Answer:
At t=T/2 the angular speed equals to its average angular speed
Explanation:
Angular Motion
Let w be the angular speed of a rotating object, [tex]\alpha[/tex] its angular acceleration, and T the time the acceleration is acting upon the object. The basic formula for the angular motion is
[tex]w=w_o+\alpha t[/tex]
We are told the initial speed is zero, so
[tex]w=\alpha t[/tex]
The average angular speed from t=0 to t=T can be found by
[tex]\displaystyle \bar w=\frac{1}{T}\int_0^T{w.dt}[/tex]
[tex]\displaystyle \bar w=\frac{1}{T}\int_0^T{\alpha t.dt}[/tex]
[tex]\displaystyle \bar w=\frac{1}{T}\frac{\alpha T^2}{2}[/tex]
[tex]\displaystyle \bar w=\frac{\alpha T}{2}[/tex]
This value is reached at a certain time we need to compute, knowing that
[tex]w=\bar w[/tex]
Or equivalently
[tex]\displaystyle \alpha t=\frac{\alpha T}{2}[/tex]
Simplifying we have
[tex]\displaystyle t=\frac{T}{2}[/tex]
At t=T/2 the angular speed equals to its average angular speed
