Question

In a park there are three concentric circular running tracks. Radius of $2^{\text{nd}}$ track is double of the first and of $3^{\text{rd}}$ track is triple of the first. Three runners are running on these tracks with constant speed. When the runner in the first track completes one round, the runner in $2^{\text{nd}}$ has completed half round and the runner in third track has completed quarter round. If the accelerations of the runners are in ratio $\alpha :\beta :\gamma$, where $\alpha$, $\beta$ and $\gamma$ are least integers, then find the value of $\frac{\alpha +\beta +\gamma }{9}$

Answer 1

Step 1: Draw a rough diagram

Here all three runners have same linear speed but are in different tracks as shown in figure
Radius of $1^{\text{st}}$ ,$2^{\text{nd}}$ and 3${}^{\text{rd}}$ tracks can be assumed as $r$,$2r$ and $3r$ respectively

Step 2: Find angular speed of runners.
Formula Used: $\omega =\frac{\theta }{t}$
$\because$ In same time interval angular displacement of $1^{\text{st}}$ runner is $2\pi$, of $2^{\text{nd}}$ runner is $\pi$ and of $3^{rd}$ runner is $\frac{\pi }{2}$
$\therefore$If we assume angular speed of $3^{\text{rd}}$ runner as $\omega$ then angular speed of $2^{\text{nd}}$ and $1^{\text{st}}$ runner can be called as $2\omega$ and $4\omega$ respectively.

Step 3:Calculate centripetal acceleration

Formula Used:$a_c={\omega }^{2}r$
Let $a_1$,$a_2$and $a_3$ be acceleration of runners
$a_1=(4\omega {)}^{2}r$,
$a_2=\left(2\omega {)}^2\right(2r)$,
$a_3={\omega }^2\left(3r\right)$
$\therefore a_1:a_2:a_3=16:8:3$
$\therefore a=16,\beta =8,\gamma =3$
From the given relation,
$\frac{\alpha +\beta +\gamma }{9}=\frac{16+8+3}{9}=3$
Final Answer: 3