**38. A quarterback fires off a pass with the football traveling at a speed of 20 m/s and witha rotational speed of 5 revolutions per second. The mass of the football is 400 g and its rota-tional inertia is 2 x 10-3 kg m2a) What is the total kinetic energy (translational plus rotational) of the football just as itis thrown?b) If the ball rises to a maximum height of 10 m above the point where it left thequarterback's hand, what will be its translational and rotational kinetic energies at thatpoint? [Neglect any effects of air resistance.]

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Asked Nov 4, 2019
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**38. A quarterback fires off a pass with the football traveling at a speed of 20 m/s and with
a rotational speed of 5 revolutions per second. The mass of the football is 400 g and its rota-
tional inertia is 2 x 10-3 kg m2
a) What is the total kinetic energy (translational plus rotational) of the football just as it
is thrown?
b) If the ball rises to a maximum height of 10 m above the point where it left the
quarterback's hand, what will be its translational and rotational kinetic energies at that
point? [Neglect any effects of air resistance.]
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**38. A quarterback fires off a pass with the football traveling at a speed of 20 m/s and with a rotational speed of 5 revolutions per second. The mass of the football is 400 g and its rota- tional inertia is 2 x 10-3 kg m2 a) What is the total kinetic energy (translational plus rotational) of the football just as it is thrown? b) If the ball rises to a maximum height of 10 m above the point where it left the quarterback's hand, what will be its translational and rotational kinetic energies at that point? [Neglect any effects of air resistance.]

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Expert Answer

Step 1

Consider the mass of the football be m, the translational speed of the football be v, the rotational speed of the football be ω, and the moment of inertia of the football be I.

 

The given values are,

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1 kg =0.4 kg m = 400 g 400 g| 1000 g v 20 m/s 2л rad 5 rev/s = 107r rad/s 1 rev I 2x 103 kg m2

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Step 2

(a)

 

The total kinetic energy of the football as just it is thrown, that is, translational and rotational kinetic energy, can be calculated as,

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KE ΚΕ: mv 2 1 (0.4 kg)(20 m/s)'(2x10 kg -m2) (107 rad/s) 1 J 80.987 kg m2/s2 1 kg m2/s2 =80.987 J

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Step 3

(b)

 

Write the expression for the potential energy of the f...

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PE = mgh (0.4 kg) (9.81 m/s2)(10 m) 1 J 39.24 kg m2 /s 1 kg m2/s2 39.24 J

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