Students will learn how to apply the kinematic equations for rotation.

### Key Equations

\begin{align*} v = r \omega \end{align*}

linear speed *v* is equal to the angular speed \begin{align*} \omega \end{align*}

\begin{align*} a = r \alpha \end{align*}

linear acceleration *a* is equal to the angular acceleration \begin{align*} \alpha \end{align*}

\begin{align*}\theta (t) = \theta_0 + \omega_0 t + 1/2 \alpha t^2\end{align*}

Rotational equivalent of \begin{align*} x (t) = x_i + v_i t +1/2at^2 \end{align*}

\begin{align*}\omega (t) = \omega_0 + \alpha t\end{align*}

Rotational equivalent of \begin{align*} v_f = v_i + at \end{align*}

\begin{align*}\omega^2 = \omega_0^2 + 2 \alpha (\triangle \theta)\end{align*}

Rotational equivalent of \begin{align*} v_f^2 = v_i^2 + 2a \Delta x \end{align*}

### Guidance

These equations work in the case of constant angular acceleration. Use them just as you would use the linear kinematic equations studied in the One-Dimensional Motion lessons. Just replace displacement with the change in angle, the velocity with the angular velocity and the acceleration with the angular acceleration.

- When something rotates in a circle, it moves through a
*position angle*\begin{align*}\theta\end{align*}θ that runs from \begin{align*}0\end{align*}0 to \begin{align*}2\pi\end{align*}2π radians and starts over again at \begin{align*}0\end{align*}0 . The physical distance it moves is called the*path length.*If the radius of the circle is larger, the path length traveled is longer.

- The angular velocity \begin{align*}\omega\end{align*}
ω tells you how quickly the angle \begin{align*}\theta\end{align*}θ changes. In more formal language, the rate of change of \begin{align*}\theta\end{align*}θ , the angular position, is called the angular velocity \begin{align*}\omega\end{align*}ω . The direction of angular velocity is either clockwise or counterclockwise. Analogously, the rate of change of \begin{align*}\omega\end{align*}ω is the angular acceleration \begin{align*}\alpha\end{align*}α .

- The linear velocity and linear acceleration of rotating object also depend on the radius of rotation, which is called the
*moment arm*(See figure below.) If something is rotating at a constant angular velocity, it moves more quickly if it is farther from the center of rotation. For instance, people at the Earth’s equator are moving faster than people at northern latitudes, even though their day is still 24 hours long – this is because they have a greater circumference to travel in the same amount of time.

#### Example 1

A 2 kg mass is attached to a .5 m long string. Starting from rest, the mass is given a constant angular acceleration of 2 rad/s^{2}. If the string breaks when the tension exceeds 50 N, how long will it be before the string breaks and what will the object's angular displacement be?

##### Solution

To start, we first want to find the object's linear speed when the string breaks.

\begin{align*}
F_c&=\frac{mv^2}{r}\\
v&=\sqrt{\frac{F_cr}{m}}\\
v&=\sqrt{\frac{50\:\text{N} * .5\:\text{m}}{2\:\text{kg}}}\\
v&=3.53\:\text{m/s}\\
\end{align*}

Now, we can find the angular speed.

\begin{align*}
\omega&=\frac{v}{r}\\
\omega&=\frac{3.53\:\text{m/s}}{.5\:\text{m}}\\
\omega&=7.06\:\text{rad/s}\\
\end{align*}

Now we can begin finding the values asked in the problem based on this information using the equations from the section above. We'll start by finding the time before the string breaks.

\begin{align*}
\omega&=\alpha*t\\
t&=\frac{\omega}{\alpha}\\
t&=\frac{7.1\:\text{rad/s}}{2\:\text{rad/s}^2}\\
t&=3.53\:\text{s}\\
\end{align*}

Lastly, we'll find the angular displacement of the mass.

\begin{align*}
\omega^2&=\omega_o^2 + 2\alpha\Delta\theta\\
\omega^2&= 0 + 2\alpha\Delta\theta\\
\Delta\theta&=\frac{\omega^2}{2\alpha}\\
\Delta\theta&=\frac{(7.06\;\text{rad/s})^2}{2*2\:\text{rad/s}^2}\\
\Delta\theta&=12.5\;\text{rad}\\
\end{align*}

### Watch this Explanation

### Time for Practice

- You spin up some pizza dough from rest with an angular acceleration of \begin{align*}5\;\mathrm{rad/s}^2\end{align*}
5rad/s2 .- How many radians has the pizza dough spun through in the first \begin{align*}10\end{align*}
10 seconds? - How many times has the pizza dough spun around in this time?
- What is its angular velocity after \begin{align*}5\end{align*}
5 seconds?

- How many radians has the pizza dough spun through in the first \begin{align*}10\end{align*}
- Driving in your truck, you notice that your speed is 30 m/s. The wheels of your truck are 0.5 meters in radius. An ant is stuck on one of your wheels.
- What is the rotating speed of the ant in rads per second?
- How many radians has the ant covered in 4 seconds?
- How many times did the ant go fully around in the 4 seconds (i.e. how many full revolutions did the ant make?

Now the truck begins to accelerate at \begin{align*} 2 m/s^2 \end{align*}

2m/s2 for the next 6 seconds.- What is the ants angular acceleration?
- After 6 seconds, what is the ants rotational speed in radians per second?
- how many full revolutions has the ant made in these 6 seconds while the truck was accelerating?

**Answers**

- a. \begin{align*}250 \;\mathrm{rad}\end{align*}
250rad b. almost \begin{align*}40 \;\mathrm{revolutions}\end{align*}40revolutions c. \begin{align*}25 \;\mathrm{rad/s}\end{align*}25rad/s - a. 60 rads/s b. 240 radians c. 38 full revolutions d. \begin{align*} 4 rads/s^2 \end{align*}
4rads/s2 e. 84 rads/s f. 432 revolutions