The figure below is of an overhead view of a thin uniform rod of length 0.600 m

Question

The figure below is of an overhead view of a thin uniform rod of length 0.600 m and mass M rotating horizontally at 81.8 rad/s counterclockwise about an axis through its center. A particle of mass M/3.00 and traveling horizontally at speed 40.5 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? m (b) In which direction do the rod and particle rotate if d is less than this value? clockwise counterclockwise

Explanation / Answer

During collision external impulse on rod+particle system is acting from axis of rotation. So angular momentum of system is conserved about center of rod.
a) If rod and particle are stationay after the collision, angular momentum of system is zero. Hence it is zero before collision also. It means angular momentum of rod and particle are equal and opposite.
Magnitude of angular momentum of rod = Iw = M L2 w /12 (counterclockwise)
Magnitude of angular momentum of particle about center of rod = (M/3) V d ( clockwise)
since two are equal we get
L2 w/12 = V d/3
d = L2 w / 4 V = 0.18 m
b) If d is less than 0.18m , net angular momentum is along couterclockwise direction. Hence rod+particle will rotate in counterclockwise direction. after collision.

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