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Conservative and Non Conservative Forces

A particle of...

Question

A particle of mass m =2 kg is suspended by a string of length l =1m is whirled in a vertical circle about point O. The particle is given a horizontal velocity of 10 m/s at lowest point. The centripetal or normal acceleration of particle at highest point is $a_{n} = 10 p m / s^{2}$ . Find the value of p.___

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Solution

$T_{L} = M g + \frac{m v_{L}^{2}}{l} v_{L} = 10 m / s M g + T_{H} = \frac{M V^{2} H}{L} \frac{1}{2} M V_{L}^{2} = 2 M g L + \frac{1}{2} M V_{H}^{2}; M V_{H}^{2} = 60 a_{n} = V_{H}^{2} = 10 \times 6 m / s^{2} P = 6$

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