Two ice skaters, both of mass 68 kg, approach on parallel paths 1.6 m apart. Both are moving at 3.5 m/s with their arms outstretched. They join hands as they pass, still maintaining their 1.6-m separation, and begin rotating about one another. Treat the skaters as particles with regard to their rotational inertia. They now pull on each other’s hands, reducing their radius to half its original value. Calculate the change in kinetic energy for this process.

A 70.0-kg person stands on a tiny rotating platform with arms outstretched. From your answer to part (d), would you expect it to be harder or easier to lift your arms when rotating or when at rest?

On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it leaves the cue stick with a center-of-mass speed v₀ and ω₀ a “reverse” spin of angular speed (see Fig. 11–41). A kinetic friction force acts on the ball as it initially skids across the table. Using conservation of angular momentum, find the critical angular speed ω_C such that, if ω₀=ω_C, kinetic friction will bring the ball to a complete (as opposed to momentary) stop.

On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it leaves the cue stick with a center-of-mass speed v₀ and ω₀ a “reverse” spin of angular speed (see Fig. 11–41). A kinetic friction force acts on the ball as it initially skids across the table. If ω₀ is 10% smaller than ω_C , i.e., ω₀ = 0.90ω_C, determine the ball’s cm velocity v_CM when it starts to roll without slipping.

(III) On a level billiards table a cue ball, initially at rest at point O on the table, is struck so that it leaves the cue stick with a center-of-mass speed v₀ and ω₀ a “reverse” spin of angular speed (see Fig. 11–41). A kinetic friction force acts on the ball as it initially skids across the table. If ω₀ is 10% larger than w_C i.e.,ω₀ = 1.10w_C, determine the ball’s cm velocity v_CM when it starts to roll without slipping. [Hint: The ball possesses two types of angular momentum, the first due to the linear speed v_CM of its cm relative to point O, the second due to the spin at angular velocity ω about its own cm. The ball’s total L about O is the sum of these two angular momenta.]

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1630 kg·m². Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. What is the new angular speed?

Suppose a 65-kg person stands at the edge of a 9.8-m-diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kg·m². The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable.

Suppose a diver spins at 8 rad/s while falling with a moment of inertia about an axis through himself of 3 kg•m². What moment of inertia would the diver need to have to spin at 4 rad/s? 

BONUS:How could he accomplish this?