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A pendulum bob swings along a circular path on a smooth inclined plane as shown in figure, where `m=3kg`, `l=0.75m`, `theta=37^@`. At the lowest point of the circle the tension in the string is `T=274N`. Take `g=10ms^-2`. The speed of the bob at the highest point on the circle isA. (a) `sqrt46ms^-1`B. (b) `sqrt26ms^-1`C. (c) `sqrt52ms^-1`D. (d) `sqrt35ms^-1` |
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Answer» Correct Answer - A Let the speed at the highest point be `v_H`. `1/2mv_H^2+mg2lsin 37^@=1/2mv_L^2impliesv_H=sqrt(46)ms^-1` |
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