Question

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second then what is the speed of ball

Answer 1

Answer:
`(\pi)/(2) ms^(-1)`

Step-by-step explanation:

Let
`T` be the time required to make one revolution.

Let
`r` be the radius of the circular path.

Let
`d` be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So,
`d=2\pi r`

Given,
`d=0.5m\\T=2sec`

`d=2* \pi * 0.5=\pi m`

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let
`s` be the speed of the ball.

`s=(d)/(T)=(\pi )/(2)ms^(-1)`

So,the speed of the ball is
`(\pi )/(2)ms^(-1)`