Answer:
Angular displacement=2π/3.9 rad
Explanation:
Given data
Time t=3.9s
Required
The angular displacement during a 1.0 s time interval
Solution
In 3.9 second the child covers a full circle=2π rad
Angular displacement after 1.0 second is given as:
[tex]=\frac{2\pi }{3.9} rad[/tex]
Angular displacement=2π/3.9 rad
Angular Displacement in one second time interval on merry-go-round = [tex]2\pi/3.9[/tex] radians
The Average angular velocity is defined as
Average angular velocity = Total Angle Covered in Radians/ Total time interval taken to cover that angle.
Time interval taken by the child to complete one round = 3.9 s
Angle covered by child in one round = [tex]2\pi[/tex]
The Angular velocity is given by equation (1)
[tex]\omega = d\theta /dt[/tex]...........(1)
where [tex]d\theta[/tex] is the instantaneous angular displacement.
and [tex]dt[/tex] is the instantaneous time interval.
By the definition of average angular velocity we can write
The angular velocity [tex]\omega=2\pi/3.9[/tex]
Putting these values in equation (1) at [tex]dt =1[/tex] we get ,
[tex]2\pi/3.9[/tex] = [tex]\dfrac{d\theta}{1}[/tex]
[tex]d\theta[/tex] = [tex]2\pi/3.9[/tex]
So the angular displacement in one second time interval = [tex]2\pi/3.9[/tex] radians
For more information please refer to the link below
https://brainly.com/question/12446100
