Problem — Diver and pirouette

Problem

A diver, fully extended, rotates at $\omega_1 = 2\;\mathrm{rad/s}$. They tuck into a ball, reducing $I$ by a factor of five. At what angular velocity do they rotate once tucked? How many full turns do they complete in a $2\;\mathrm{s}$ dive?

Solution

No external torque during the dive: angular momentum is conserved. Reducing $I$ by a factor of five ($I_2 = I_1/5$): $I_1\omega_1 = I_2\omega_2 \;\Rightarrow\; \omega_2 = \frac{I_1}{I_2}\omega_1 = 5\cdot 2 = \ev{10\;\mathrm{rad/s}}$ Turns in $2\;\mathrm{s}$ (while tucked): angle $\theta = \omega_2\,t = 10\cdot 2 = 20\;\mathrm{rad}$, i.e. $n = \frac{20}{2\pi} \approx \ev{3\;\text{turns}}$ This is the secret of acrobatic dives: tucking into a ball multiplies the rotation speed, allowing more twists in the same flight time.

Links

Topics: Rotational dynamics Concepts: Conservation of angular momentum · Moment of inertia