A Child Playing with a Twirling Ball: Centripetal Force Calculation

A child is twirling a 0.0154-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around the circle in 0.639 s.

(a) Determine the centripetal force acting on the ball.

(b) If the speed is doubled, by what factor does the centripetal force increase?

1. a) Fcp = 0.23 N
2. b) As Fcp = 0.93 N, it increases 4 times when the speed is doubled

Explanation:

Let's explain what the centripetal force is. It is the force applied to an object moving on a curvilinear path, directed towards the center of rotation.

The formula for centripetal force is given by Fcp * r = m * V^2, where:

Fcp is the Centripetal Force

r is the horizontal circle radius

m is the ball mass

V is the tangential speed, same as the rotational speed

We are given that the ball travels once around the circle in 0.639 s. To calculate the angular speed, we convert 1 cycle to 2π radians:

ω = (1 cycle / 0.639 s) * (2π / 1 cycle)

ω = 9.83 rad/s

Next, we find the tangential speed using the angular speed and circle radius:

V = ω * r

V = 9.83 rad/s * 0.149 m = 1.5 m/s

Now, we can calculate the Centripetal Force:

Fcp = (m * V^2) / r

Fcp = ((0.0154 kg) * (1.5 m/s)^2) / 0.149 m

So, the Centripetal Force is Fcp = 0.23 N

When the tangential speed is doubled to 3 m/s, the new Centripetal Force is:

Fcp = ((0.0154 kg) * (3 m/s)^2) / 0.149 m

Thus, the Centripetal Force becomes Fcp = 0.93 N

Upon comparison, we see that when the speed is doubled, the Centripetal Force increases four times.