- #1
ArticMage
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Homework Statement
A uniform stick d = 1 m long with a total mass of 249 g is suspended vertically from a frictionless pivot at its top end. A wad of putty of mass 29 g is moving horizontally with speed v. It strikes and sticks to the stick at a distance 0.75 m from the pivot.
What is the minimum speed of the putty such that the stick, with the putty stuck to it, will rise to the horizontal position?
Homework Equations
L=m*v*r
L=I*w
.5*I*w^2=m*g*h
The Attempt at a Solution
Conservation of momentum applies so first I have the angular momentum of just the putty with respect to the axis of the rod.
L=m*v*r
then just after the collision the rod has a angular momentum of
L=I*w
this gives me m*v*r/I=w which I use in a bit
The I = (1/3)M*d^2+m*r^2
Next, after the collision i use conservation of energy.
.5*I*w^2=(M+m)*g*h
h=cm=(M*d+m*r)/(M+m)
solving the above for w i get sqrt((2*g(M*d+m*r)/I)
now back to w=m*v*r/I gets me I/(m*r)*sqrt((2*g(M*d+m*r)/I) = v
this gets me 33.39 m/s which is wrong.
I tried without the cm stuff and just had h= 1 but that gets just about the same answer just a littler higher and also wrong.