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mariarad [96]
3 years ago
15

During a trampoline routine, a gymnast is tumbling in the air at 20 rad/s in a tuck position. He then extends into a layout posi

tion and doubles his radius of gyration just before landing on the trampoline bed. How fast is his angular velocity at this instant, just before landing on the trampoline bed? Group of answer choices 5 rad/s 10 rad/s 5 m/s 4 rad/s 80 rad/s
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

\omega = 5 rad/s

Explanation:

As we know that the gymnast has no external torque on it

so here we can say that angular momentum of the system will be conserved

so here we have

L_1 = L_f

I_1\omega_1 = I_2\omega_2

m_1r_1^2\omega_1 = m_1r_2^2\omega

r_1^2\omega_1 = r_2^2\omega

here we know that

r_2 = 2r_1

r_1^2\omega_1 = (2r_1)^2\omega

20 = 4\omega

\omega = 5 rad/s

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