AP Physics 1 · Question of the Day

AP Physics 1 Question of the Day

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Sunday, September 6, 2026

A rigid carousel rotates counterclockwise with constant angular speed $\omega$ . Two riders stand on the platform at distances $r_1$ and $r_2$ from the center, with $r_2>r_1$ . Assume both riders rotate with the platform without slipping. Which statement about their linear speeds is correct?

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$v_1>v_2$ because the inner rider has a shorter path and must move faster to keep up
$v_1=v_2$ because both riders have the same angular speed
$v_2>v_1$ because $v=\omega r$ and $r_2>r_1$ (correct answer)
$v_2=v_1$ because linear speed depends only on $\omega$ , not on $r$

Explanation: This problem tests connecting linear and rotational motion for a carousel system. The linear speed v of any point on a rigid rotating body is given by v = ωr, where ω is the angular speed and r is the distance from the rotation axis. Since both riders are on the same rigid carousel rotating at angular speed ω, they share this angular speed. The rider at r_2 (where $r_2$ > $r_1$ ) has linear speed v_2 = ωr_2 > ωr_1 = v_1. Choice B incorrectly claims equal linear speeds, confusing the shared angular speed with linear speed. The strategy is to recognize that farther points on a rotating rigid body move faster linearly, even though all points complete rotations in the same time.