## Revolving Ball

## Part A

A 270 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.43 m. The ball makes 1.3 revolutions in one second.

What is the speed of the ball as it goes around in the circle?

What is the speed of the ball as it goes around in the circle?

**3.51 m/s**## Approach

Speed v is distance traveled divided by the time taken.

## Solution

The distance traveled in one second is (2πr)×1.3 = 3.51 m. The speed...

V= 3.51m/1s = 3.51m/s

Alternatively, you can find the speed by first finding the period, T. (T is the time it takes to complete one circle.) The distance traveled in one circle is 2πr. Since the ball makes 1.3 revolutions per second, the time it takes to complete one revolution (one circle) is 1/1.3 s = 0.769 s. This gives...

v = (2*pi*r)/T = (2 *pi *0.43)/0.769 = 3.51 m/s

V= 3.51m/1s = 3.51m/s

Alternatively, you can find the speed by first finding the period, T. (T is the time it takes to complete one circle.) The distance traveled in one circle is 2πr. Since the ball makes 1.3 revolutions per second, the time it takes to complete one revolution (one circle) is 1/1.3 s = 0.769 s. This gives...

v = (2*pi*r)/T = (2 *pi *0.43)/0.769 = 3.51 m/s

## Part B

## Approach

We know that the centripetal acceleration ac = v2/r. Since we just found the speed in part (a) and we are given the radius, it is simple to calculate the centripetal acceleration.

## Solution

The centripetal acceleration is the...

aR= V^2/r = 3.51 ^2/0.43 = 28.7 m/s^2

aR= V^2/r = 3.51 ^2/0.43 = 28.7 m/s^2