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A uniform disk of mass m and radius R is projected horizontally with velocity v_(0) on a rough horizontal floor so that it starts off with a purely sliding motion at t=0. After t_(0) seconds it acquires a purely rolling motion as shown in figure. (i) Calculate the velocity of the centre of mass of the disk at t_(0). (ii) Assuming the coefficient of friction to be mu calculate t_(0). Also calculate the work done by the frictional force as a function of time and the total work done by it over a time t much longer than t_(0). |
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Answer» (a) `t le t_(0) v = v_(0) - mu g t` …(1) For Rotational motion, torque is being provided by fricitonal FORCE. `mu mgR = I alpha rArr mu mg R = (1)/(2) mR^(2) alpha` `alpha = (2 mu g)/(R)` so `omega = at rArr omega = (2 mu g)/(R) t` at `t = t_(0), v = R omega` form (1) & (2) `v_(0) - mug t_(0) = 2 mu g t_(0) rArr t_(0) = (v_(0))/(3 mu g)` by elimination `t_(0)` from EQUATION (1) `v = (2 v_(0))/(3)` For `t lt t_(0)` Work done by friction `= DELTA K = K_(f) - K_(i)` =` ((1)/(2) mv^(2) + (1)/(2) I omega^(2)) -(1)/(2) mv_(0)^(2)` =`(1)/(2) m(v_(0) - mu g t)^(2) + (1)/(2) ((1)/(2) mR^(2)) ((2mu g)/(R) t)^(2) - (1)/(2) mv_(0)^(2)` =`(1)/(2) [3 mu^(2) mg^(2) t^(2) - 2 mu mgt v_(0)]` For `t gt t_(0)` `W_("friction") = K_(f) -K_(i)` `W_("friction") = (1)/(2) mv^(2) (1+(K^(2))/(R^(2))) -(1)/(2) mv_(0)^(2)` `W_("friction") = (1)/(2) mv^(2) XX(3)/(2) -(1)/(2) mv_(0)^(2)` By putting `v = (2v_(0))/(3)` `W_("friction") = - (mv_(0)^(2))/(6)`. |
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