Posted by **Wendy** on Wednesday, July 17, 2013 at 11:24pm.

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.490 kg·m2.

(b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity drops to 1.35 rev/s.

(c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 14.0 seconds?

## Answer This Question

## Related Questions

- Physics - . (a) Calculate the angular momentum of an ice skater spinning at 6.00...
- Physics - An ice skater with an initial moment of inertia of 1.44 kg*m^2 is ...
- Physics - A skater spinning with angular speed of 1.5 rad/s draws in her ...
- Physics - A spinning disk of mass 5.0 kg and radius 0.25m is rotating about an ...
- physics - A spinning disk of mass 5.0 kg and radius 0.25m is rotating about an ...
- Physics - A woman with a mass of 50kg is standing on the rim of a large disk ...
- Physics - A 5.0 kg skater begins a spin with an angular speed of 4 rad/s. By ...
- physics - A skater spinning with angular speed of 1.5 rad/s draws in her ...
- physics - An ice skater has a moment of inertia of 2.3 kg * m^2 when his arms ...
- Physics - The Spinning Figure Skater. The outstretched hands and arms of a ...

More Related Questions