InterviewSolution
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InterviewSolution
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A solid cylinder rolls down an inclined plane. Its mass is `2 kg and radius 0.1 m`. It the height of the inclined plane is `4m`, what is its rotational `K.E.` when it reaches foot of the plane ? Assume that the surfaces are smooth. Take `M.I.` of solid cylinder about its axis = `mr^(2)//2`. |
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Answer» `K_(T) + K_(R) = K` `(1)/(2)mv^(2) + (1)/(2)I omega^(2) = K` But `(1)/(2)I omega^(2) = (1)/(2) ((1)/(2)mr^(2)) omega^(2) = (1)/(4)mv^(2)` `:. (1)/(2)mv^(2) + (1)/(4)mv^(2) = K or (3)/(4)mv^(2) = K` Rotational `KE, (1)/(2)I omega^(2) = (1)/(4)mv^(2) = (K)/(3)` `=(mgh)/(3) = (2 xx 9.8 xx 4)/(3) = 26 .13 J` |
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