# Introduction to Rotational Variables Uniform Circular Motion, being as useful as it is to create rough models, has its own shortfalls. To plug in the gaps created by it, we'll introduce **Rotational Motion**, an all new, all expenses included chapter (or two, because this is really important regardless of engineering/physics field) which will be sure to bring some tears of joy to you. Let's go over some of the basics and lay some groundwork first. Mind you - this is a student-created mini physics textbook - so this foundation will consist of crushed gravel and rusted rebar at best. In Rotational Motion, we use intrinsic variables called "Rotational Variables". Intuitive, huh? They're listed below for yours and my own convenience. >[!Info]- Angular Units >$s_{\theta}$, Angular Displacement > >$\omega$, Angular Velocity > >$\alpha$, Angular Acceleration With the variables defined we can now go use common kinematics to give further definitions for angular velocity and displacement: >[!Info]- Angular Velocity and Acceleration >$\omega = \frac{d\theta}{dt}$ >$\alpha = \frac{d^2\theta}{dt^2}$ Angular units are given in radians, meaning that we can define one revolution as the value $2\pi$. This means that angular velocity is given by the units $rad/s$ while angular acceleration is given by $rad/s^2$. # Rotational Identities When prompted, physicists usually don't want to measure angular units directly - it typically requires special equipment and is a hassle to explain to non-physics people. As a result, we've sought out ways to convert common variables, like linear velocity, to their angular counterparts. Gyroscope measure manufacturers are reeling! >[!Abstract]- Common Rotational Identities >Using our knowledge of circles we should now that, given an angle in radians: >$s = \theta r$ > >Where $s$ is the arc-length of the circle, which is our linear displacement. > >This means that we can also write the linear velocity as a function of the angular displacement by differentiating both sides by $dt$: >$v = \omega r$ >Remember in Uniform Circular Motion that the period of revolution (the time taken for a single revolution) can be given by: >$T = \frac{2\pi r}{v}$ >All we have to do is substitute $v$ with the variables in equation 2 to get: >$T = \frac{2\pi}{\omega}$ >We can differentiate $v = \omega r$ over $t$ one more time to get an expression for the acceleration: >$a_{t} = \alpha r$ >Note the subscript $t$. This is as we are calculating the *tangential* acceleration of the object - remember in Uniform Circular Motion that an object have both tangential and normal (centripetal) acceleration components! > >To get the centripetal acceleration, use the equation introduced in Uniform Circular Motion: >$a_{n} = \frac{v^2}{r}$ >Substitute the identity from Equation (2): >$a_{n} = \omega ^2 r$ >[!Abstract]- Angular SUVAT Equations >If the angular acceleration of an object is constant, we can revise the SUVAT equations to help us model the motion of the object: >$\omega = \omega_{0} + \alpha t$ >$\Delta s_{\theta} = \omega_{0}t + \frac{1}{2}\alpha t^2$ >$\omega^2 = \omega_{0}^2 + 2\alpha s_{\theta}$ >$\alpha = \frac{\omega - \omega_{0}}{t}$ >Where $\omega_{0}$ is an analog to $u$. # Moments of Inertia %%stick a diagram somewhere - after the first integral!%% During an experiment, a magnetic spinner is used. A chemist might want to measure the energy it may exert on his system to subtract redundancies, meaning that he'll have to find the rotational kinetic energy of the object. As an object rotates, it has kinetic energy. However, if this object is spinning in the same place (like a big merry-go-round), we cannot apply the kinetic energy equation we are familiar with, as the kinetic energy at the centre of mass is going to equal zero. Oh no! Instead, we're going to have to do a little trick - we can treat the object as a collection of particles with different masses. Before we start, let's take a detour to kinetic energy - which we can find as a sum of particles (in the large object): $K = \sum\frac{1}{2}m_{i}v_{i}^2$ To get our moment of inertia, $K = \sum \frac{1}{2}m_{i}(\omega r_{i})^2$ $K = \sum \frac{1}{2}(m_{i}r_{i}^2)\omega_{i}^2$ Therefore, the expression for $I$, the moment of inertia, is: $I = \sum \frac{1}{2}m_{i}r_{i}^2$ The expression for rotational kinetic energy can be shown as: $K = \frac{1}{2}I\omega ^2$ When it comes to dealing with the moment of inertia, we can rewrite how we express it as a sum of all particles as a Riemann sum, or integral: $I=\int r^2 \, dm $ Note that most objects also have their own *translational* kinetic energy - meaning to find the total kinetic energy we must add the original $\frac{1}{2}mv^2$ onto the total: $K_{tot} = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$ >[!Tip]- Moment of Inertia Trends >The further away from the centre of mass an object's mass is concentrated, the higher its moment of inertia. We can see this when we double back to the moment of inertia formula - it scales with the distance $r$ from the centre of mass! > >Just look at the difference between the moment of inertia of a ring and a disc! > >![[Pasted image 20231202191437.png]] > >This "ring" has much of its mass concentrated past the distance $r$. A disc doesn't have this problem - the disc has its mass concentrated closer to its centre of mass. That's why the moment of inertia of a disc is half of a ring's. > >It's a great way to create comparisons of moment of inertias without doing any volume integrals. ## Parallel Axis Theorem When we look at cars, we typically have long, big axles which permit us to attach rings (or tyres, if you're THAT person). These axles rotate when the car moves. Perhaps you'd want to find the moment of inertia of the object wrapped around said axle, perhaps to make sure it won't make the axle snap. The parallel axis theorem defines a cool shortcut we can use to find this. However, this requires that we know the moment of inertia of the axle - or $I_{com}$, as the centre of mass of the system is going to be somewhere within the axle. The theorem is given as: $I = I_{com} + Mh^2$ Where $h$ is the perpendicular distance from the object around said axle to the centre of mass of the axle and $M$ is the mass of the entire body. >[!Abstract]- Proving the Parallel Axis Theorem >Let's imagine that we have an axis and an irregular object wrapped around this disk. We can define coordinate axes so that the origin point of these axes intersect the centre of mass of the axle. We can then imagine an axis perpendicular to the spinning ring and the plane be placed through an arbitrary point $P$ on said ring with coordinates $(a,b)$. > >*Diagram for clarity:* > >![[Pasted image 20231109163747.png]] > >So we can define the radius of the "object" as being a function of the distance between the axle and the point, so we can rewrite the moment of inertia equation like: > >$I = \int [(x-a)^2 + (y-b)^2] \, dm $ >Expand to put like coordinates together: > >$I = \int (x^2+y^2) \, dm - 2a\int x \, dm - 2b\int y \, dm + \int (a^2+b^2) \, dm$ > >Since $(x^2+y^2) = R^2$ is the distance between the axle through the [[Newton's Laws and SUVAT#Centre of Mass|centre of mass]] and the arbitrary object's edge, we'll treat the first integral as denoting the moment of inertia relative to the centre of mass - the $I_{com}$. The integrals in the centre yield us negligible values and the last one denotes the position of the mass element relative to the secondary axis P - so the extra moment of inertia term, $MH^2$. We'll know go over a couple cool examples of rotational inertia in objects. Try not to stress over memorising them --> a "simple" volume integral will yield the same result! >[!Warning]- Moment of Inertia of a Cylinder + Derivations! >Given a cylinder around an axle (which is on an axis itself): > >![[CYLINDER MOI.png]] > >We can define the moment of inertia using the formula: >$I_{cylinder} = \frac{1}{2}MR^2$ >Where $M$ is the mass of the cylinder and $R$ is the radius of the cylinder. > ><h3><u>Derivations!</u></h3> >Let's take a bird's eye view of a cylinder. Since a moment of inertia is an integral over the surface of an object, brute-forcing this shape will require some sort of volume integral. > >However, there's a way to completely circumvent this, keeping us away from the prying eyes of the computer sims. To do this, we're going to use limits and the virtue of concentric rings. > >![[Pasted image 20231109225032.png]] >This cylinder also has a height $h$. To clarify, we're taking the axis the cylinder's aligned with - so no [[Angular Identities - Intro to Rotational Motion (Mechanics)#Extended - The Perpendicular Axis Theorem|perpendicular axis theorem]] just yet! > >The area of each concentric circle is given by: >$dA = \pi(r+dr)^2 - \pi r^2$ >$dA = \pi(r^2+2r)dr + $ >$dA = 2r\pi \ dr$ >The mass element of the object also changes slightly, resulting in a change of $dm$. As $\rho = M/V$ where $\rho$ is the [[Newton's Laws and SUVAT#Density|density]] we can represent this changing mass element as: >$dm = \rho dV$ >We can find $dV$ - it's just the area times the height $h$, as shown: >$dV = \pi h(2rdr)$ >Integrate both sides of the original equation: >$\int \, dm = \int 2\rho\pi hr \, dr$ >$m = 2\rho \pi h\frac{r^2}{2}$ >Remember the general formula for finding the moment of inertia - the one with the integral? Let's use it now that we have our mass $m$! >$I = mr^2 = 2\rho \pi h\frac{r^2}{2} r^2$ >$I = \rho \pi h {r^4}$ >This is a bit of a roadblock we've hit! Time to figure out if any of our variables can be canceled out. Remember - $\rho = m/V$, so we can find the volume using the expression for a cylinder's volume: >$I = 2\frac{m}{\pi r^2h} \pi h r^4$ >Now $h$ and $\pi$ cancels out - great! >$I = \frac{2mr^4}{r^2}$ >$I = \frac{1}{2}mr^2$ >Really fun side project regardless of the object - 10/10 recommend! >[!Warning]- Moment of Inertia of a Sphere >A sphere rotating around an axis has a moment of inertia: > > ![[SPHERE MOI.png]] > >$I_{sphere} = \frac{2}{5}MR^2$ >Where $M$ is the mass of the sphere and $R$ is the radius of the sphere. >[!Warning]- Moment of Inertia of a Rod >The moment of inertia of a rod rotating through (so the axis is pointed vertically) an axis is given by: > >![[ROD MOI.png]] > >$I_{rod} = \frac{1}{2}ML^2$ >Where $M$ is the mass of the rod while $L$ is the length (not cross-sectional area!) of the rod. ## Extended - The Perpendicular Axis Theorem Take a look at the cylinder MOI derivation if you haven't already - there are certain insights we can glean from that one that will really, really help us! For example, look at how we divided the cylinder into rings to mimic an integral. A couple of you may have already jumped up to apply the same logic to the sphere - it's easy to think that we can divide a sphere into equal rings too! However, if you try doing so, you'll have to change the height of the discs - the very axis the sphere is rotating around. ![[Pasted image 20231113211910.png]] Very fun. Since the radius (unless you're taking it from the center) is going to be a three-dimensional vector, all these quantities change when the radius itself changes. Now what? Time to use a bit of a cheat! We'll use the **perpendicular axis theorem**, which is: $I_{z} = I_{x} + I_{y}$ >[!Success]- Proving the Theorem - Perpendicular Axis >For the proof, let's double back to the interpretation of $r$ we have for our sphere. For a perfectly two-dimensional disc, we can get: >$r = \sqrt{ x^2+y^2 }$ >If this sphere rotates around the $z$ axis, we can just use the general formula for moment of inertia around one axis: >$I_{z} = \int r^2 \, dm $ >Now we just have to substitute our expression for $r$! >$I_{z} = \int x^2 \, dm + \int y^2 \, dm $ >Or... >$I_{z} = I_{x} + I_{y}$ >Which proves the perpendicular axis theorem! > >![[Pasted image 20231113213238.png]] Finally, here's me trying to prove the MOI of a sphere, though I admittedly have no idea if it's right or not. >[!Example]- Problem Taster # Torque (Moments) Try opening with your door with a hinge. Easy, right? Now try pushing it open closer to the hinge. Notice that the closer you get to the hinge, the easier it is to push the door. ![[Pasted image 20231109230735.png]] Idealised in this context means the force is perpendicular to the radius $r$. Just simplifies things. In physics, this phenomenon is called **torque**. Although the art of door hinges is one that is slowly being phased out to give way for "smart" homes and electric sensors, it is still the most well-known example of this phenomenon in action, followed closely by old-fashioned sinks. The formula for torque is given by: $\tau = rF\sin \theta$ Where: - $r$ is the distance from the pivot - where the object's rotational velocity is zero when it turns - $F$ is the force exerted on the object The unit of the Torque is the **NewtonMeter** - or Nm. Make yourself understand why! This is the **angular equivalent of force**. As a result, the phrase "**no net force and no net moment**", where moment refers to the turning of the object >[!Success]- More Torque - An Intuitive Dive >Let's make our torque an example - say, a really, really bad door hinge! > >![[Pasted image 20230724103709.png]] > >Like we said before, the torque here is: $\tau = rF$ > >Mind you, this is only for perpendicular forces. Say your > **Equilibrium with Torques** >For a system to be in equilibrium, there must not be any **net torque** - so $\tau_{net} = 0$. As a result, we can make an identity: >$\tau_{left} = \tau_{right}$ >Where $\tau$ left and right denote the torque at opposite sides of the pivot. ## Newton's Second Law for Rotation We can use the knowledge of the moment of inertia to create an expression for Newton's Second Law in Angular Units! $\tau_{net} = I\alpha$ This feels just like a straight transition from the original second law - so why is that? >[!Abstract]- Proof Format 1 - Using Identities! >Remember Newton's Second Law - $F_{t} = ma_{t}$. We need to use the **tangential** components of the object, given that the force exerted is tangential to the circle the rotating object will create. > >Let's set the angle $\sin \theta$ to 90, allowing us to use our simplified equation. This gives us an expression for the net torque: > >$\tau_{net} = ma_{t}r$ > >We can substitute the angular identity $\alpha_t = \alpha r$ to give us: >$\tau_{net} = m(\alpha r)r = mr^2 \alpha$ > >Remember that the moment of inertia I is given by the identity $\int r^2 \, dm$. This is the same as the first section of the equation, $mr^2$. As a result, we have proved the equation! >$\tau_{net} = I\alpha$ #Work Done & Kinetic Energy - Rotational Motion Ver. # Rotational Energies ## Work Done and Power Remember that one of the equations behind [[Power, Work Done and Energy|Work Done]] revolves around integrating the force function across a linear dimension $x$ . This works well when we deal with fluctuating forces: $W = \int_{x_{i}}^{x_{f}} F(t) \, \ dx $ As a result, we can generalise this into its angular form, given that torque is actually our angular equivalent for force: $W = \int_{\theta_{i}}^{\theta_{f}} \tau \, d\theta $ >[!Success]- Proof - Identities Ver. > >Remember the [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)|Work-Kinetic Energy Formula]]. This allows us to denote the work done as a function of the change in kinetic as follows: >$W = \frac{1}{2}I\omega_{f}^2 - \frac{1}{2}I\omega_{i}^2$ >When a particle moves a distance - say, a circular distance $dS$, there is a change in the work done on the object given as $dW$. However, since the arc-length (the linear distance we travel as we rotate) is given by the formula $s = r\theta$, we can replace the distance $S$ by this variable: >$dW = F_{t}r d\theta$ >Mind you, this formula is for the tangential, momentarily translational component of the object. Remember that $Fr$ is equal to the torque $\tau$. This allows us to create an equation: >$dW = \tau d \theta$ >Now we just have to [[Calculus Basics (Maths)|integrate]] both sides (with respect to an arbitrary variable). Remember that we are measuring the angular displacement of the object, allowing us to define lower and upper limits in terms of $\theta$. This proves it! Well done if you got here on your own. >$W = \int_{\theta_{i}}^{\theta_{f}} \tau \, d\theta $ > Remember that we can generalise work done into $W = Fd$ if force is constant. We can therefore treat the distance $d$ as the angular displacement, $s_{\theta}$, as shown: $W = \tau s_{\theta}$ Given that power is just the unit change in work done per unit time $P = \frac{dW}{dt}$ we can rewrite the equation above as shown: $P = \tau \omega$ Likewise, the change in potential energy of an object can therefore be written as: $\Delta U = -\int_{\theta_{i}}^{\theta_{f}} \tau \, d\theta$ ## Rotational Kinetic Energy Now that we've generalised work done into its rotational counterparts, we can continue onwards! To find kinetic energy using angular variables, let's first write down the formula for kinetic energy. This should hopefully be common sense at this point! $E_{k} = \frac{1}{2}mv^2$ Remember the identity $v = \omega r$. We can substitute this into the equation above to get: $E_{k} = \frac{1}{2}m(\omega r)^2 = \frac{1}{2}m\omega ^2r^2$ The general formula for rotational inertia is $I=\int r^2 \, dm$. Notice how this is the same as the component $mr^2$ in the above equation. As a result, we can rewrite the formula to (completely) give it in terms of angular elements! $E_{k} = \frac{1}{2}(mr^2)(\omega^2)$ $E_{k} = \frac{1}{2}I\omega^2$ It's oddly simple and elegant - you can imagine us just replacing all the linear variables with their angular counterparts! # BONUS: The Coriolis Force *An article within an article! How exciting.* Ever notice that hurricanes spin differently depending on the hemisphere you're on, or that a football might leave you in an arc if you're spinning *really* fast when holding it? That's the **Coriolis Force** in action, the fictitious false that arises from a *rotating reference frame!* In other words, it's the 'force' objects on a rotating object experience. Easy, huh? %%COMIC%% ## Proving the Coriolis Force *Take a look at the maths modules before you go here.* First, let's take a look at the velocity mapped to the stationary frame: $v_{s} = v_{r} + (\omega \times r) \tag{1}$ Now we differentiate it - using the operator of $[(\frac{d}{dt})_{r} + (\omega \times \mathrm{transposed \ quantity})]$, where the 'transposed quantity' for this one is formula 1 (no pun intended >:(). By taking the cross product of said quantity, we get to know the magnitude of the fictitious force pushing it 'vertically' relative to the spin of the object - as would a football on one of those carnival rides. So, let's expand our equation by differentiating both sides of (1): $a_{s} = \frac{d}{dt}( v_{r} )+ \frac{d}{dt} (\omega \times r) + (w \times v_{r})$ $a_{s} = a_{r} + (\alpha \times r) + 2(\omega \times v_{r})$ Multiplying this with the mass $m$ (assuming that it's constant) gives us: $F_{eff } = F + m\left( \frac{d\omega}{dt} \times r \right) + 2m(\omega \times v_{r})$ If we think about it, this fictitious force arises due to the inertia of the object within the rotating reference frame - which means we have to minus these two terms! (In other words, the object will move OPPOSITE to the direction of the body's rotation, so we'll have to flip these signs.) $F_{eff} = F - m( \frac{d\omega}{dt} \times r) - 2m(\omega \times v_{r}) $ This $F_{eff}$ term refers to the effective force that actually acts on it, and $F$ is the force that acts on the object in the rotating reference frame (aka the force we give the object). $-m \left( \frac{d\omega}{dt} \times r \right )$ is the *Euler Acceleration*, and is non-zero only when the object is spinning at an increasing/decreasing rate, and $-2m(\omega \times v_r)$ is the **Coriolis Force**. ...and from the cross product alone, can you guess why objects that are 'stationary' in a rotating reference frame tend to move towards the rotational axes? # Problems >[!Example]- Module 10-1 Problem 1 >A good baseball pitcher can throw a baseball toward home plate at 85 km/h with a spin of 30 rev/sec. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 metre path is a straight line. > >**Solution:** >We can first find the time it takes for the baseball to get to the finish line: >$0.06 \div \frac{85\times1000}{3600} \approx 0.00254s$ > >This gives us a value for the time taken for the ball to travel the path. >$30 \times 0.00254 = 0.0762 \ revs$ >[!Example]- Module 10-2 Problem 9 >A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions? > >**Solution:** >a) Given that the flywheel has turned 40 times before coming to a stop, we can find the angular displacement of the object as $40 \times 2\pi = 80\pi$. >This means that the time taken for the object to come to rest is given by: >$\omega^2 = \omega_{0}^2 + 2\alpha 80\pi$ >$0 = 2.25 + 2\alpha 80\pi$ >$0 = 2.25 + 160\pi \left( -\frac{1.5}{t} \right) $ >$-2.25t = -240\pi$ >$t = 335.1s$ > >b) This is just a simple division - divide the angular velocity by the time value we get! >$\alpha = \frac{\Delta \omega}{t} = \frac{-1.5}{335.1}$ >$\alpha = 4.48 \times 10^{-3} rad/s^2$ > >c) This time our displacement is $40\pi$. We can use the equation: >$s_{\theta} = \omega t + \frac{1}{2}\alpha t^2$ >Substituting numbers gives us: >$40\pi = 1.5t + 8.96 \times 10^{-3}t^2$ >$(t + 228.7)(t - 61.317) = 0$ >Take the positive value: >$t = 61.317s$ >[!Example]- Module 10-5 Problem 39 >Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 200$\pi$ rad/s. Suppose that one such flywheel is a solid, uniform cylinder with a mass of 500 kg and a radius of 1.0 m. >(a) What is the kinetic energy of the flywheel after charging? >(b) If the truck uses an average power of 8.0 kW, for how many minutes can it operate between chargings? > >**Solution a)** >Remember that the formula for Rotational Kinetic Energy (since the flywheel doesn't have translational motion) is: >$E_{k} = \frac{1}{2}I\omega^2$ >We can find the moment of inertia of a cylinder with the identity: >$I_{cylinder} = MR^2$ >$I = 500$ >We now have everything we need to substitute and get a value for the flywheel's kinetic energy! >$E_{k} = \frac{1}{2}500 (200\pi)^2$ >$E_{k} = 493480.2201 \approx 493000 J$ > >**Solution b)** > >8 kilowatts translates to 8000 joules of potential-kinetic energy conversions per second. Just divide our kinetic energy value by the power to get our time value in seconds! >$\frac{493000}{8000} = 61.625 \approx 62s \approx 1.03min$ >[!Example]- Module 10-6 Problem 48 >The length of a bicycle pedal arm is 0.152 m, and a downward force of 111 N is applied to the pedal by the rider.What is the magnitude of the torque about the pedal arm’s pivot when the arm is at angle: > >(a) 30$^\circ$ >(b) 90$^\circ$, and >(c) 180$^\circ$ >with the vertical? >[!Example]- Module 10-7 Problem 57 >A pulley, with a rotational inertia of $1.0 \times 10^{-3}$ kgm^2 about its axle and a radius of 10 cm, is acted on by a force applied tangentially at its rim.The force magnitude varies in time as $F = 0.50t + 0.30t2$, with F in newtons and t in seconds.The pulley is initially at rest.At $t = 3.0$ s what are its: >*(a)* angular acceleration and >*(b)* angular speed? > >**Obligatory Diagram:** > >![[Pasted image 20231203162754.png]] > >**Solution:** >*a)* This is pretty easy. Common torque equivalence: >$I\alpha = rF\sin \theta$ >At $t=3$, $F = 4.2N$. Now all we have to do is plug it in - being tangential, the angle is 90. Fun! >$1.0 \times 10^{-3} \alpha = 0.1 \times 4.2$ >$\alpha = 420 rad s^{-2}$ > >*b)* Now comes the hard part. Force can be expressed in terms of momentum, which in turn can be expressed through velocity - one identity away from angular speed. Let's start! >$F = \frac{dp}{dt}$ >Integrate both sides by $dt$ and solve! >$\int F \, dt = \int \, dp$ >$p = [0.25t^2 + 0.1t^3]^3_{0} = 0.25 \times 9 + 0.1 \times 27 = 4.95$ >Divide by the radius and mass to get your answer! Remember - $p = mv$. >$\frac{4.95}{0.1^2} = 495 rad s^{-1}$ # What's Next? For the next instalment in our physics series, go to part 2 of angular mechanics! Have fun... [[Angular Momentum - Rotational Motion Extended (Mechanics)]] To go back to the physics home page, here: [[The (Incomplete) Physics Almanac]]