# Problem: Rotational Kinematics; Moment of Inertia

1. Apr 11, 2010

### m2010

1. The problem statement, all variables and given/known data
A mass (M) is dropped from height (H) onto one end of a stick of mass (M) and of length (L) pivoted around the opposite end. Upon collision the mass adheres to the stick. Respond to the following in terms of M, L, H, and g.

a. Find speed of mass just before impact
b. Find angular speed of the system immediately after impact
c. Find linear speed of the mass M at its lowest point (when stick is vertical).
d. Determine the mechanical energy lost as a result of the collision.

2. Relevant equations
The moment of inertia of the stick is 1/3ML^2.
conservation of momentum
conservation of energy

3. The attempt at a solution
a. v=sqrt(2gH)
b. w= sqrt(2gH)/2L
I need help with c and d.

2. Apr 11, 2010

### reaiy

a) use kinematic equations
you know that it falls a distance (H - L) so use the (v_final)^2 = (v_initial)^2 + 2*a*d where v_initial is 0 and a is acceleration which equals g (gravity), d distance
(v_final) = sqrt(2*g*(H-L))

b) use the the eq L = I*(omega, which i will use as 'w')
so w = L/I
I = 1/3 * M * L^2
also use L = r x p (r crossproduct p)
in this case, theta is 90 degrees so cos(90deg) = 1
L = L * M * v_final
so w = L*M*sqrt(2*g*(H-L))/(1/3)*M*L^2
w = 3sqrt(2g(H-L))/L

c)for part C, you would need to find the center of mass of the stick. the stick weighs M and the mass weighs M, so the center of mass is 3/4 out toward the end of the stick. therefore, the center of mass falls 3/4 * L

shoot im not 100% sure how to do the next part ... i had this on a test but now i forgot it -.-

Last edited: Apr 11, 2010
3. Apr 11, 2010

### m2010

I actually think I had part a right...I attached the diagram given. Can anyone give a more detailed explanation of parts c and d? (L is the length of the stick; M is the mass of the stick)

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