A diver performing a double somersault spins at an angular speed of 4.5 π rad/s precisely 0.70s after leaving the platform.

Assuming the diver begins with zero initial angular speed and accelerates at a constant rate, what is the diver’s angular acceleration during the double somersault?
Answer in units of rad/s^2.

1 answer

ω = ε•t
ε = ω/t= 4.5•π/0.7=20.2 rad/s^2