Question

Just after leaving the platform, the diver’s fully extended 90-kg body has a rotational speed of 0.32 rev/s about an axis normal to the plane of the trajectory. Estimate the angular velocity N later in the dive when the diver has assumed the tuck position. Model the diver as a uniform rod in the first position and as a uniform sphere in the second.

Answer 1

Answer:

$\omega_{f} \approx 47.659\,\frac{rad}{s}\,(7.585\,\frac{rev}{s} )$

Explanation:

The diver can be modelled by using the Principle of Angular Moment Conservation:

$I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}$

The final angular velocity is:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{\frac{1}{12}\cdot m\cdot L^{2} }{\frac{2}{5}\cdot m \cdot R^{2} }\cdot \omega_{o}$

$\omega_{f} = \frac{5\cdot L^{2}}{24\cdot R^{2}} \cdot \omega_{o}$

Let assume that L = 1.60 m and R = 0.15 m.

$\omega_{f} = \frac{5\cdot (1.60\,m)^{2}}{24\cdot (0.15\,m)^{2}}\cdot (0.32\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$\omega_{f} \approx 47.659\,\frac{rad}{s}\,(7.585\,\frac{rev}{s} )$