You stand in front of a door with mass m(D) and width L(D) as it swings toward y

Question

You stand in front of a door with mass m(D) and width L(D) as it swings toward you with rotational speed omega(initial). You throw a ball of mass (ball) and speed V(ball) toward the door perpendicularly at a distance d from the hinged edge (see figure). The ball bounce directly backward toward you with one fourth the initial speed.

A) What is the final rotational speed of the door after the collision?

B) What must the speed of the ball be at impact if you want the door to swing the door away from you after the ball hits?

Explanation / Answer

let m(D) = M

and m(ball) = m

L(D) = L

and V(ball) = V

then

initial angular momentum of the system is Li = [(1/3)*M*L^2*wi]+[(m*V*L)

final angular momentum Lf = [(1/3)*M*L^2*wf] - (m*(V/4)*L)

according to law of conservation of angular momentum

Li = Lf

[(1/3)*M*L^2*wi]+[(m*V*L)] = [(1/3)*M*L^2*wf] - (m*(V/4)*L)

(m*V*L)(5/4) = (1/3)*M*L^2(wf-wi)

(wf-wi) = [(m*V*L)(5/4)]/[(1/3)*M*L^2]

(wf-wi) = (15/4)*(m*V*L)/(M*L^2) = (15/4)*(m/M)*(v/L)

wf = [(15/4)*(m/M)*(v/L)] +wi

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then again using law of conservation of momentum

Li = [(1/3)*M*L^2*wi]+[(m*V*L)]

Lf = [-(1/3)*M*L^2*wi]+[(m*Vf*L)]

Li = Lf

[(1/3)*M*L^2*wi]+[(m*V*L)] = [-(1/3)*M*L^2*wi]+[(m*Vf*L)]

[(2/3)*M*L^2*wi] = (m*L)*(Vf-V)

Vf-V = [(2/3)*M*L^2*wi]/(m*L)

Vf = [(2/3)*M*L^2*wi] + V

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