Question
(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is $0.400 \textrm{ kg} \cdot \textrm{m}^2$. (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 decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?

a) $15.1 \textrm{ kg} \cdot \textrm{m}^2 \textrm{s}$

b) $1.92 \textrm{ kg}\cdot \textrm{m}^2$

c) $-0.503 \textrm{ kg} \cdot \textrm{m}^2 \textrm{/s}^2$

Solution Video