Question

Suppose a 0.350 kg ball is thrown at 20.0 m/s to a motionless person standing on ice who catches it with an outstretched arm.

Calculate the final linear velocity (in m/s) of the person, given his mass is 65.0 kg.

(Already have, 0.107 m/s)

What is his angular velocity (in rad/s) if each arm is 4.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.800 m) and the rest of his body as a uniform cylinder of radius 0.190 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center.

Answer 1

m_a = mass of each arm = 4 kg

L = length of each arm = 0.8 m

I_a = moment of inertia of the arms = 2 m_a L²/3 = 2 (4) (0.8)²/3 = 1.71 kgm²

m_r = mass of the rest of the body = total mass - 2 m_a = 65 - 2 (4) = 57 kg

r = radius = 0.190 m

I_r = moment of inertia of rest of the body = (0.5) m_r r² = (0.5) (57) (0.190)² = 1.03 kgm²

v = velocity of the ball = 20 m/s

M = mass of the ball = 0.350 kg

using conservation of momentum

M v L = (ML² + I_a + I_r) w

(0.350) (20) (0.8) = ((0.350) (0.8)² + (1.71) + (1.03)) w

w = 1.89 rad/s