angular momentum of system before collision about pivot point
= m v R = m Vo (d/3) = m Vo d/3
Moment of inertia of rod about pivot = (1/12)(4m) d^2 +(4m)(d/6)^2
= 4 m d^2 (1/12 + 1/36) =4 m d^2(4/36) = (4/9)m d^2
Moment of inertia of m about pivot = m (d/3)^2
Total moment of inertia after crash
= (4/9 + 1/9)m d^2 = (5/9) m d^2
if at angular velocity w
total angular momentum = I w = (5/9) m d^2 w
so
(5/9) m d^2 w = m Vo d/3
solve for w
for part b, v = w times distance from pivot which is d/6
I'm reposting this one because I'm still stuck...
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 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. 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.
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?
b. Using again M=4m. What is the speed of the center of mass of the rod right after collision?
I know I need conservation of angular momentum, but how do I set up the equation -- i.e what is the initial moment of inertia expression, and what is the final?
1 answer