An ice skater is preparing for a jump with turns and has his arms extended. his moment of inertia is 1.7 kg · m2 while his arms are extended, and he is spinning at 0.7 rev/s. if he launches himself into the air at 8.8 m/s at an angle of 45° with respect to the ice, how many revolutions can he execute while airborne if his moment of inertia in the air is 0.7 kg · m2?

2 revolutions.

Explanation:

We need to obtain the angular velocity when he is in the air.

The angular momentum is given by:

Because of angular momentum conservation L1=L2, so the final angular velocity is given by:

We need to calculate the time the skater is in the air, so we need to use the formula of parabolic motion:

so the time taken is 1.26s

the angular displacement is given by:

the number of revolutions is given by:

the skater execute two complete revolutions in the air.

it about 600 mph from what i calculated

true tan speed = ang vel x radius