A uniform rod of mass M and length d is initially at rest on a horizontal and frictionless table contained in the xy plane, the plane of the screen. The figure is a top view, gravity points into the screen. The rod is free to rotate about an axis perpendicular to the plane and passing through the pivot point at a distance d/3 measured from one of its ends as shown. A small point mass m, moving with speed v0, hits the rod and stick to it at the point of impact at a distance d/3 from the pivot.

Question

(a) If the mass of the rod is M=4m, what is the magnitude of the angular velocity of the rod+small mass system after the collision?

1)ω=2v0/(3d)
2)ω=3v0/(2d)
3)ω=4v0/(3d)
4)ω=3v0/(4d)
5)ω=8v0/(3d)
6)ω=3v0/(8d)
7)ω=5v0/(3d)
8)ω=3v0/(5d)

Using again M=4m. What is the speed of the center of mass of the rod right after collision?

1)vcm=v0
2)vcm=v0/2
3)vcm=v0/3
4)vcm=v0/5
5)vcm=v0/10
6)vcm=v0/20

Answer 1

distance of Center of rod from pivot = 3d/6 - 2d/6 = d/6
so
I about pivot = (1/12)M d^2 + M d^2/36
= (1/9)Md^2 = (4/9)m d^2

distance of mass trajectory from pivot = d/3

angular momentum of m about pivot before collision = m vo r = m vo d/3
= total ang momentum of system about pivot before and after collision

angular momentum of m about pivot after collision = m (d/3)^2 w = m d^2 w/9

angular momentum of rod about pivot after collision = [(1/12)M d^2 + M d^2/36] w
=(Md^2 /9)w
so
m vo d/3 = M d^2 w/9 + m d^2 w/9
if M = 4 m

m vo /3 = 4 m d w/9 + m d w/9

vo = 4 d w/3 + d w/3 = 5 d w/3

w = 3 vo / 5d

Answer 2

w r = (3 v0/5d)(d/6) = vo/10

Answer 3

Thank you for giving the steps, I was lost in this question.

Answer 4

A mokey of mass 15 kg is climbing on a rope with an acceleration of 1 m/s square, how much force should it apply to the rope ? If the rope is 5 m long an the monkey starts from rest, how time will it take to reach the ceiling?