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 $2.95m/s^2$.

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

$\text{Linear speed}=v=\frac{distance}{time}=\frac{2\pi r}{t}=\frac{2\times 3.14\times 1.2 m}{4.0 s}=\frac{7.536 m}{4.0 s}=1.884 m/s$

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

b) Acceleration

Linear velocity ,v = 1.884 m/s

$a_c=\frac{v^2}{r}$

$=\frac{1.884 m/s\times 1.884m/s}{1.2 m}=2.95m/s^2$

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