Question

A ball of clay with a mass of 0.250 kg is traveling horizontally to the left with a speed of 12.0 m/s when it hits a rod that can rotate about an axis through its center. The rotational inertia of the rod is 0.800 kg middot m^2 and the rod has a length 1.20 m. If the ball sticks to the rod, what is the angular speed of the rod and ball after the collision?

Answer 1

Given: mass of ball m =0.250 kg, speed of ball v= 12 m/s , I_rod= 0.800 kg.m², L=1.20m

I_rod = 1/12 ML²

Mass of rod M= (12×0.800)/1.20² = 6.67 kg

Angular momentum conservation equation about rod's pivot point is:

L_i = L_f =mvr= I_ball+rod$\dpi{120} \omega$

Note, r = L/2 = 0.6 m

I_ball+rod = mr² +I_rod = .89 kg.m²

If v_f is the final velocity of clay ball, $\dpi{120} \omega$ = v_f/r .

Angular momentum L = mvr = I_ball+rod$\dpi{120} \omega$

$\dpi{120} \omega$ = mvr /I_ball+rod = (0.250×12×0.6)/0.89

  $\dpi{120} \omega$ = 2.02 rad/s and v_f = 2.02×0.6 = 1.21 m/s