In: Physics
A thin rod (mass 8.28 kg, length 8.04 m) is sitting on a horizontal, frictionless table. There is a 0.626 kg frog sitting on the very end of the rod; you can treat the frog as a point particle. Suddenly the frog jumps off at speed 4.75 m/s, moving horizontally and perpendicular to the rod. Find ω, the angular speed of the rod after the frog jumps off, in rad/s.
Initially the frog and rod are stationary. Hence initial momentum and angular momentum of the frog-rod system is zero.
When the frog (0.626 kg) jumps off with a speed of 4.75 m/s then the centre of mass of the rod moves in opposite direction such that the momentum of rod is equal and opposite to that of frog.
The velocity of centre of rod is V = 0.626 x 4.75 / 8.28 = 0.36 m/s
The rod not only translates but also rotates about centre of the
rod with angular velocity
.
The rod rotates about centre such that the angular momentum of rod with respect to centre is equal and opposite to that of the frog.
The angular momentum of the frog with respect to centre is L = mvr = 0.626 x 4.75 x 4.02
L = 11.95 kg-m2/s
(as length of the rod is 8.04m and frog jumps from one end then distance of frog from centre is 4.02 m)
The angular momentum of the rod with respect to centre is L = I
= (Md2/12)
L = (8.28 x (8.04)2 / 12)
= 44.6
Then 44.6
= 11.95 =>
= 0.268 rad/s
The angular speed of the rod after the frog jumps off is 0.27 rad/s.