InterviewSolution
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InterviewSolution
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The pulley shown in figure has a moment of inertias I about its xis and mss m. find the tikme period of vertical oscillastion of its centre of mass. The spring has spring constant k and the string does not slip over the pulley. |
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Answer» For rotatiobnal equation the tension in spring on both sides must be same . Let T be the tension in the string in equilibrium position and `l_(0)` the extension of spring . Then for equilibrum of any part of spring on RHS. `2 T = mg implies2 k l_(0) = mg or l_(0) = (mg)/(2 K)` For translational equation of pulley `l_(0) = (mg)/(2 K)` Thus when pulley attains equilibrium , the spring is streched by a distance `l_(0) = (mg)/(2 K)` (i) Now suppose that the pulley is down a little and released. The pulley starts up and down oscillation . Let y be instant neous displacement of center of pulley from equilibrium. Then total increase in length of (string + spring) is `2y` (y to the left of pulley and y to the right). As steing is inextensible, total extension of spring is `2 y`. As pulley also ratates energy of system is `E = (1)/(2) l omega^(2) + (1)/(2) m v^(2) - mgy + (1)/(2) K (l_(0) + 2 y)^(2)` But `omega = (v)/( r)` `E = (1)/(2) l (v^(2))/( r^(2)) + (1)/(2) K (l_(0) + 2 y)^(2)` `= (1)/(2) ((l)/r^(2) + m) v^(2) = mgy + (1)/(2) K (l_(0) + 2 y)^(2)` Asd total energy is constant (system is conservative) `(dE)/(dt) = 0` ` (1)/(2) ((l)/r^(2) + m) 2v (dv)/(dt) - mg (dv)/(dt) + (1)/(2) K 2 (l_(0) + 2 y) , 2 (dy)/(dt) = 0` and `(dy)/(dt) = v and (dv)/(dt) = acceleration a` and `l_(0) = (mg)/(2K)` we get `((l)/r^(2) + m) a - mg + 2 K ((mg)/(2 k) + 2y) = 0` `a prop - y` (ii) As `a prop - y`, motion of pulley is SHM. Standard equiation of SHM `a = - omega^(2) y` Comparing Eqs, (ii) and (iii), `omega^(2) = (4 k)/((l)/r^(2) + m)` time period. `T = (2 pi)/(omega) = 2 pi sqrt({{(l)/(r^(2)) + m)/(4K)}} = pi sqrt({{m + (l)/(r^(2)))/(K)}}` |
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