Question

A rotating merry-go-round makes one complete revolution in 4.0s a) what is the linear speed of a child seated 1.2 meters from the center? b) what is her acceleration (give components)

Answer 1

Answer:a) The linear speed of the rotating merry-go-round is 1.884 m/s.

b) The acceleration of the rotating merry-go-round device is
`2.95m/s^2`.

Step-by-step explanation:

a) linear speed of the device:

Distance =
`2\pi (r)`

Child seated 1.2 meters from the center, which means that radius ,r = 1.2 meters

Time taken to complete one revolution = 4.0 seconds

`\text{Linear speed}=v=(distance)/(time)=(2\pi r)/(t)=(2* 3.14* 1.2 m)/(4.0 s)=(7.536 m)/(4.0 s)=1.884 m/s`

The linear speed of the rotating merry-go-round is 1.884 m/s.

b) Acceleration

Linear velocity ,v = 1.884 m/s

`a_c=(v^2)/(r)`

`=(1.884 m/s* 1.884m/s)/(1.2 m)=2.95m/s^2`

The acceleration of the rotating merry-go-round device is
`2.95m/s^2`.