How Do You Solve These Classic Meter Stick Rotational Motion Problems?

AI Thread Summary
The discussion focuses on solving two classic meter stick problems involving rotational motion and collisions. The first problem involves a perfectly inelastic collision where a wad of putty strikes a vertical rod, requiring the application of conservation of momentum and rotational dynamics to find the translational and rotational motions. The second problem concerns the angular velocity of a meter stick falling to the ground, with calculations involving gravitational potential energy and rotational inertia. Participants express confusion over the application of equations and constants, particularly regarding angular acceleration and the moment of inertia. Progress is noted in solving both problems, with specific equations derived for angular velocity and translational motion.
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I'm having trouble with these two "classic meter stick problems"

1: A rod of length L and mass M stands vertically on a flat frictionless surface. A wad of putty of mass m and initial velocity v strikes the stick at a right angle at height 3/4 L. The collision is perfectly inelastic. Find the translational and rotational motions of the object.

2: A meter stick stands vertically at rest on a frictionless level surface. If it falls, what angular velocity will it have when it contacts the floor?

Heres what I've got, not sure where to go from here:

1: Momentum Initial = mv
Momentum final = (m+M)vf

L = mV x r = I w

Dont know what to do with the equations

2: I assumed a and alpha were constant, don't know if they are:

a = L/t^2 alpha = pi/t^2 L/a = pi/alpha

L w=pi v

w = pi * v / L

mg(L/2) = 1/2 mv^2 + 1/2 I w^2 >> I = 1/12 mR^2

gL = v^2 + 1/12L^2(pi * v/L)^2

v = (pi/L)sqrt(gL/(1+pi/12))
 
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I think I've oslved the second one now, was off by a lot before, let me know, also, still lost on the first:

2: mgh = 1/2 I w^2

mg(L/2)=1/2 (1/3mL^2)w^2

g=(1/3)Lw^2

w^2=3g/L

w=sqrt(3g/L)=5.422 radians/second

v = w * r = 5.422 * 1m = 5.422 m/s
 
Again not sure but here's some progress I made on the first:

mv = (m+M)vf

mv x (3L/4) = Ix

I = 1/3 ML^2 + m (3L/4)^2

I = (1/3 M + 9/16 m)L^2

vf (center of mass) = mv./(m+M)

w = (3/4 mv.)/(1/3 M + 9/16 m)
 
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