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

Velocity along a straight line is measured in units of length divided by time. In modern physics that's usually meters per second (m·s^{-1}). By now you should have had plenty of experience working with linear velocities and the vectors that represent them.

Things are different on the *rotational world*. In the rotational world, we measure velocity in a circular motion in terms of angle traversed divided by time, usually in units of degrees per second or radians per second (rad·s^{-1}).

We usually denote angular velocity with the Greek lower case letter omega, **ω**

$$\omega = \frac{\text{angle}}{\text{time}}$$

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 |

The **average angular velocity** of an object traveling rotating about an axis is

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

where **Δθ** is the change in angle over the change in time, **Δt**. Recall that a bar over a quantity in this context means "mean" or "average." In such a calculation, we have no details about acceleration during the time period **Δt**, so, just like we saw for linear velocity, this is an average between two endpoints:

$$\bar{\omega} = \frac{\theta_f - \theta_i}{\Delta t}$$

where **θ _{f}** and

We can also write the average value of two known instantaneous angular velocities, **ω _{1}** and

$$\bar{\omega} = \frac{\omega_1 + \omega_2}{2}$$

The instantaneous angular velocity is found when **Δt** becomes infinitesimally small, or

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

This is read "the limit of **Δθ**/**Δt** as **Δt** approaches zero."

In **calculus** (which you don't need to know to work through the rest of this section), the angular velocity is the first derivative of the angle with respect to time:

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

We'll need to be clear about when we're working with average and instantaneous angular velocities.

You might have watched a bicycle tire rotate and noticed that the tire moves much faster than the hub in the center. Your observation was correct. While the angular velocity of the hub and tire were the same, their linear velocities were different.

Consider the car tire show here. Let's think about two points on the tire. One on the surface, at a radius of **r _{1}** from the center, and one at a shorter radius,

Now if the tire is rotating at **2π** radians per second (that would be 60 rpm), then both points on the more-or-less rigid tire would be going around at the same rate; they would both have the same angular velocity.

But we could also ask how far each point travels in the same time. The point at **r _{1}** travels the full circumference of the tire,

The linear velocity of the point at **r _{1}** is

For an object rotating around an axis like the wheel, every point on the object has the same **angular velocity**, but not necessarily the same **linear velocity**.

I've left off the units of the radii here, but that's not important to the discussion. Points farther from the center of the rotating body have the same angular velocity as any other point, but a greater linear velocity.

Consider the figure below, describing the rotation of an object about an axis.

The arc **S** can be determined very 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:

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

or

$$v = r \cdot \omega$$

The two linear velocity vectors, tangents to the circle in the last figure, are given by **v = r·ω**.

**angular velocity**, but not necessarily the same **linear velocity**.

Like linear velocity, angular velocity is a vector quantity. But because a rotating body is in continuous motion, how do we decide the direction of the angular velocity vector?

We overcome that problem by choosing to have the angular velocity vector to be aligned with the axis of rotation, as shown.

One problem remains, however: The axis of rotation has two ends. Which way does the vector point?

We decide to use what is called the **right-hand rule** to decide. The right hand rule works as shown in the picture.

If I imagine wrapping the fingers of my right hand around the axis of rotation, fingers pointing in the direction of the rotation, then my thumb can only point one way. That's the direction of the angular velocity vector.

Calculate the angular velocity and the linear velocity of a child revolving at the edge of a playground merry-go-round that is 3.6 m in diameter, if the child completes one revolution every 2.1 seconds.

**Solution**

$$ \begin{align} \bar{\omega} &= \frac{\Delta \theta}{\Delta t} = \frac{2\pi \; rad}{2.1 \; s} \\[5pt] &= 3 \; \frac{rad}{s} \end{align}$$

(That's about 171˚ per second.) Now the linear velocity should just be **v = r·ω**. That's

$$ \begin{align} v &= r \cdot \omega \\[5pt] &= (1.8 \; m)(3.0 \, \frac{rad}{s}) \\[5pt] &= 5.4 \, \frac{m}{s} \end{align}$$

The circumference of the merry-go-round is

$$ \begin{align} C &= 2\pi r \\[5pt] &= 2\pi \, (1.8 \, m) = 11.3 \, m \end{align}$$

So 5.4 m/s seems about right for a rotational speed of 171˚/s.

Calculate the angular velocity of Earth (a) as it rotates about its axis, and (b) in its orbit around the sun.

**Solution**

$$ \require{cancel} \begin{align} \omega &= \left( \frac{2\pi \; rad}{24 \cancel{h}} \right) \left( \frac{1 \cancel{h}}{3600 \; s} \right)\\[5pt] &= 0.0000727 \; \frac{rad}{s} \\[5pt] &= 73 \; \frac{\mu \, rad}{s} \end{align}$$

In the last step, an awkward number of radians has been converted to a nicer number om *micro*-radians. It's good to know your metric prefixes and be able to do that when you want.

For part (b), we know that Earth completes one revolution about the sun (in a nearly circular orbit) every 365.25 days, so it's a similar calculation:

$$ \begin{align} \omega &= \left( \frac{2\pi \; rad}{365.25 \cancel{d}} \right) \left( \frac{1 \cancel{d}}{24 \cancel{h}} \right) \left( \frac{1 \cancel{h}}{3600 \; s} \right) \\[5pt] &= 0.000000119 \, \frac{rad}{s} \\[5pt] &= 0.2 \, \frac{\mu \, rad}{s} \end{align}$$

It's interesting to go a little further here and calculate the *linear* velocities of the surface of Earth as it spins and of Earth as it revolves about the sun.

For the first, the radius of Earth is about 6.4 x 10^{6} m, so

$$ \begin{align} v &= r \cdot \omega \\[5pt] &= (6.4 \times 10^6 \; m)(7.27 \times 10^{-5} \, \frac{rad}{s} \\[5pt] &= 465 \; \frac{m}{s} \end{align}$$

That's pretty fast. Let's put it in terms of miles per hour to see just how fast.

$$\left( \frac{465 \cancel{m}}{\cancel{s}} \right) \left( \frac{1 \; mi}{1609 \cancel{m}} \right) \left( \frac{3600 \cancel{s}}{1 \; h} \right) \\[5pt] = 1041 \, \frac{mi.}{h}$$

Sot that's surprising. If you're standing on the surface of Earth at the equator, "not moving," you're already moving at more than 1000 mi./h !

We can repeat the calculation for the linear velocity of Earth as it orbits the sun. The mean Earth-Sun distance is 1.496 x 10^{11} m:

$$ \begin{align} v &= r \cdot \omega \\[5pt] &= (1.496 \times 10^{11} \; m)\left(1.99 \times 10^{-7} \, \frac{rad}{s}\right) \\[5pt] &= 29,770 \, \frac{m}{s} \end{align}$$

... and 29,770 m/s in miles per hour is:

$$\left( \frac{29,770 \cancel{m}}{\cancel{s}} \right) \left( \frac{1 \; mi}{1609 \cancel{m}} \right) \left( \frac{3600 \cancel{s}}{1 \; h} \right) \\[5pt] = 66,609 \, \frac{mi.}{h}$$

1. |
Let's say you are designing a wind turbine to generate electricity. the tower height is 200 ft to safely accommodate a 105 ft. rotor diameter. If the shaft of the rotor is connected directly to a generator that needs to spin at 1000 revolutions per minute (rpm), how fast would the ends of the turbine blades be turning in order to acheive this? Is this feasible? If not, what's your solution? ## SolutionThe relationship between angular and linear velocity is $\omega = \frac{v}{r}.$ First we'll have to convert 1000 rpm to radians per second: $$ \require{cancel} 1000 \frac{\cancel{rev}}{\cancel{min}} \left( \frac{2\pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) = 33.3 \pi \, \frac{rad}{s}$$ Now the radius of the tower in meters is $$r = \frac{105 \cancel{ft}}{2} \left( \frac{1 \, m}{3.28 \cancel{ft}} \right) = 16 \, m$$ So the linear velocity at the rotor tips is $$v = \omega r = 33.3 \pi \, \frac{rad}{s} \cdot 16 \, m = 1675 \, \frac{m}{s}$$ That seems pretty fast. Let's put it into miles per hour and see: $$1675 \frac{m}{s} \left( \frac{3600 \cancel{s}}{1 \, h} \right) \left( \frac{1 \, mi}{1609 \cancel{m}} \right) = 3749 \, \frac{mi}{h}$$ Ok, that's not feasible. The materials just couldn't withstand that kind of stress for long. The way that wind-turbine engineers overcome this is to place gears between the rotor and the generator that allow the rotor to spin more slowly (but against more resistance) to spin the generator at the best rate. |

2. |
A typical road bicycle has a wheel diameter of 678 mm diameter wheel (that's a "700c" rim with 28mm tires – who knows why some things are named the way they are?) How many times does such a wheel rotate in covering 25 miles on the road? Calculate the angular velocity of such a wheel when the bike is rolling at 18 mi/h. (1 mi = 1609 m) ## SolutionFor the first part, we just need to divide the circumference of the tire into the total distance. The circumference of a circle is $c = \pi d,$ where d is the diameter. $$c = \pi (0.678 \, m) = 2.1299 \, m$$ Now we just divide that into the 25 mile distance, rememberting to convert the distance to meters first, to find $n,$ the number of revolutions: $$ \begin{align} n &= \frac{\left( 25 \cancel{mi.} \frac{1609 \, m}{1 \, \cancel{mi.}} \right)}{2.1299 \, m} \\[5pt] &= 18,885 \, \text{revolutions} \end{align}$$ For the second part, 18 mi./h is $$18 \frac{\cancel{mi}}{\cancel{h}} \left( \frac{1609 \, m}{1 \cancel{mi}} \right)\left( \frac{1 \cancel{h}}{3600 \, s} \right) = 18.045 \frac{m}{s}$$ Dividing by the circumference gives us the number of cycles per second: $$\frac{18.045 \, \cancel{m}/s}{2.1299 \cancel{m}} = 3.77 \frac{rev}{s}$$ Now converting revoluions (rev) to radians gives us $$3.77 \, \frac{\cancel{rev}}{s} \left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) = 23.7 \frac{rad}{s}$$ |

3. |
A sling (like what David slew Goliath with) is rotated at a rate of 90 rpm (revolutions per minute). - Calculate the length of the sling that would be required to give the stone being thrown to have a linear velocity of 30 m/s.
- If the sling is 1m long, what linear velocity would a thrown stone have?
## SolutionFirst convert revolutions per minute (rpm) into radians per second: $$ \begin{align} \frac{90 \cancel{rev}}{\cancel{min}} &\left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] \omega &= 9.42 \, \frac{rad}{s} \end{align}$$ Now for part (a) recall that $v = r \omega,$ so $$r = \frac{v}{\omega} = \frac{30 \frac{m}{s}}{9.42 \frac{1}{s}} = 3.18 \, m$$ Notice above that we drop the unit radians when doing these calculations. For (b) we have $$ \begin{align} v &= r \omega \\[5pt] &= 1 \, m \left( 9.42 \frac{rad}{s} \right) = 9.42 \frac{m}{s} \end{align}$$ Just for perspective, if we convert that speed to miles per hour we get $$ \begin{align} 9.42 \frac{\cancel{m}}{\cancel{s}} &\left( \frac{1 mi}{1609 \cancel{m}} \right) \left( \frac{3600 \cancel{s}}{1 \, h} \right) \\[5pt] &= 21 \, \frac{mi.}{h} \end{align}$$ |

4. |
The wheels of a car have a diameter of 23 in (58 cm). If they are rotating at a speed of 560 rpm while the car is in motion, calculate its linear velocity. ## SolutionFirst convert revolutions per minute (rpm) into radians per second: $$ \begin{align} \frac{560 \cancel{rev}}{\cancel{min}} &\left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] \omega &= 58.64 \, \frac{rad}{s} \end{align}$$ Now recall that $v = r \omega,$ so $$v= 0.29 \, m \left( 58.64 \frac{rad}{s} \right) = 17.0 \frac{m}{s}$$ Just for perspective, if we convert that speed to miles per hour we get $$ \begin{align} 17.0 \frac{\cancel{m}}{\cancel{s}} &\left( \frac{1 mi}{1609 \cancel{m}} \right) \left( \frac{3600 \cancel{s}}{1 \, h} \right) \\[5pt] &= 38 \, \frac{mi.}{h} \end{align}$$ |

5. |
A windmill has three blades that are 9 m in length. The group of three blades is rotating at 4 rpm. Calculate the linear velocity of the tips of the blades in miles/hour. ## SolutionFirst convert revolutions per minute (rpm) into radians per second: $$ \begin{align} \frac{4 \cancel{rev}}{\cancel{min}} &\left( \frac{2 \pi \, rad}{1 \cancel{rev}} \right) \left( \frac{1 \cancel{min}}{60 \, s} \right) \\[5pt] \omega &= 0.4189 \, \frac{rad}{s} \end{align}$$ Now recall that $v = r \omega,$ so $$v= 9 \, m \left( 0.4189 \frac{rad}{s} \right) = 3.77 \frac{m}{s}$$ If we convert that speed to miles per hour we get $$ \begin{align} 3.77 \frac{\cancel{m}}{\cancel{s}} &\left( \frac{1 mi}{1609 \cancel{m}} \right) \left( \frac{3600 \cancel{s}}{1 \, h} \right) \\[5pt] &= 8.4 \, \frac{mi.}{h} \end{align}$$ |

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