A thin rod of mass m and length l is suspended vertically from a frictionless pi

Question

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

A mass of putty traveling horizontally with a speed v strikes the rod at its center of mass and sticks there.

How high does the bottom of the rod swing?

Express your answer in terms of the variables m, M, v and appropriate constants

**please explain and show work very clearly on how they get to answer boxed below, please explain especially when they get to find the final angular momentum, to understand***

A thin rod of mass m and length l is suspended vertically from a frictionless pivot at its upper end. A mass of putty traveling horizontally with a speed v strikes the rod at its center of mass and sticks there. How high does the bottom of the rod swing? Express your answer in terms of the variables m, M, v and appropriate constants **please explain and show work very clearly on how they get to answer boxed below, please explain especially when they get to find the final angular momentum, to understand*** A thin rod of mass m and length l is suspended ver

Explanation / Answer

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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