Question

Calculate the angular momentum (in kg·m2/s) of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.370 kg·m2. kg·m2/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·m2) if his angular velocity drops to 2.05 rev/s. kg·m2 (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 (in n·m) was exerted if this takes 10.0 seconds?

Answer 1

The angular momentum of a spinning ice skater is calculated using their moment of inertia and angular velocity. When the skater's spin rate changes, the moment of inertia can be found using the conservation of angular momentum. The average torque exerted by friction is calculated through the change in angular momentum over time.

Step-by-step explanation:

To calculate the angular momentum of a spinning ice skater, we use the formula L = I ∙ ω, where L is angular momentum, I is the moment of inertia, and ω is the angular velocity. For part (a), the skater's angular momentum with a moment of inertia of 0.370 kg·m2 and angular velocity of 6.00 revolutions per second (rev/s) is found using:

L = 0.370 kg·m2 ∙ (6.00 rev/s ∙ 2π rad/rev)

For part (b), to find the new moment of inertia after the skater slows down to 2.05 rev/s, we assume conservation of angular momentum since no external torques are acting. If the initial angular momentum L is conserved, then:

Linitial = Lfinal

Iinitial ∙ ωinitial = Ifinal ∙ ωfinal

For part (c), the average torque exerted while friction slows the skater can be found by using the angular acceleration formula derived from the rotational form of Newton's second law:

τ = ΔL / Δt

where τ is the torque, ΔL is the change in angular momentum, and Δt is the time over which the change occurs.