Calculate the angular momentum, in kg $$ m$2/$s$,$ of an ice skater spinning at $6\mathrm{.}00$ rev$/$s given his moment of inertia is $0\mathrm{.}450$ kg $$ m$2.$

kg $$ m$2/$s

$($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, in kg $$ m$2,$ if his angular velocity drops to $1\mathrm{.}55$ rev$/$s$.$

kg $$ m$2$

$($c$)$

Suppose instead he keeps his arms in and allows friction with the ice to slow him to $3\mathrm{.}00$ rev$/$s$.$ What average torque, in N m$,$ was exerted if this takes $25\mathrm{.}0$ seconds? $($Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.$)$

N $$ m