# 6.5: Angular Velocity and Angular Acceleration

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## . Angular Velocity

We shall always choose a right-handed cylindrical coordinate system. If the positive z - axis points up, then we choose θ to be increasing in the counterclockwise direction as shown in Figures 6.6.

For a point object undergoing circular motion about the z -axis, the angular velocity vector $$\vec{\omega}$$ is directed along the z -axis with z -component equal to the time derivative of the angle θ,

$\vec{\omega}=\frac{d \theta}{d t} \hat{\mathbf{k}}=\omega_{z} \hat{\mathbf{k}} \nonumber$

The SI units of angular velocity are $$\left[\mathrm{rad} \cdot \mathrm{s}^{-1}\right]$$ Note that the angular speed is just the magnitude of the z -component of the angular velocity,

$\omega \equiv\left|\omega_{z}\right|=\left|\frac{d \theta}{d t}\right| \nonumber$

If the velocity of the object is in the $$+\hat{\boldsymbol{\theta}}$$-direction, (rotating in the counterclockwise direction in Figure 6.7(a)), then the z -component of the angular velocity is positive, $$\omega_{z}=d \theta / d t>0$$ The angular velocity vector then points in the $$+\hat{\mathbf{k}}$$-direction as shown in Figure 6.7(a). If the velocity of the object is in the $$-\hat{\boldsymbol{\theta}}$$-direction, (rotating in the clockwise direction in Figure 6.7(b)), then the z -component of the angular velocity angular velocity is negative, $$\omega_{z}=d \theta / d t<0$$. The angular velocity vector then points in the $$-\hat{\mathbf{k}}$$-direction as shown in Figure 6.7(b).

The velocity and angular velocity are related by

$\overrightarrow{\mathbf{v}}=\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}}=\frac{d \theta}{d t} \hat{\mathbf{k}} \times r \hat{\mathbf{r}}=r \frac{d \theta}{d t} \hat{\boldsymbol{\theta}} \nonumber$

Example 6.2 Angular Velocity

A particle is moving in a circle of radius R. At t = 0 , it is located on the x -axis. The angle the particle makes with the positive x -axis is given by $$\theta(t)=A t-B t^{3}$$ where A and B are positive constants. Determine (a) the angular velocity vector, and (b) the velocity vector. Express your answer in polar coordinates. (c) At what time, $$t=t_{1}$$ is the angular velocity zero? (d) What is the direction of the angular velocity for 1. $$t<t_{1}$$ 2. $$t>t_{1} ?$$?

Solution:

The derivative of $$\theta(t)=A t-B t^{3}$$ is

$\frac{d \theta(t)}{d t}=A-3 B t^{2} \nonumber$

Therefore the angular velocity vector is given by

$\overrightarrow{\boldsymbol{\omega}}(t)=\frac{d \theta(t)}{d t} \hat{\mathbf{k}}=\left(A-3 B t^{2}\right) \hat{\mathbf{k}} \nonumber$

The velocity is given by

$\overrightarrow{\mathbf{v}}(t)=R \frac{d \theta(t)}{d t} \hat{\mathbf{\theta}}(t)=R\left(A-3 B t^{2}\right) \hat{\boldsymbol{\theta}}(t) \nonumber$

The angular velocity is zero at time $$t=t_{1}$$ when

$A-3 B t_{1}^{2}=0 \Rightarrow t_{1}=\sqrt{A / 3 B} \nonumber$

For $$t<t_{1}, \frac{d \theta(t)}{d t}=A-3 B t_{1}^{2}>0$$ hence $$\overrightarrow{\boldsymbol{\omega}}(t)$$ points in the positive $$\hat{\mathbf{k}}$$-direction.

For $$t>t_{1}, \frac{d \theta(t)}{d t}=A-3 B t_{1}^{2}<0$$ hence $$\overrightarrow{\boldsymbol{\omega}}(t)$$ points in the positive $$\hat{\mathbf{k}}$$-direction.

## Angular Acceleration

In a similar fashion, for a point object undergoing circular motion about the fixed z -axis, the angular acceleration is defined as

$\vec{\alpha}=\frac{d^{2} \theta}{d t^{2}} \hat{\mathbf{k}}=\alpha_{z} \hat{\mathbf{k}} \nonumber$

The SI units of angular acceleration are $$\left[\mathrm{rad} \cdot \mathrm{s}^{-2}\right]$$ The magnitude of the angular acceleration is denoted by the Greek symbol alpha,

$\alpha \equiv|\vec{\alpha}|=\left|\frac{d^{2} \theta}{d t^{2}}\right| \nonumber$

There are four special cases to consider for the direction of the angular velocity. Let’s first consider the two types of motion with $$\overrightarrow{\boldsymbol{\alpha}}$$ pointing in the $$\hat{\mathbf{k}}$$-direction: (i) if the object is rotating counterclockwise and speeding up then both $$d \theta / d t>0$$ and $$d^{2} \theta / d t^{2}>0$$ (Figure 6.8(a)) (ii) if the object is rotating clockwise and slowing down then $$d \theta / d t<0$$ but $$d^{2} \theta / d t^{2}>0$$ (Figure 6.8(b). There are two corresponding cases in which $$\vec{\alpha}$$ pointing in the $$-\hat{\mathbf{k}}$$-direction (iii) if the object is rotating counterclockwise and slowing down then $$d \theta / d t>0$$ but $$d^{2} \theta / d t^{2}<0$$ (Figure 6.9(a), (iv) if the object is rotating clockwise and speeding up then both $$d \theta / d t<0$$ and $$d^{2} \theta / d t^{2}<0$$ (Figure 6.9(b).

## Example 6.3 Integration and Circular Motion Kinematics

A point-like object is constrained to travel in a circle. The z -component of the angular acceleration of the object for the time interval $$\left[0, t_{1}\right]$$ is given by the function

$\alpha_{z}(t)=\left\{\begin{array}{l} b\left(1-\frac{t}{t_{1}}\right) ; 0 \leq t \leq t_{1} \\ 0 ; t>t_{1} \end{array}\right. \nonumber$

where b is a positive constant with units rad$$\cdot \mathrm{S}^{-2}$$.

a) Determine an expression for the angular velocity of the object at $$t=t_{1}$$.

b) Through what angle has the object rotated at time $$t=t_{1}$$?

Solution:

a) The angular velocity at time $$t=t_{1}$$ is given by

$\omega_{z}\left(t_{1}\right)-\omega_{z}(t=0)=\int_{t^{\prime}=0}^{t^{\prime}=t_{1}} \alpha_{z}\left(t^{\prime}\right) d t^{\prime}=\int_{t^{\prime}=0}^{t^{\prime}=t_{1}} b\left(1-\frac{t^{\prime}}{t_{1}}\right) d t^{\prime}=b\left(t_{1}-\frac{t_{1}^{2}}{2 t_{1}}\right)=\frac{b t_{1}}{2} \nonumber$

b) In order to find the angle $$\theta\left(t_{1}\right)-\theta(t=0)$$ that the object has rotated through at time $$t=t_{1}$$, you first need to find $$\omega_{z}(t)$$ by integrating the z-component of the angular acceleration

$\omega_{z}(t)-\omega_{z}(t=0)=\int_{t^{\prime}=0}^{t^{\prime}=t} \alpha_{z}\left(t^{\prime}\right) d t^{\prime}=\int_{t=0}^{t^{\prime}=t} b\left(1-\frac{t^{\prime}}{t_{1}}\right) d t^{\prime}=b\left(t-\frac{t^{2}}{2 t_{1}}\right) \nonumber$

Because it started from rest, $$\omega_{z}(t=0)=0$$, hence $$\omega_{z}(t)=b\left(t-\frac{t^{2}}{2 t_{1}}\right) ; 0 \leq t \leq t_{1}$$

Then integrate $$\omega_{z}(t)$$ between t = 0 and $$t=t_{1}$$ to find that

$\theta\left(t_{1}\right)-\theta(t=0)=\int_{t^{\prime}=0}^{t^{\prime}=t_{1}} \omega_{z}\left(t^{\prime}\right) d t^{\prime}=\int_{t^{\prime}=0}^{t^{\prime}=t_{1}} b\left(t^{\prime}-\frac{t^{\prime 2}}{2 t_{1}}\right) d t^{\prime}=b\left(\frac{t_{1}^{2}}{2}-\frac{t_{1}^{3}}{6 t_{1}}\right)=\frac{b t_{1}^{2}}{3} \nonumber$

This page titled 6.5: Angular Velocity and Angular Acceleration is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.