Question

CENTRIPETAL ACCELERATION
A ball is whirled at the end of a string in a horizontal circle 60cm in radius at the rate of 1 revolution every 2s. Find the balls centripetal acceleration.

Answer 1

Answer:

Frequency f=2514Hz

Angular velocity ω=2πf=252×14πrad/s=3.52rad/s

Acceleration is a=ω2r=3.522×0.8=9.9m/s2

The direction is along the radius at every point towards the centre.

Answer 2

Given : A ball is whirled at the end of a string in a horizontal circle 60cm in radius at the rate of 1 revolution every 2s.

To find : centripetal acceleration of the ball.

solution : r = 60 cm = 0.6 m

The ball is whirled at the end of a string in a horizontal circle at the rate of one revolution every 2s.

so angular velocity, ω = angular displacement/time taken = 2π/2s = π rad/s

now centripetal acceleration, α = ω²r

= (π)² × 0.6

= 0.6π² m/s²

Therefore the centripetal acceleration of the ball is 0.6π² m/s²