A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s.The magnitude of its angular momentum is:

Question

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The angular momentum of the paper clip is given by:$$L = Iω$$where:* I is the moment of inertia of the paper clip* ω is the angular velocity of the paper clipThe moment of inertia of a point mass is given by:$$I = mr^2$$where:* m is the mass of the paper clip* r is the distance from the axis of rotation to the paper clipIn this case, the paper clip is attached to the rim of the phonograph record, so the distance from the axis of rotation to the paper clip is equal to the radius of the record.Therefore, the moment of inertia of the paper clip is:$$I = mr^2 = (0.015 kg)(0.15 m)^2 = 3.375 × 10^{-4} kg.m^2$$The angular velocity of the paper clip is given by:$$ω = 3.5 rad/s$$Therefore, the angular momentum of the paper clip is:$$L = Iω = (3.375 × 10^{-4} kg.m^2)(3.5 rad/s) = 1.18 × 10^{-3} kg.m^2/s$$Therefore, the correct answer is A.

A 15-G Paper Clip Is Attached to the Rim of a Phonograph