Question

A particle moves with constant speed $v$ along a circular path of radius $r$ and completes the circle in time $T$. The acceleration of the particle is

Answer 1

Step 1: Given Data,

$v$ is the constant speed of the particle.

$r$ is the radius

$T$ is the time taken for the particle to complete the circle

Step 2: Formula used,

$\omega$ = angular velocity of the particle

$v=r\omega$

We can write acceleration $a$ in terms of linear velocity $v$ and angular velocity $\omega$,

$a=v\times \omega .....eqn\left(1\right)$

$a=r{\omega }^2.....eqn\left(2\right)$ since, ($v=r\omega$)

$\omega =\frac{2\pi }{T}.....eqn\left(3\right)$

$T$ is the time taken for the particle to complete the circle

$v$ is the constant speed of the particle.

$r$ is the radius of the circular path

Step 3: Calculating the acceleration,

We can write angular velocity in terms of time$T$,

$\omega =\frac{2\pi }{T}$

Putting the value of $\omega$ in equation $2$from the above equation $3$,

$$\begin{gathered} a=r{\omega }^2 \\ a=r{\left(\frac{2\pi }{T}\right)}^2 \end{gathered}$$

Now from the equation $1$ we know,

or, $$\begin{gathered} a=v\times \omega \\ a=\frac{2\pi }{T}v \end{gathered}$$

Hence the acceleration of the particle is $\frac{2\pi v}{T}$as well as $r{\left(\frac{2\pi }{T}\right)}^2$