Question

Q20 (15 points): A bullet of mass m X102 kg is moving with a speed of 100 m/s when it collides with a rod of mass mR 5 kg and length L m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating. L/4 Pivot a) Find the angular velocity of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass. b) How much kinetic energy is lost in the collision? (The rotational inertia of a thin rod of mass M and length L rotating about central axis is 1/12 ML2) Answer a) 0.6 rad/s b) 49.9J

Answer 1

(a) impulse J = m_B v_B = 1 times 10^-2 times 100 kg m/s The effective improve on the pivot = distance of impact times impulse(J) = L/4 times J = L/4 times 10^-2 times 100 This will equal to angular momentum doublerightarrow Iw = L/4 times J, I = 1/12 ML^2 1/12 ML^2 w = LJ/4 w = 12/ML times 1/4 times 10^-2 times 100 = 3 times 10^-2 times 100/5 times 1 rad/s w = 0.6 rad/s (b)Energy of a rotating object = 1/2 Iw^2 previously the energy of moving buffer = 1/2 mb^2 therefore loss in energy, E_L = 1/2 mv^2 - 1/2 Iw^2 = [1/2 mv^2 - 1/2 times 1/12 times ML^2 w^2] = [1/2(10^-2)(100)^2 - 1/24 times 5 times (1)^2 times (0.6)^2]J = (50 - 0.075)J E_L = 49.925J