Question

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m/s and the speed is increasing at a rate of $2m/s^2$. The magnitude of net acceleration at this instant is then
$A.\text{ 5 m/}{\text{s}}^2$
$B.\text{ 2 m/}{\text{s}}^2$
$C.\text{ 3}\text{.2 m/}{\text{s}}^2$
$D.\text{ 4}\text{.3 m/}{\text{s}}^2$

Answer 1

Hint: We are given the tangential acceleration of the particle moving in a circular path. We need to calculate its centripetal acceleration which is equal to the ratio of square of the velocity of the particle and the radius of the path in which the particle is moving. The net acceleration is given as the magnitude of the resultant of the tangential acceleration and the centripetal acceleration.

Formula used: The net acceleration of a particle moving in a circular path is given as
$$a_{net}=\sqrt{a_t^2+a_c^2}...\left(i\right)$$
Here $a_t$ is known as the tangential acceleration of the particle while $a_c$ is known as the centripetal acceleration of the particle.
The centripetal acceleration of a particle can be calculated by the following formula.
$a_c={\displaystyle \frac{v^2}{r}}...\left(ii\right)$
Here v is the velocity of the particle while r is the radius of the circular path in which the particle moves.

Complete step-by-step answer:
We are given that a particle is moving on a circular path of 10 m radius.
$r=10m$
The speed of the particle at any instant is given as
$v=5m/s$
We are also given the tangential acceleration of the particle which as the value
$a_t=2m/s^2$
We need to calculate the net acceleration of the particle which is equal to the magnitude of the resultant of the tangential acceleration and the centripetal acceleration of the particle moving in the circular path. We already have the value of the tangential acceleration. The value of the centripetal acceleration can be calculated by using equation (ii) in the following way.
$a_c={\displaystyle \frac{v^2}{r}}={\displaystyle \frac{{\left(5\right)}^2}{10}}=2.5m/s^2$
Now we can calculate the value of the net acceleration using the equation (i) by substituting the values of two accelerations in the following way.
$$\begin{gathered} a_{net}=\sqrt{a_t^2+a_c^2}=\sqrt{{\left(2\right)}^2+{\left(2.5\right)}^2} \\ =\sqrt{4+6.25}=\sqrt{10.25}=3.2m/s^2 \end{gathered}$$

So, the correct answer is “Option C”.

Note: It should be noted that we call the net acceleration as the resultant of the two accelerations. This is because the two accelerations are the vector quantities and the add up according to the vector algebra. But in this question we just need the magnitude of the resultant of the two vectors which can be calculated easily by equation (i).