InterviewSolution
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InterviewSolution
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A solid cylinder rolls up an inclined plane of inclination `theta` with an initial velocity `v`. How far does the cylinder go up the plane ?A. `(3 v^2)/(2 g sin theta)`B. `(v^2)/(4 g sin theta)`C. `(3 v^2)/(g sin theta)`D. `(3 v^2)/(4 g sin theta)` |
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Answer» Correct Answer - D (d) Let the cylinder go up the plane upto a height `h`. Let `M` and `R` be the mass and radius of cylinder respectively. According to law of conservation of mechanical energy, we get. `(1)/(2) Mv^2 + (1)/(2) I omega^2 = Mgh` `(1)/(2) Mv^2 + (1)/(2) (MR^2)/(2) omega^2 = Mgh` (because For a solid cylinder, `I = (1)/(2) MR^2`) `(1)/(2) Mv^2 + (1)/(4) MR^2 omega^2 = Mgh` `(1)/(2) Mv^2 + (1)/(4) Mv^2 = Mgh` `(because v = R omega)` `(3)/(4) Mv^2 = Mgh` `h = (3 v^2)/(4 g)` ...(i) Let `s` be distance travelled by the cylinder up the plane. `sin theta (h)/(s)`or `s = (h)/(sin theta) = (3 v^2)/(4 g sin theta)` (Using (i)). |
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