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Brrunno [24]
2 years ago
7

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

elocity of 2rev/s. During the dismount she stretches out into the straight position, with a length of 1.5m, (assume she is a uniform rod through the center) for her landing. The gymnast has a mass of 50kg. What is her angular velocity in the straight position?
Physics
1 answer:
kifflom [539]2 years ago
6 0

Here we will say that there is no external torque on the system so we will have

L_i = L_f

here we know that

L_i = I_1\omega_1

where we know that

I_1 = \frac{2}{5}mr^2

Also we know that

I_2 = \frac{1}{12}mL^2

initial angular speed will be

\omega_1 = 2\pi(2rev/s) = 4\pi rad/s

now from above equation

\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega

0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega

0.452 = 0.1875 \omega

now we have

\omega = 2.41 rad/s

so final speed will be 2.41 rad/s

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