Question

In given figure a= 15 m/s^2 represent the total acceleration of a particle moving in clockwise direction a circle radius R=2.5 m at a given instant time. the speed of the particle is ​

Answer 1

Answer:

Component along radius direction will be balanced by centrifugal force .

$ma \cos(30) = \frac{m {v}^{2} }{r} \\ {v}^{2} = arcos30 = 15 \times 2.5 \times \frac{ \sqrt{3} }{2} \\ {v}^{2} = \frac{15 \times 2.5 \times 1.73}{2} = 32.43 \\ v \approx \: 5.7 \: \frac{m}{s}$

Answer 2

Answer:

Option (3) 5.7 m/s

Explanation:

[Note; The given case is of Non Uniform Accelerated Motion]

Given;-

[Refer to attached image 1 for diagram]

       Acceleration, a= 15 m/s²

        Radius, R = 2.5 m

[Refer to attached image 2 to visualize the case better]

Component of Acceleration on radial length;-

           Ac = a cos∅___  (1) [in this case] & [Ac= centripetal acceleration]

But,     Ac = V²/ R___ (2) [By formula]

From (1) and (2), we obtain;-

              V²/ R = a cos∅

              V² = a cos∅ R

              V² = 15 × cos 37° × 2.5

              V² = 15 × 4/5 × 5/2   [since, cos 37°= 4/5]

              V² = 30

               V = √30

               V ≈ 5.7 m/s (approx)

Hence, the speed of the particles is 5.7 m/s.

Hope it helps! ;-))