Question

A gymnast is in a tucked position to complete her somersaults. While tucked her moment of inertia about an axis through the center of her body is 16.0 kg-m² and she rotates at 2.5 rev/s. When she kicks out of her tuck into a straight position, her moment of inertia becomes 19.5 kg-m². What is her rate of rotation after she straightens out?

Answer 1

Answer:

$\omega_s=12.8886\ rad.s^{-1}$

Explanation:

Given that:

• moment of inertia of tucked body, $I_t=16\ kg.m^2$

• rotational speed of the body, $N_t=2.5\ rev.s^{-1}$

• i.e. $\omega_t=2\pi\times 2.5=15.708\ rad.s^{-1}$

• moment of inertia of  the straightened body, $I_s=19.5\ kg.m^2$

Now using the law of conservation of angular momentum:

angular momentum of tucked body=angular momentum of straight body

$I_t.\omega_t=I_s.\omega_s$

$16\times 15.708=19.5\times \omega_s$

$\omega_s=12.8886\ rad.s^{-1}$