A particle of mass m is released from a heigh | vedclass

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(D) By conservation of energy,the velocity $v$ of the particle at an angle $	heta$ is given by: $mgH = mgR(1 + \cos 	heta) + \frac{1}{2}mv^2$,which

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) By conservation of energy,the velocity $v$ of the particle at an angle $	heta$ is given by: $mgH = mgR(1 + \cos 	heta) + \frac{1}{2}mv^2$,which simplifies to $v^2 = 2g(H - R - R \cos 	heta)$.Applying Newton's second law in the radial direction: $N - mg \cos 	heta = \frac{mv^2}{R}$.Substituting $v^2$: $N = mg \cos 	heta + \frac{m}{R}[2g(H - R - R \cos 	heta)] = mg \cos 	heta + \frac{2mgH}{R} - 2mg - 2mg \cos 	heta = \frac{2mgH}{R} - 2mg - mg \cos 	heta$.At $	heta = 0$,$N = \frac{2mgH}{R} - 3mg$,which is the maximum value.At $	heta = \pi$,$N = \frac{2mgH}{R} - mg$,which is the minimum value (if $H > 5R/2$,$N > 0$ at the top).Thus,all statements are correct.

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