Find Sweet Spot for Elastic Collision w/ Rod of Mass M & Length L

[syang9]

finding "sweet spot"

"a ball of mass m makes an elastic collision with rod of mass M and length L. the ball hits a distance y from the bottom of the rod. find the value of y corresponding to the "sweet spot" for the collision, where the bottom of the rod remains instantaneously at rest after the collision."

condition for hitting sweet spot: I (moment of inertia) for rod is instantaneously 1/3ML^2 (for rotation about stationary axis).

angular momentum is conserved if we consider the system to be the rod and the ball. therefore:

L_i = L_f; m*(v_i)*y = 1/3(ML^2)wf (omega final, final angular velocity of rod after collision)

at this point, unknowns are y and wf. to find wf, we consider the fact that energy and linear momentum are conserved:

E_i = E_f

1/2*m*v_i^2 = 1/2*I*wf^2 + 1/2*m*v_f^2

eliminate 1/2, use linear momentum to eliminate v_f

p_i = p_f

m*v_i = M*v_CM + m*v_f

we know that the center of mass will be rotating, so we can use the relation v_CM = r*w, where r is the location of the center of mass (L/2). therefore, solving for v_f, we obtain

v_f = v_i - M/2m*(L*wf)

this is where i run into problems. my strategy was to use this v_f and solve for wf in the energy conservation equation, and then consequently solve for y in the angular momentum conservation equation. however, upon subsituting the above formula for v_f into the energy conservation equation, it becomes an extremely convoluted quadratic equation. what have i done wrong? is there an easier way to do this (preferably, without using impulses?)

thanks so much for any help.

 

[siddharth]

L_i = L_f; m*(v_i)*y = 1/3(ML^2)wf (omega final, final angular velocity of rod after collision)

What about the angular momentum of the ball after collision?

 

[Doc Al]

is there an easier way to do this (preferably, without using impulses?)

Actually, using impulses is the way to go. Forget about energy conservation, just consider that the collision imparts some impulse to the rod. That (linear) impulse = Mv, where v is the speed of the rod's CM after the collision. That same collision also imparts an angular impulse that equals Iw. (If the linear impulse is Mv, what's the angular impulse?) The condition for pure rotation about one end is that the velocity of the CM (v) must equal Lw/2. Combine these statements and you can solve for the point of impact that is the "sweet spot".