expression for vv into the equation.

L=m(kmR)RL = m \left(\sqrt{\frac{k}{m}}R\right) R L=mk R2L = \sqrt{mk}\,R^2

Answer: The correct statements are v=kmRv = \sqrt{\dfrac{k}{m}}R and L=mk R2L = \sqrt{mk}\,R^2.

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