Answer:
(a) t=5.384s
(b) t=2.23s
Explanation:
Given data
Initial angular speed of merry go round ω₀=0 rad/s
Angular acceleration of merry go round α=1.30 rad/s²
For Part (a)
Using the equation of angular motion for constant angular acceleration
Θ=ω₀t+(1/2)at²
[tex]3*2\pi=0+(1/2)(1.30)t^{2}\\ 6\pi=0.5(1.30)t^{2}\\ t=\sqrt{\frac{6\pi }{0.65} }\\ t=5.384s\\[/tex]
The time that the merry takes to complete 3 revolutions is 5.384 seconds
For Part (b)
Using the equations of angular motion for constant acceleration to find the time for the round to complete 6 revolutions
Θ=ω₀t+(1/2)at²
[tex]6*2\pi=0+(1/2)(1.30)t^{2}\\ 12\pi=0.65t^{2}\\ t=\sqrt{\frac{12}{0.65} }\\ t=7.614s[/tex]
The time that the round takes to finish the second three revolutions is
[tex]t=7.614-5.384\\t=2.23s[/tex]
