Suppose a 0.250 kg ball is thrown at 16.0 m/s to a motionless person standing on

Question

Suppose a 0.250 kg ball is thrown at 16.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure 9.31.

(a) Calculate the final linear velocity of the person, given his mass is 72.0 kg.

(b) What is his angular velocity if each arm has a 5.00 kg mass? You may treat his arms as uniform rods of length 0.9 m and the rest of his body as a uniform cylinder of radius 0.170 m. Neglect the effect of the ball on his rotational inertia and on his center of mass, so that it remains in his geometrical center.

(c) Compare the initial and final total kinetic energy.

Explanation / Answer

The diagram is important.

(a) Here you must conserve linear momentum. If the person catches the ball while it's velocity is w/r/t horizontal, then
0.250kg * 16.0m/s * cos = 72.25kg * v

v=0.055m/s

(b) Again, it's the horizontal component of the velocity that's important.
Here you'll conserve angular momentum:
mvr = I

I_body = ½mr² = ½ * 62.0 * (0.170m)² = 0.8959 kg·m²
I_arms = mL²/3 = 10kg * (0.9m)² / 3 = 2.7 kg·m²
I_ball = mr² = 0.250kg * (0.9m)² = 0.2025 kg·m²
total I = 3.798 kg·m²

0.250kg * 16.0m/s * cos * 0.9m = 3.798kg·m² *

That's the best I can do without the diagram. Of course, if = 0º, then cos = 1.

=0.948rad/s

(c) initial KE = ½ * 0.250kg * (14.5m/s)² = 32J
final KE = ½ * 3.798kg·m² * (0.948rad/s)² = 1.71J

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