Question

Find the magnitude of angular acceleration of a particle moving in a circle of a radius 10 cm with uniform speed completing the circle in 4 seconds

Answer 1

Answer:

The distance covered in completing the circle 2πr=2π×10cm. 

The linear speed is v=2πr/t

=2π×10cm/4s

=5πcm/s

The acceleration is a=v^2/r

=(5πcm/s)2/10cm

=2.5π^2cm/s^2.

Hence, the magnitude of its acceleration is =2.5π^2cm/s^2.

Answer 2

Answer:

24.68 cm/s2

Explanation:

angular acceleration (a) = radius x square of angular velocity

a=rω²

angular velocity = 2 x Pie / period (T)

ω=2π/T

=2 x 22/7 / 4

ω= 1.571 rad/s

a = 10 x 1.571^2

=10 x 2.468

a=24.68 cm/s2