Question

Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 4.9 times that of particle B. The period of particle B is 2.4 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

Answer 1

Answer:

Step-by-step explanation:

Given

acceleration of A is 4.9 times of acceleration of B

$a_a=4.9\cdot a_b$

since it is a uniform acceleration therefore there is only centripetal acceleration

$a=\frac{v^2}{r}$

$\frac{v_a^2}{r_a}=4.9\times \frac{v_b^2}{r_b}$

$\sqrt{4.9\times \frac{r_a}{r_b}}=\frac{v_a}{v_b}----1$

also Time-period of B is 2.4 times of A

$T_b=2.4\cdot T_a$

$2.4\times \frac{2\pi r_a}{v_a}=\frac{2\pi r_b}{v_b}$

$2.4\times \frac{r_a}{v_a}=\frac{r_b}{v_b}$

$2.4\times \frac{r_a}{r_b}=\frac{v_a}{v_b}----2$

substitute the value of 1 in 2

$2.4\times \frac{r_a}{r_b}=\sqrt{4.9\cdot \frac{r_a}{r_b}}$

$\sqrt{\frac{r_a}{r_b}}=\frac{\sqrt{4.9}}{2.4}$

$\frac{r_a}{r_b}=\frac{4.9}{2.4^2}$

$\frac{r_a}{r_b}=0.85$