An ice skater is spinning on frictionless ice with her arms extended outward. She then pulls her arms in toward her body, reducing her moment of inertia. Her angular momentum is conserved, so as she reduces her moment of inertia, her angular velocity increases and she spins faster. Compared to her initial rotational kinetic energy, her final rotational kinetic energy is (a) the same (b) larger, because her angular speed is larger (c) smaller, because her moment of inertia is smaller.

Question

10-03-2020
Physics

contestada

An ice skater is spinning on frictionless ice with her arms extended outward. She then pulls her arms in toward her body, reducing her moment of inertia. Her angular momentum is conserved, so as she reduces her moment of inertia, her angular velocity increases and she spins faster. Compared to her initial rotational kinetic energy, her final rotational kinetic energy is (a) the same (b) larger, because her angular speed is larger (c) smaller, because her moment of inertia is smaller.

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therealabefe therealabefe · Answer 1 · 2020-03-15T04:54:53+01:00

Answer: option B

Explanation: The formulae that defines the rotational kinetic energy of an object is given as

K.E = (1/2) Iω^2

Where I = momemt of inertia

ω = angular velocity

The formulae simply implies that a large value of angular velocity (ω) produces a large rotational kinetic energy.

Back to our question , as she ( the ice skater) reduces her moment of inertia, her angular velocity will increase thus increasing her rotational kinetic energy.

Option B is correct