Uniform circular motion: Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 8.5 times that of particle B. The period of particle B is 2.0 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to A) r_A/r_B = 2.1. B) r_A/r_B = 4.3. C) r_A/r_B = 18. D) r_A/r_B = 0.24. E) r_A/r_B = 17.

Question

Quizplus · Accepted Answer

The acceleration in uniform circular motion is given by $a = \frac{v^2}{r}$ or $a = \omega^2r$, where $v$ is the linear velocity, $r$ is the radius of the circle, and $\omega$ is the angular velocity. The angular velocity $\omega$ is related to the period $T$ by $\omega = \frac{2\pi}{T}$. Given that the acceleration of A is 8.5 times that of B and the period of B is 2 times that of A, we can use the relation $a = \omega^2r$ to find the ratio of the radii. Since $T_B = 2T_A$, it follows that $\omega_A = 2\omega_B$, and thus $a_A = 8.5a_B$. Substituting the relation for acceleration in terms of angular velocity and radius, we get $4\omega_B^2r_A = 8.5\omega_B^2r_B$, leading to $r_A/r_B = 8.5/4 = 2.125$, which is closest to 2.1.

Uniform Circular Motion: Two Particles, a and B, Are in Uniform