Question

Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?

Answer 1

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, $\alpha$ its angular acceleration, and T the time the acceleration is acting upon the object. The basic formula for the angular motion is

$w=w_o+\alpha t$

We are told the initial speed is zero, so

$w=\alpha t$

The average angular speed from t=0 to t=T can be found by

$\displaystyle \bar w=\frac{1}{T}\int_0^T{w.dt}$

$\displaystyle \bar w=\frac{1}{T}\int_0^T{\alpha t.dt}$

$\displaystyle \bar w=\frac{1}{T}\frac{\alpha T^2}{2}$

$\displaystyle \bar w=\frac{\alpha T}{2}$

This value is reached at a certain time we need to compute, knowing that

$w=\bar w$

Or equivalently

$\displaystyle \alpha t=\frac{\alpha T}{2}$

Simplifying we have

$\displaystyle t=\frac{T}{2}$

At t=T/2 the angular speed equals to its average angular speed