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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 velocity 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?

Respuesta :

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

[tex]L_i = L_f[/tex]

here we know that

[tex]L_i = I_1\omega_1[/tex]

where we know that

[tex]I_1 = \frac{2}{5}mr^2[/tex]

Also we know that

[tex]I_2 = \frac{1}{12}mL^2[/tex]

initial angular speed will be

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

now from above equation

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

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

[tex]0.452 = 0.1875 \omega[/tex]

now we have

[tex]\omega = 2.41 rad/s[/tex]

so final speed will be 2.41 rad/s

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