A thin rod of mass M and length l is suspended vertically from a frictionless pivot at its upper end.?

Question

11-35) A thin rod of mass M and length l is suspended vertically from a frictionless pivot at its upper end. A mass, m, of putty traveling horizontally with a speed , v, strikes the rod at its CM and sticks here. How high does the bottom of the rod swing?

please help, thanks!

Answer 1

When the putty strikes the CM it does so in-elastically, so momentum is conserved

m*v=(M*l^2/3+m*l^2/4)*w
solve for w

w=m*v/(M*l^2/3+m*l^2/4)
lets us P to denote (M*l^2/3+m*l^2/4)
so
w=m*v/P

after the collision the energy is conserved as it swings upward
(M+m)*g*h=.5*P*w^2
where h is the height of the CM
solving for h
h=.5*P*w^2/((M+m)*g)
the height of the bottom of the rod is related to the CM height, let's call it H, as

(l/2-h)/(l/2)=(l-H)/l
l/2-h=l/2-H/2
H=2*h
Therefore
H=P*w^2/((M+m)*g)

Recall that w=m*v/P
so w^2=m^2*v^2/P^2
H=m^2*v^2/(P*g*(M+m))
recall that
P=(M*l^2/3+m*l^2/4)
H=m^2*v^2/(l^2*g*(M/3+l/4)*(M+m))

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