The sections on rotational motion are best tackled in this order:

- Introduction to rotational motion (this section)
- Moment of inertia
- Angular velocity
- Angular acceleration
- Angular momentum
- Centripetal force

You might also want to refresh your memory about units of angle measurement, degrees and radians.

**Angular acceleration** is any change in the angular velocity vector. That means an increase in angular velocity, a decrease in angular velocity, or a change of direction of the angular velocity vector.

Recall that the angular velocity vector lies along the axis of rotation (at a right angle to the rotational plane) and, by convention, follows the right hand rule.

X
#### The Greek alphabet

alpha | Α | α |

beta | Β | β |

gamma | Γ | γ |

delta | Δ | δ |

epsilon | Ε | ε |

zeta | Ζ | ζ |

eta | Η | η |

theta | Θ | θ |

iota | Ι | ι |

kappa | Κ | κ |

lambda | Λ | λ |

mu | Μ | μ |

nu | Ν | ν |

xi | Ξ | ξ |

omicron | Ο | ο |

pi | Π | π |

rho | Ρ | ρ |

sigma | Σ | σ |

tau | Τ | τ |

upsilon | Υ | υ |

phi | Φ | φ |

chi | Χ | χ |

psi | Ψ | ψ |

omega | Ω | ω |

Angular acceleration is the change in angular velocity over time.

$$\alpha = \frac{\Delta \omega}{\Delta t}$$

As defined, the angular acceleration is the average angular acceleration, which we denote by placing a bar over the alpha:

$$\bar{\alpha} = \frac{\omega_f - \omega_i}{\Delta t}$$

Of course, the average accleration can also be determined by simply taking the numerical average of two accelerations, assuming a smooth transition from one to the next:

$$\bar{\alpha} = \frac{a_1 + a_2}{2}$$

The instantaneous angular acceleration is the limit of **Δω/Δt** as **Δt** approaches zero:

$$\alpha = \lim_{\Delta t \to 0} \frac{\Delta \omega} {\Delta t}$$

If you know a little differential calculus, this is just the first derivative of the angular velocity, **Δ** with respect to time.

$$\alpha = \frac{d\omega}{dt}$$

And finally, because angular velocity is the first derivative of the angle with respect to time, angular acceleration is the second derivative.

$$\alpha = \frac{d^2 \theta}{dt^2}$$

X
### SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units

length | meter | (m) |

mass | Kilogram | (Kg) |

time | second | (s) |

force | Newton | N |

energy | Joule | J |

To find the **linear acceleration** of an object traveling in a circle (also called **tangential acceleration** because the vector is always tangent to the circle), we can step back to an earlier development of linear velocity from **angular velocity**.

The arc **S** can be determined simply if we work in **radians**. We simply set up the proportion: "The angle **θ** is to a full circle (**2π**) as the arc **S** is to the full circumference (**2πr**):

$$\frac{\theta}{2\pi} = \frac{S}{2\pi r}$$

The **2π**'s cancel, giving the angle in terms of the arc length and radius.

$$\theta = \frac{S}{r}$$

Now the average angular velocity is:

$$\bar{\omega} = \frac{\Delta \theta}{\Delta t} = \frac{\Delta S}{r \Delta t}$$

In the last step we substituted **S/r** for **θ**. Now we showed that

$$\frac{\Delta S}{\Delta t} = \bar{v}$$

Therefore the angular velocity is related to the velocity (average or instantaneous) by:

$$\omega = \frac{v}{r}$$

Now angular acceleration is the change in angular velocity over time,

$$\alpha = \frac{\Delta \omega}{\Delta t}$$

So we can combine that with the previous result to get:

$$\omega = \frac{v}{r} = \frac{1}{r} \frac{\Delta v}{\Delta t}$$

Note that **r** is constant, so we can exempt it from the *delta* (it doesn't change). Substituting linear acceleration for **Δv/Δt**, we get the relationship between angular and linear or tangential acceleration:

$$\alpha = \frac{a}{r}$$

Let's insert here a few comments about **centripetal acceleration**, the only acceleration needed to keep a turning body turning.

You are familiar with the feeling of going around a corner as a passenger in a car, and feeling like if the door wasn't there, you'd fly out of the car (you would). That's often confused with an outward force (away from the axis of rotation), but no such force exists. Such a "centrifugal" force is a **pseudo-force**. It's not a real force, only the impression of one that really stems from the **centripetal** or "center-seeking" force.

What you're feeling when you're going around that corner is the car pushing you into the turn. Without the car there to push on you, you'd continue in the path of your inertia, which is a straight line tangent to the curve. The figure here illustrates that.

A body, like the green circle in the figure, has centripetal acceleration just because it is moving in a circle, because its velocity vector is always changing direction.

If its velocity vector is also changing in length, then it is accelerating tangentially, too. Otherwise it has no linear or tangential acceleration.

1. |
Starting from rest, the rotors of a helicopter undergo an angular acceleration of 0.75 rad/s - How fast will they be traveling in 60 seconds?
- How many radians will they have rotated through?
- Calculate the linear velocity of the rotor tips (r = 7.25 m) after 60 s of this acceleration.
## SolutionThe speed after t seconds of acceleration is $\omega = \alpha t,$ giving a final angular velocity (speed) of $$\omega = \alpha t = 0.75 \, \frac{rad}{s^2} (60 \, s) = 45 \, \frac{rad}{s}$$ For (b), in order to find the number of rotations, we'll need the $$\bar \omega = \frac{0 + 45}{2} = 22.5 \, \frac{rad}{s}$$ Now the number of radians is $$ \require{cancel} 22.5 \frac{rad}{\cancel{s}} (60 \, \cancel{s}) = 1350 \, rad$$ If we divide 2π radians into that, we get the number of full rotations: $$1350 \, \cancel{rad} \left( \frac{1 \, rev}{2\pi \cancel{rad}} \right) = 215 \, rev.$$ Now for the linear velocity at the rotor tips: $$v = r\omega = (7.25 \, m)\left( 45 \frac{rad}{s} \right) = 326 \, \frac{m}{s}$$ That's just below the speed of sound (~340 m/s). Generally, helicopters keep the blade-end speed below the speed of sound, where strange things happen and the machine can lose a lot of power. |

2. |
How long will it take for a basketball spinning on someone's finger to stop if it undergoes an angular acceleration (deceleration) of -0.13 rad/s ## SolutionThe inital angular velocity in rad/s is $$ \begin{align} 258 \, &\frac{\cancel{rev}}{\cancel{min}} \left( \frac{2\pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] &= 27 \, \frac{rad}{s} \end{align}$$ Now the definition of angular acceleration can be rearranged to find the time: $$\alpha = \frac{\omega_f - \omega_i}{\Delta t} \: \color{magenta}{\longrightarrow} \Delta t = \frac{\omega_f - \omega_i}{\alpha}$$ Then $$\Delta t = \frac{(0 - 27)\frac{rad}{s}}{0.13 \frac{rad}{s^2}} = 208 \, s$$ |

3. |
A washing machine spins clothes to near dryness. If it begins spinning at 20 rad/s and slows down to 7 rad/s while turning through 500 revolutions, - calculate the angular acceleration, and
- how much time will it take to slow down?
## SolutionWe'll need the average angular velocity, which is just the numerical average of 20 and 7 rad/s: $$\bar{\omega} = \frac{20 + 7}{2} = 13.5 \, \frac{rad}{s}$$ We can convert this to revolutions per second: $$\bar{\omega} = 13.5 \, \frac{\cancel{rad}}{s} \left( \frac{1 \, rev}{2 \pi \, \cancel{rad}} \right) = 2.1486 \, \frac{rev}{s}$$ Now if we divide 500 revolutions by that number of revolutions per second (just pay attention to the units), we get the time that it takes for 500 revolutions: $$\frac{500 \cancel{rev}}{2.1486 \frac{\cancel{rev}}{s}} = 232.7 \, s$$ That's 3:53 in minutes:seconds. Now we have all we need to calculate the angular accleration: $$\alpha = \frac{(20 - 7) \frac{rev}{s}}{232.7 \, s} = 0.056 \, \frac{rad}{s^2}$$ |

4. |
A centrifuge starts from rest and reaches an angular velocity of 30,000 rpm in 5 seconds. Calculate the angular acceleration (in rad/s ## SolutionFirst convert revolutions per minute (rpm) into radians per second: $$ \begin{align} \frac{30,000 \cancel{rev}}{\cancel{min}} &\left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] \omega &= 3141 \, \frac{rad}{s} \end{align}$$ Now calculate the angular accleration: $$\alpha = \frac{(3141 - 0) \frac{rev}{s}}{5 \, s} = 628 \, \frac{rad}{s^2}$$ The linear velocity is $v = \omega r,$ $$v = 0.01 \, m \left( 3141 \frac{rad}{s} \right) = 31.41 \, \frac{m}{s}$$ Finally, convert to mi./h, where 1 mile = 1609 meters and 1 hour = 3600 s. $$ \begin{align} 31.41 &\frac{\cancel{m}}{\cancel{s}} \left( \frac{mi}{\cancel{m}} \right)\left( \frac{\cancel{s}}{h} \right) \\[5pt] &= 70 \, \frac{mi.}{h} \end{align}$$ |

5. |
A flywheel rotating at 2.0 rpm undergoes an angular acceleration of 2.0 rad/s ## SolutionRecall that for linear motion, $d = \frac{1}{2} at^2.$ We can make the same analogy for rotational motion, and rearranging for time gives: $$\theta = \frac{1}{2} \alpha t^2 \: \color{magenta}{\longrightarrow} \: t = \sqrt{\frac{2 \theta}{\alpha}}$$ The time to spin up to the final angular velocity is then $$t = \sqrt{\frac{2 \theta}{\alpha}} = \sqrt{\frac{2 (10 \pi)}{2}} = 5.605 \, s,$$ where $10 \pi$ is the angle measure of 5 revolutions. Now by further analogy with the linear world, we know that $\omega_f = \alpha t + \omega_i,$ so we have our final angular velocity: $$ \begin{align} \omega_f &= 2 \frac{rad}{s^{\cancel{2}}} (5.605 \, \cancel{s}) + 2 \, \frac{rad}{s} \\[5pt] &= 13.2 \, \frac{rad}{s} \end{align}$$ |

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