Question

4. Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 7.3 times that of particle B. The period of particle B is 2.5 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:

Explanation:

Acceleration of particle A is 7.3 times the acceleration of particle B.

Let the acceleration of particle B is a, then the acceleration of particle A is

7.3 a.

Let the period of particle A is T and the period of particle B is 2.5 T.

Let the radius of particle A is RA and the radius of particle B is RB.

Use the formula for the centripetal force

$a=r\omega ^{2}=r\times \frac{4\pi^{2}}{T^{2}}$

So, $r = a\frac{T^{2}}{4\pi^{2}}$

The ratio of radius of A to the radius of B is given by

$\frac{R_{A}}{R_{B}}=\frac{a_{A}\times T_{A}^{2}}{a_{B}\times T_{B}^{2}}$

$\frac{R_{A}}{R_{B}}=\frac{7.3 a\times T^{2}}{a\times 6.25T^{2}}$

RA : RB = 1.17

Answer 2

Answer:

The ratio of he radius of the motion of particle A to that of particle B is closest to 1.16.

Explanation:

Let $a_A,a_B\ and\ r_A,r_B$ are accelerations of particle and radius of A and B respectively. It is given that :

$a_A=7.3\times a_B\\\\\dfrac{a_A}{a_B}=7.3\\\\and\\\\T_B=2.5\times T_A\\\dfrac{T_B}{T_A}=2.5$

The centripetal acceleration is given by the formula as :

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

r is the radius of motion

Since, $v=\dfrac{2\pi r}{T}$

$a=\dfrac{4\pi ^2r}{T^2}$

For A

$a_A=\dfrac{4\pi^2 r_A}{T^2_A}$.........(1)

For B

$a_B=\dfrac{4\pi^2 r_B}{T^2_B}$...........(2)

Dividing equation (1) and (2) we get :

$\dfrac{a_A}{a_B}=\dfrac{T^2_B\times r_A}{T^2_A\times r_B}$

$\dfrac{r_A}{r_B}=\dfrac{a_A\times T^2_A}{a_B\times T^2_B}$

Now using given conditions :

$\dfrac{r_A}{r_B}=\dfrac{7.3}{(2.5)^2}\\\\\dfrac{r_A}{r_B}=1.16$

So, the ratio of he radius of the motion of particle A to that of particle B is closest to 1.16.