Respuesta :
Answer:
Angular velocity of merry-go-round is πk - 1 at t= T
Explanation:
From the question it is given that
[tex]\theta(t) = \pi k(t+k_e-\frac{t}{k} )[/tex] ..........................(1)
since mathematically, angular velocity is defined as
[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] ........................(2)
on substituing the value of θ(t) from equation 1 in equation (2) we get
[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] = [tex]\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}[/tex] ............................(3)
on differentiating equation (3) with respect to time we get
ω(t) = πk(1 -[tex]\frac{1}{k}[/tex]) = πk - 1 angular velocity of merry-go-round
Therefore, angular velocity of merry-go-round is πk - 1 at t= T
The angular velocity of the merry-go-round at t=T is πk - 1. The pace at which angular displacement changes is defined as angular velocity.
What is angular velocity?
The rate of change of angular displacement is defined as angular velocity.
It is stated as follows:
ω = θ t
The equation for the angular displacement is given as;
[tex]\theta (t) = \pi k (t+k_e-\frac{t}{k} )[/tex]
The angular velocity is found by the differentiation of the angular displacement.
[tex]\rm \omega(t) = \frac{d(\theta)}{dt} \\\\ \rm \omega(t) = \frac{d (\pi k (t+k_e-\frac{t}{k} )}{dt} \\\\ \rm \omega(t) = \pi k-1[/tex]
Hence the angular velocity of the merry-go-round at t=T is πk - 1.
To learn more about the angular speed refer to the link;
https://brainly.com/question/9684874
