Rotational Kinematics

Rotational Kinematics and Linear Kinematics are very closely related. In fact, many of the same equations we used with Linear Kinematics will be used here at we look at rotational motion. To be sure that since we're moving in circles rather than going in a straight line, there will be some differences. Those differences, however, will be very minor. Most of the concepts we investigate here in this chapter should be extremely familiar to you.

Centripetal Force

Thinking back to Newton's 1st Law of Motion, if an object is moving in a circular path, then there MUST be a force acting on it. Left on its own, an object will move with constant velocity, same speed in a straight line. The force that keeps an object moving in a circular path is called a Centripetal Force, and is always directed toward the center of the circular path. The more massive the object, the faster the velocity, and the tighter the turning radius, the stronger the force that is needed to maintain that circular path.

(1)
\begin{align} \vec{F_c} = - {{m v^2} \over {r}} \hat {r} \end{align}

Recalling from Newton's 2nd Law that $\vec{F_{net}} = m \vec{a}$, we can divide the above equation to define what we will call the centripetal acceleration:

(2)
\begin{align} \vec{a_c} = - {{v^2} \over {r}} \hat {r} \end{align}

Example:
A pilot in the Red Bull Air Races can experience accelerations as high as 11g during hard turns. If the aircraft is moving at 220 mph, what is the radius of it's turn?
$a = 11 g = 11 * 9.8 = 107.8 m/s^2, v = 220 mph = 98.3 m/s$
${a_c} = {{v^2} \over {r}}$
$=> {r} = {{v^2} \over {a_c}} = {{98.3^2} \over {107.8}} = 89.6 m$

Exercise:
A car moving at 35 mph is going around a flat corner whose radius of curvature is 120 m. What is it's acceleration?

The centripetal force isn't any one specific force, but its what we call the net force, the sum of all forces acting on a body, that results in a circular motion. When solving circular motion problems, the set-up is the same as it is with any other force problem. Start by identifying the object of interest, draw a free-body diagram for that object, define a coordinate system, determine the components of all the forces in that coordinate system, and set their sum equal to the net force. In circular motion problems, the net force will be of the form shown in Eq.(1) Let's examine the previous exercise a bit more and see what the coefficient of friction must be in order to keep the car on the clover leaf and prevent it from sliding off.

Example:
What is the minimum coefficient of friction necessary to keep a car moving at 35 mph (15.6 m/s) is going around a clover leaf on-ramp whose radius of curvature is 120 m assuming the corner is flat?

First, let's draw a free-body diagram for the object of interest, the car. (With my amazing art budget, I've rendered the car as a box) From this perspective, the car is moving away from us and turning to the right. Gravity is pulling the car down, the normal force between the car and the road is pushing upwards, and the static frictional force is pushing the car toward the center of the curve.
Here we use a standard +x-axis to the right and +y-axis is up. Keep in mind that the Universe doesn't care what coordinate system we choose. You can't choose wrong, but you can choose a coordinate system that creates more work for yourself. Here it is simple to align the coordinate axes with our three forces.
The acceleration of the car is wholly in the +x-direction with no acceleration in the y-axis. This allows us to create the following equations of motion.
$\Sigma F_x = \mu_s F_N = {{m v^2} \over {r}}$
$\Sigma F_y = F_N - mg = 0$
Like with most friction problems, our task now is to solve the y-axis equation for $F_N$ and substitute it into the x-axis equation.
$F_N - mg = 0$
=> $F_N = m g$
$=> \mu_s m g = {{m v^2} \over {r}}$
Now we can solve for our coefficient of static friction, $\mu_s$, plug in our data, and do the arithmetic.
$\mu_s = {{m v^2} \over {m g r}}$
=> $\mu_s = {{v^2} \over {g r}}$
=> $\mu_s = {{(15.6 m/s)^2} \over {9.8 m/s^2 \cdot 120 m}}$
=> $\mu_s = 0.207$
This is well within the limit of most passenger car tires whose coefficient of static friction is typically in the 0.5 to 0.7 range.

Uniformly Accelerated Rotational Motion

When we discussed linear motion, we started with looking at uniform accelerated motion and derived four equations of motion. We'll do the same here, but with a couple of additions. We will not only look at the rotational position, velocity, and acceleration, but also their linear counterparts along the circular path.

Displacement and Position

Like with linear motion, there are some fundamental and derived quantities that we can use to describe motion. The two fundamental quantities are of course distance and time. With those two we can derive the quantities of velocity and acceleration. Our definition of distance when we're talking about rotational motion will be a bit different as we'll use the angle through which the object rotates. There are many different ways of measuring angles, though, including number of rotations, number of degrees, or what we'll use by default, number of radians. The radians measure of angle is simply the ratio of the arc length, s, distance along the circular path, and the radius of the circular path, r.

(3)
\begin{align} \theta = {{s} \over {r}} \end{align} Notice here that measuring angle in this way results in a value that has no unit since both the arc length, s, and the radius, r, are measured in units of length. We use the label radians as a reminder of how the angle is being defined, but radians isn't genuinely a unit in the traditional sense.

Constant Velocity Motion

When we look at something spinning with a constant rate of rotation, there are a number of quantities we can use to help describe that motion. First, we can look at the rate of change of the angular position which we'll call the angular velocity, $\omega$. The angular velocity is simply the change in angular position divided by the change in time, measured in units of rad/s.

(4)
\begin{align} \omega = {{ \Delta \theta } \over { \Delta t }} \end{align}

If we incorporate our definition of $\theta$ from earlier, we see there is an interesting relationship between the angular velocity, $\omega$ and the linear velocity around the circular path.

(5)
\begin{eqnarray} \omega &=& {{ \Delta \left( s/r \right) } \over { \Delta t }} \\ \omega &=& {{ \Delta s } \over {r \Delta t }} \\ \omega &=& {{ v_t } \over { r }} \\ v_t &=& r \omega \end{eqnarray}

Even though we're talking about an object moving with a constant angular velocity, keep in mind that its still moving in a circle and therefore experiences a centripetal acceleration. Using our definition of centripetal acceleration from above, and our newly derived way of determining the tangential velocity, $v_t$, we can express the magnitude of the centripetal acceleration in terms of the radius and the angular velocity.

(6)
\begin{align} a_c = r \omega^2 \end{align}

Constant Acceleration

If the angular velocity changes over time, then we can define an angular acceleration as the time rate of change in angular velocity measured in units of $rad / s^2$.

(7)
\begin{align} \alpha = { {\Delta \omega} \over {\Delta t}} \end{align}

Like we did in relating the angular velocity to the tangential velocity, we can also relate the angular acceleration to the tangential acceleration, the acceleration along the circular path.

(8)
\begin{align} a_t = t \alpha \end{align}

Since the tangential acceleration and the centripetal acceleration are perpendicular to each other, we can use the Pythagorean theorem to determine the magnitude of the total acceleration of the object.

(9)
\begin{align} a_{total} = \sqrt{ {a^2_c} + {a^2_t} } \end{align}

Equations of Motion

Looking at our fundamental definitions of angular velocity and angular acceleration, you'll notice that these are defined the same exact way as we defined linear velocity and acceleration. Therefore, we can do the same derivation to arrive at the same four kinematic equations of motion that we had before. We simply swap $x$ for $\theta$, $v$ for $\omega$, and $a$ for $\alpha$.

(10)
\begin{eqnarray} \theta_f &=& \theta_i + \omega_i t + {\scriptstyle \frac{1}{2}} \alpha t^2 \\ \theta_f &=& \theta_i + \left( { {\omega_f + \omega_i } \over {2} } \right) t \\ \omega_f &=& \omega_i + \alpha t \\ \omega_f^2 &=& \omega_i^2 + 2 \alpha \left( {\theta_f - \theta_i} \right) \\ \end{eqnarray}

We can use these kinematic equations of motion in the same way we used their linear counterparts. Let's take a look at some examples.

Example:
A good hard drive drive can spin up the platters from rest to their operating speed of 7200 rpm in 6 ms.
If the radius of the drive is 3.5 inches, what is the angular acceleration of the drive, and through what
angle do the platters turn before being at their final speed?
$\omega_i = 0, \omega_f = 7200 rpm * (1 min / 60 sec) * (2 \pi rad / 1 rev) = 754 rad/s, t = 0.006 s$
$\omega_f = \omega_i + \alpha * t$
$\omega_f = \alpha * t (\omega_i = 0)$
$\alpha = \omega_f / t = { { 754 rad/s } \over {0.006 s} } = 126000 rad/s^2$

Now to find the angle through which the platters rotate.
$\theta_f = \theta_i + \left( { {\omega_f + \omega_i} \over {2} } \right) t$
$\theta_f = 0 + \left( { { 754 rad/s + 0 } \over {2} } \right) 0.006 s = 2.26 rad$

Example:
A merry-go-round, which is initially at rest, experiences an angular acceleration of 0.783 rad/s2. How
many rotations does the merry-go-round move through after 5.7 seconds?
$\theta_f = \theta_i + \omega_i t + {\frac{1}{2}} \alpha t^2$
$\theta_f = {\frac{1}{2}} * 0.783 * 5.7^2 = 12.7 rad$
$\theta_f = 12.7 rad * \left( { { 1 rev } \over { 2 \pi rad } } \right) = 2.02 rev$

page revision: 21, last edited: 14 Nov 2018 19:52