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 \(2.95m/s^2\).