InterviewSolution
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InterviewSolution
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A wheel of rdius r and moment of inertia I about its axis is fixed at top of an inclined plane of inclination `theta` as shownin figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. INitialy, tehwheel is rotating at a speed `omega` in direction such that the block slides pu the plane. How far wil the block move before stopping? |
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Answer» Let `a` be the deceleration of the block moving up the incline. Therefore, liner deceleration of the rim of the wheel would also be a. Angular deceleration of wheel `alpha = (a)/(r )` ..(i) It `T` is tension in the string, then equation of motion of block would be `Mg sin theta - T = Ma` ...(i) Also, `tau = T xx r = I alpha` `:. T = (I alpha)/(r ) = (Ia)/(r^(2))` using (i) Put in (ii) `Mg sin theta = Ma + T = Ma + (Ia)/(r^(2))` `= a(M+ (I)/(r^(2))) = a ((Mr^(2)+ I)/(r^(2)))` `a = (Mg r^(2) sin theta)/(I + M r^(2))` Initial velocity of the block up the incline `v = r omega` `:.` Distance moved by the block before stopping `s = (v^(2))/(2a) = (r^(2)omega^(2)(I + Mr^(2)))/(2Mg r^(2) sin theta)` `s = ((I + Mr^(2)) omega^(2))/(2Mg sin theta)` |
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