# Mechanics 8: Rotational Motion

This chapter, we will go back through the equations of motion that we developed in Chapter 3. Those equations were for linear (straight-line) motion; this time we’ll develop them for rotational (around a circle) motion. The set of equations are very similar — they are related to each other by a factor of \(r,\) the radius of rotation. They also use different symbols for position, velocity and acceleration. It’s not too hard — let’s get started.

### Equations of Motion

In Chapter 3, we showed how the three equations of motion are related: position, velocity, and acceleration. Now we will go through these same quantities for rotational motion.

Before, we called the position variable either \(x\) or \(y,\) depending on the direction. In a rotation sense, position is angle, or how far around the circle has the “wheel” turned? Not surprisingly, this measure of rotational distance (or rotational angle) is noted by the variable theta, \(\theta,\) just like in trig. Since \(\theta\) is an angular measurement, its units are *degrees* or *radians.* Remember from trig that a circle has \(360^\circ\) or \(2\pi\,\)rad (radians).

How can we relate this to the linear position? Remember from geometry that arc length is angle times radius, so we can write

$$\theta r = x$$

or

$$\theta = \frac x r$$

We also need to introduce the rotational version of velocity and acceleration. Rotational velocity is noted by a lower-case omega \(( \omega ),\) and rotational

acceleration uses a lower-case alpha \(( \alpha )\) for its symbol. The “conversion factor” from linear to rotational is to divide each by the radius

of rotation.

Linear | Rotational | Conversion | Unit |
---|---|---|---|

\(x\) | \( \theta \) | \(\theta = \frac x r\) | rad (radians) |

\( v = \frac{dx}{dt} \strut\) | \(\omega = \frac{d\theta}{dt}\) | \(\omega = \frac v r\) | rad/s |

\(a = \frac{dv}{dt}\) | \(\alpha = \frac{d\omega}{dt}\) | \(\alpha = \frac a r\) | rad/s\(^2\) |

So you can think of \(\theta\) (distance) as “how far has the object rotated,” \(\omega\) (velocity) as “how fast is it rotating,” and \(\alpha\) (acceleration) as “at what rate is the velocity changing” (speeding up or slowing down).

So let’s write the equivalent three equations of motion for rotational motion. Assume that we have a constant acceleration \((\alpha).\) As before, we

integrate once to get velocity and a second time to get position. This gives us

$$\alpha = {\rm constant}$$

$$\omega = \alpha t + \omega_0$$

$$\theta = \frac 1 2 \alpha t^2 + \omega_0 t + \theta_0$$

Where the constant terms have been set to the initial velocity, \(\omega(0) = \omega_0,\) and initial position, \(\theta(0) = \theta_0.\) Let’s look at some examples.

#### Examples

- We have a wheel that is initially at rest and at a rotational position of \(90^\circ\) or \(\pi/2\) radians. It is being accelerated at \(\alpha = 2\,\)rad/s\(^2.\) Calculate the (a) rotational velocity and the (b) rotational position as functions of time.
(a) First, we have \(\alpha = 2.\) Since the wheel is stopped, its initial velocity is \(\omega_0 = 0,\) and so the velocity equation is

$$\omega(t) = 2t + 0$$

or just \(\omega(t) = 2t.\)(b) Next, we are told that the initial rotation is \(\theta_0 = \pi/2,\) which gives

$$\theta(t) = \frac 1 2 (2) t^2 + \pi/2$$

which simplifies to \(\theta(t) = t^2 + \pi/2.\)

- We have a wheel on a car that has radius \(r = 0.3\,\)m. It rotates 6 full revolutions. How far (in a linear sense) did it travel?
One revolution is \(2\pi\) radians, so six revolutions is \(12\pi\) radians. Linear displacement is related to rotational displacement by \(\theta = x/r,\) or rearranging gives \(x = r\theta.\) So our distance is \(x = (0.3\,{\rm m})(12\pi)\) or \(x = 11.3\,\)m.

### Kinetic Energy & Moment of Inertia

Energy is work being done. Remember that we talked about one kind of energy,

*kinetic energy.*Recall that for linear motion, kinetic energy is \(K = \frac 1 2 m v^2.\) Then, above we said that \(v = \omega r.\) Substituting, then rearranging, we get

$$K = \frac 1 2 m (\omega r)^2 = \frac 1 2 m r^2 \omega^2$$

Now, we define a new quantity, the

*Moment of Inertia.*It is defined as \(I = m r^2.\) Using this definition, we have the rotational version of kinetic energy,

$$K = \frac 1 2 I \omega^2$$

where \(I\) is the rotational equivalent of mass. Moment of inertia is a calculation of how much mass is rotating, and how it’s distributed — how far each “piece” is from the axis of rotation. So if we have four masses that are all different distances from the axis of rotation, the moment of inertia calculation is

$$I = \sum_{i=1}^4 m_i r_i^2$$

That is, the sum of each mass times it distance-squared (from the axis). This works for individual (point) masses; if we have an object that is bigger than a point mass, we need some calculus (we have to do an integral). That calculation looks like

$$I = \int r^2\, dm$$

So each little mass \(dm\) is some distance \(r\) from the axis, and the integral “adds them up.”

These calculations require a fair bit of calculus and can get tricky, so we’ll leave it at that for now. There are tables that show the moment of inertia for most common shapes, and we can use those for now.

Think of the illustration of a figure skater who starts spinning with her arms outstretched. As she folds her arms, she spins faster and faster. Why? She does not add energy nor accelerate. What’s changing is her *moment of inertia* \((I)\). As her arms fold in, the \(r\) decreases, which makes \(I\) decrease even faster (proportional to \(r^2\)). So if the kinetic energy, \(K,\) stays constant, then if \(I\) decreases, \(\omega,\) her rotational speed has to increase.

#### Example

- Say we have two masses connected by a rod, which rotates about its center (lengthwise). It rotates at a rotational velocity of \(\omega = 7\) rad/s. The masses are both 3 kg, and the rod is 0.5 m long. Neglect the mass of the rod. (a) Calculate the moment of inertia for the rod-with-masses system. (b) Calculate it kinetic energy.
(a) Start with the definition of \(I = \sum_{i=1}^2 m_i r_i^2.\) Each mass is a distance of half the rod from the axis of rotation, and they are both 3 kg. This gives

$$I = m_1 r_1^2 + m_2 r_2^2$$

or

$$I = (3\,{\rm kg})(0.25\,{\rm m})^2 + (3\,{\rm kg})(0.25\,{\rm m})^2 $$

giving \(I = 0.375\,\)kg-m\(^2.\)(b) Next, use the definition of rotational kinetic energy,

$$K = \frac 1 2 I \omega^2$$

so we have

$$K = \frac 1 2 (0.375) (7)^2$$

or \(K = 9.19\,{\rm J}\).

### Torque

Torque is the rotational equivalent of force. It is noted with a lower case tau \((\bf\tau)\) and is noted boldface because it is a vector. It is defined as the

vector cross-product of radius and force, like so

$$\bf\tau = \bf r \bf\times \bf F$$

From a previous chapter, remember that (1) in a vector cross product, the product is perpendicular to each vector in the product, and (2) we can write the magnitude as

\(\tau = rF \sin\phi.\)

The angle \(\phi\) is the angle between the force \((F)\) and the radius \((r)\). Why is there a \(\sin \phi\) term? Think about using a wrench to turn a bolt. The wrench is the radius, and the force applied is \(F.\) Only the part of the force that is perpendicular to the wrench can make it rotate. If the force is completely perpendicular to the wrench, then we have \(\sin (90^\circ) = 1,\) and we can forget about it. If the force is at another angle, then \(\sin\phi\) calculates the part (component) of the force that will cause the wrench to move.

Now, let’s work out a rotational equivalent for Newton’s Second Law, \(F=ma.\) Start with the definition of torque, \(\tau = rF \sin \phi.\) Here, we’re only looking at the part of the force that is perpendicular to \(r\) (the lever arm). So in this case, \(\phi = 90^\circ,\) so \(\sin\phi = 1,\) and we can drop that term. In other words, with a force perpendicular to the radius, \(\tau = r F.\) Substituting for \(F,\) we get

$$\tau = r(ma)$$

Remember above that we said \(a = \alpha r\); substitute for this and we get

$$\tau = r(m \alpha r)$$

rearranging gives

$$\tau = mr^2\alpha$$

substituting for the moment of inertia, \(I = mr^2,\) gives

$$\tau = I\alpha$$

which is the rotational equivalent of Newton’s Second Law.

### Work

Next, we need to work out the rotational equivalent of work. Remember that work is defined as (force) \(\times\) (displacement). So we can start with

$$W = Fx$$

Only the force in the direction of rotation does work, so that component of the force is \(F \sin\phi,\) and from above, \(x = r \theta.\) Substituting, we get

$$W = (F \sin\phi)(r\theta)$$

Rearranging gives

$$W = (Fr \sin \phi) \theta = \tau \theta$$

which is the rotational equivalent of work, and equals (force) \(\times\) (displacement).

#### Example

(a) From the definition of torque,

$$\tau = r F \sin\phi$$

or

$$\tau = (0.25\,{\rm m}) (8\,{\rm N}) \sin(70^\circ)$$

giving a torque of \(\tau = 1.88\,\)N-m.

(b) From the rotational definition of work, \(W = \tau \theta,\) so we get

$$W = (1.88\,{\rm N\cdot m}) \bigg(\frac \pi 2 \bigg) = 2.95\,{\rm J}$$

### Angular Momentum

Our final quantity is the rotational (or “angular”) equivalent to momentum. Linear momentum is a vector quantity, \(\bf p = m \bf v,\) and so is angular momentum. Angular momentum is noted as \(\bf L,\) and like torque, is defined as a vector cross product, like so

$$\bf L = \bf r \bf\times \bf p$$

Again, the scalar magnitude can be written as \(L = p r \sin \phi = m v r \sin \phi.\)

How does this relate to torque? Recall earlier we showed that, instead of \(\bf F = m \bf a,\) we can write \({\bf F} = d{\bf p}/dt.\) If we substitute this into the definition of torque, we get

$${\bf \tau} = {\bf r} {\bf\times} \frac{d\bf p}{dt}$$

Since \(r\) is a constant, it can be moved inside the derivative (interchange the order of operations), and we get

$${\bf \tau} = \frac d {dt} ({\bf r} {\bf\times} {\bf p})$$

which becomes

$${\bf \tau} = \frac {d {\bf L}} {dt}$$

So, just as force equals the time-derivative of momentum, torque equals the time-derivative of angular momentum.

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