The figure is an overhead view of a thin uniform rod of length 0.680 m and mass
ID: 1501001 • Letter: T
Question
The figure is an overhead view of a thin uniform rod of length 0.680 m and mass M rotating horizontally at 90.0 rad/s counterclockwise about an axis through its center. A particle of mass M/2.90 and traveling horizontally at speed 35.0 m/s hits the rod and sticks. The particle's path is perpendicular to the rod at the instant of the hit, at a distance d from the rod's center. (a) At what value of d are rod and particle stationary after the hit? (b) In which direction do rod and particle rotate if d is greater than this value?
Explanation / Answer
a)
I of rod = M*L^2/12
For system to eb at rest, initial angular momentum should be 0
Use conservation of moomentum:
Initial momentum of rod = initial momentum of particle
M*L^2/12 * w = m*v*d
M*0.68^2/12 * 90 = (M/2.9)*35*d
0.68^2/12 * 90 = (1/2.9)*35*d
d = 3.48 m
Answer: 3.48 m
b)
For distance more than this
Momentum of particle will be more and hence final momentum will be in direction opposite to rod rotation. Rod will rotate opposite to initial direction
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