Answer:
a) The angular speed of the child is approximately 1.257 rad/s
b) The angular speed of the teenager is approximately 1.257 rad/s
c) The tangential speed of the child is approximately 1.257 m/s
d) For the child, r = 2 m
The tangential speed of the teenager is approximately 2.513 m/s
Step-by-step explanation:
The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)
The radius of the merry-go-round = 2 m
The location of the child = 1 m from the axis
The location of the teenager = 2 m from the axis
1 revolution = 2·π radians
Therefore, we have;
The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)
∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second
a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s
b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s
c) The tangential speed, v = r × The angular speed, ω
Where;
r = The radius of rotation of the object
For the child, r = 1 m
The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s
d) For the child, r = 2 m
The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s