A 200 g, 20 cm long thin rod is initially at rest on an icy surface (frictionles

Question

A 200 g, 20 cm long thin rod is initially at rest on an icy surface (frictionless surface). An incoming puck strikes the rod at one of its ends, as depicted in the figure below. The puck can be treated as a point particle that has a mass of 100 g, moving with a velocity of (-20 m/s) ˆj before the collision and (-10 m/s) ˆj after the collision. 2 seconds after the collision, how far has the rod’s center of mass traveled? How many revolutions will the rod undergo in the 2 seconds after the collision?

Explanation / Answer

Since there is no external force is applied therefor the linear momentum remain conserved,
initial momentum = mv (where m is the mass of puck and v is the velocity of puck)
Initial momentum = 100*20
Final momentum = mv2 +MV where v2 is the velocity of puck after collision , M and V is the mass and velocity of the rod .
Conserving the linear momentum
100*20 = 100*10 + (200*V)
V = 5 m/s
Hence in 2 seconds the Centre of mass of rod will cover = 2*5 = 10 m
Now since the external torque is also zero therefore we can also apply the conservation of angular momentum therefore .
mv1r = mv2r + Iw
Where r is the distance of the puck from the Centre of the rod i.e 20 cm
I is the moment of inertia of rod = mL2/12  , L is length of the rod
W is the angular velocity
100*20*(0.1) = 100*10*(0.1) + (200*(0.2)2)/12) *w
On solving we get
w = 150 rad /s
w = 2PiN where N is the rotation per second
N = 150/2Pi = 23.87 rps
Therefore in 2 seconds
Revolutions = 23.87*2 = 47.74 approx 48 revolutions

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