Suppose a 0.250-kg ball is thrown at 12.5 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in the fiqure below.(a)(b)(a) Calculate the final linear velocity of the person, given his mass is 65.0 kg.1923Consider conservation of linear momentum. m/s(b) What is his angular velocity if each arm is 5.00 kg in mass? You may treat his arms as uniform rods of length 0.9 m (measured from the center axis of his body) and the restof 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.rad/s(c) Compare the initial and final total kinetic energy.KEKE,

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Answered: Suppose a o.250-kg ball is thrown at…

Suppose a o.250-kg ball is thrown at 12.5 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in the fiqure below.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter13: Rotation Ii: A Conservation Approach

Section: Chapter Questions

Problem 50PQ: CHECK and THINK Our results give us a way to think about how a person might steer a unicycle....

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Question

Suppose a 0.250-kg ball is thrown at 12.5 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in the fiqure below.
(a)
(b)
(a) Calculate the final linear velocity of the person, given his mass is 65.0 kg.
1923
Consider conservation of linear momentum. m/s
(b) What is his angular velocity if each arm is 5.00 kg in mass? You may treat his arms as uniform rods of length 0.9 m (measured from the center axis of his body) 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.
rad/s
(c) Compare the initial and final total kinetic energy.
KE
KE,

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CHECK and THINK Our results give us a way to think about how a person might steer a unicycle. Consider the person, the unicycle, and the Earth as the system. No net torque acts on the system. Ignoring the motion of the Earth, Figure 13.37A shows the initial angular momentum of the system with the unicycle in motion: Li=Ltire. The person leans to his left so that the angular momentum of the tire rotates downward. The total angular momentum must still point to the right (Fig. 13.37B), so the angular momentum of the person must be upward to compensate. The persons angular velocity therefore points upward (parallel to his own angular momentum), and he is able to make a turn. Another way to analyze this situation is to exclude the Earth from the system and calculate the torque done by gravity. This approach is left as a homework problem. FIGURE 13.37 C Reanalyze the unicycles motion in Example 13.15 (page 382). This time, leave the Earth out of the system and explain how the torque exerted by gravity causes the unicycle to turn. Your explanation should involve a diagram.
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