Chapter Summary

8.1 Angular Acceleration

Uniform circular motion is the motion with a constant angular velocity [latex]\omega = \frac{Δ \theta}{Δ t}[/latex].
In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is [latex]\alpha = \frac{Δ \omega}{Δ t}[/latex].
Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as [latex]a_{\text{t}} = \frac{Δ v}{Δ t}[/latex].
For circular motion, note that [latex]v = rω[/latex], so that
[latex]a_{\text{t}} = \frac{\Delta \left(rω\right)}{Δ t} .[/latex]
The radius r is constant for circular motion, and so [latex]\Delta \left(rω\right) = r Δ \omega[/latex]. Thus,
[latex]a_{\text{t}} = r \frac{Δ \omega}{Δ t} .[/latex]
By definition, [latex]Δ \omega / Δ t = \alpha[/latex]. Thus,
[latex]a_{\text{t}} = rα[/latex]

or

[latex]\alpha = \frac{a_{\text{t}}}{r} .[/latex]

Kinematics is the description of motion.
The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
Starting with the four kinematic equations we developed in Kinematics, we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in Table 8.1.
In these equations, the subscript 0 denotes initial values (_{[latex]x_{0}[/latex]} and _{[latex]t_{0}[/latex]} are initial values), and the average angular velocity [latex]\overset{-}{\omega}[/latex] and average velocity [latex]\overset{-}{v}[/latex] are defined as follows:
[latex]\bar{\omega} = \frac{\omega_{0} + \omega}{2} \textrm{ }\text{and}\textrm{ } \bar{v} = \frac{v_{0} + v}{2} .[/latex]

The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
If we exert a force [latex]F[/latex] on a point mass [latex]m[/latex] that is at a distance [latex]r[/latex] from a pivot point and because the force is perpendicular to [latex]r[/latex], an acceleration [latex]\text{a }=\text{ F}/\text{m}[/latex] is obtained in the direction of [latex]F[/latex]. We can rearrange this equation such that
[latex]F = ma ,[/latex]

and then look for ways to relate this expression to expressions for rotational quantities. We note that
[latex]a = rα[/latex], and we substitute this expression into [latex]F=ma[/latex], yielding

[latex]F=mrα[/latex]
Torque is the turning effectiveness of a force. In this case, because [latex]F[/latex] is perpendicular to [latex]r[/latex], torque is simply [latex]\tau = \mathit{rF}[/latex]. If we multiply both sides of the equation above by [latex]r[/latex], we get torque on the left-hand side. That is,
[latex]\text{rF} = \text{mr}^{2} \alpha[/latex]

or

[latex]\tau = \text{mr}^{2} \alpha .[/latex]
The moment of inertia [latex]I[/latex] of an object is the sum of [latex]\text{MR}^{2}[/latex] for all the point masses of which it is composed. That is,
[latex]I = \sum \text{mr}^{2} .[/latex]
The general relationship among torque, moment of inertia, and angular acceleration is
[latex]\tau = Iα[/latex]

or

[latex]\alpha = \frac{\text{net }\tau}{I} \cdot \\[/latex]

The rotational kinetic energy [latex]\text{KE}_{\text{rot}}[/latex] for an object with a moment of inertia [latex]I[/latex] and an angular velocity [latex]\omega[/latex] is given by [latex]\text{KE}_{\text{rot}} = \frac{1}{2} Iω^{2} .[/latex]
Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
Work and energy in rotational motion are completely analogous to work and energy in translational motion.
The equation for the work-energy theorem for rotational motion is [latex]\text{net } W = \frac{1}{2} Iω^{2} - \frac{1}{2} I \omega_{\text{0}}^{2} .[/latex]

Every rotational phenomenon has a direct translational analog , likewise angular momentum [latex]L[/latex] can be defined as [latex]L = Iω .[/latex]
This equation is an analog to the definition of linear momentum as [latex]p = \text{mv}[/latex]. The relationship between torque and angular momentum is [latex]\text{net } \tau = \frac{\Delta L}{\Delta t} .[/latex]
Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.

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