Question

2. In the given figure, a = 15 m/s2 represents the
total acceleration of a particle moving in the
clockwise direction in a circle of radius R = 2.5 m
at a given instant of time. The speed of the particle
[NEET (Phase-2) 2016]
30°​

Answer 1

Answer: 5.7 m/s

Explanation:

The Centripetal acceleration of a particle moving on a circular path is given by

$a_{C}$ = $\frac{V^2}{R}$   ...................... Equation 1

where V is velocity of the particle at that given instance and R is the radius of the circular path about which particle is moving.

Now, we know that centripetal acceleration is always towards the center and the angle subtended is given as 30°.

The total acceleration at any instant can be broken into two components, one component towards the center (cos) and one along the circular path (sin)

So, $a_{C}$ = a cos 30° where a is the total acceleration.

$a_{C}$ = 15 cos 30° = 15 X $\frac{\sqrt{3} }{2}$

But we know that  $a_{C}$ = $\frac{V^2}{R}$   (from equation 1)

Hence, $\frac{V^2}{R}$  = 15 X $\frac{\sqrt{3} }{2}$

$V^{2}$ = 2.5 X 15  X $\frac{\sqrt{3} }{2}$

$V^{2}$ = 32.476

V = 5.7 m/S

Hence, The Velocity of the particle at that given instant of time is 5.7 m/S.