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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 position 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

User Mfaerevaag
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1 Answer

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Answer:


\omega = 5 rad/s

Step-by-step 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

User Peycho Dimitrov
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