A rigid massless rod of length L = 2.5m pivotscounterclockwise about its center

Question

A rigid massless rod of length L = 2.5m pivotscounterclockwise about its center O in the horizontal plane at aconstant angular speed of = 18.6 rad/s. A ball of mass m =9.3 is attached to both ends of the rod. At a particular instant,the rod is oriented in the i direction, andanother ball of mass m = 9.3 strikes the left end of the rod with avelocity of v = 22.3j m/s. If the ball sticks tothe rod at impact, what is the rod's angular velocity immediatelyafter collision (counterclockwise is positive and clockwise isnegative)? Specify your answer in rad/s to 2 decimal places.

Explanation / Answer

initial angular momentum =   final angularmomentum . m (L/2)2    +   m (L/2)2    -  m v (L/2)   =    2m(L/2)2  f   +   m (L/2)2 f . Eliminate m everywhere and one factor ofL and simplify... .      L / 4   + L / 4     -   v /2        =   Lf / 2      +  L f / 4 . Multiply everything by 4... .           L + L     - 2 v = 2 L f + Lf . Solve for f .           2L - 2 v = 3 L f . f =  (2/3) - (2/3) v / L    = (2/3) * 18.6 - (2/3) * 22.3 / 2.5 = .           =   6.45 rad/sec     (positive, so it iscounterclockwise) . f =  (2/3) - (2/3) v / L    = (2/3) * 18.6 - (2/3) * 22.3 / 2.5 = .           =   6.45 rad/sec     (positive, so it iscounterclockwise)

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