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 $a = \frac{v^2}{r} = \frac{4\pi^2r}{T^2}$, where $v$ is the velocity, $r$ is the radius of the motion, and $T$ is the period of the motion. Given that the acceleration of particle A is 8.5 times that of particle B and the period of particle B is 2 times the period of particle A, we can set up the following relationship using the formula for acceleration in terms of period and radius: $a_A = 8.5a_B$ and $T_B = 2T_A$. Substituting the acceleration formula, we get $\frac{4\pi^2r_A}{T_A^2} = 8.5\frac{4\pi^2r_B}{T_B^2}$. Since $T_B = 2T_A$, we can substitute $T_B$ with $2T_A$ in the equation, leading to the simplification and solving for the ratio $\frac{r_A}{r_B}$, which gives a value close to 2.1.

Two Particles,A and B,are in Uniform Circular Motion About a Common