# Circular Motion Intro Uniform Circular Motion is used to describe bodies that move in a circle in constant speed. This doesn't happen often in real life, but it's a great introductory prerequisite for rotational motion, where the modelling aspect really opens up. # Centripetal Acceleration Even though the magnitude of a body's speed may be constant, the body still accelerates as the velocity changes direction. As this velocity vector always points outwards tangentially to the circle, there **must** be a changing vector pointing towards the centre of the circle to prevent the object from simply flying away. This is called the **Centripetal Acceleration**. >[!Info] Formula >$a = \frac{v^2}{r}$ >[!Abstract]- Proving the formula >Let the object be in question be considered a variable $p$. >At any point, the coordinates of $p$ can be given as $x_{p}, y_p$ respectively. >We can thus write the velocity vector as a component of the two: >$\overrightarrow{v} = v_{x}\hat{i} + v_{y}\hat{j}$ >We can imagine the circle of motion is a unit circle, allowing us to expand the expression for velocity: > >![[proving UCM.png|400]] > >^[Unit circles are incredibly useful in many more aspects of both maths and physics! To discover more go here - [[The Unit Circle (Maths)]]] >Giving us $\overrightarrow{v} = (-v\sin{\theta})\hat{i} + (v\cos {\theta})\hat{j}$ as the object moves counterclockwise. >We can then define these coordinates in terms of their $p$ component using the unit circle. $\sin{\theta}$ and $\cos{\theta}$ can be replaced by $\frac{vy_{p}}{r}$ and $\frac{vx_{p}}{r}$ respectively, creating: >$\overrightarrow{v} = \frac{vy_{p}}{r}\hat{i} + \frac{vx_{p}}{r}\hat{j}$ >[[Calculus Basics (Maths)#The Derivative|Differentiate]] both sides to get an expression for the acceleration: >$\frac{d\overrightarrow{v}}{dt} = -\frac{v}{r} \frac{dy_{p}}{dt} + \frac{v}{r} \frac{dx{p}}{dt}$ >Notice how $v$ and $r$ remain constant. This is as the object must fulfil the prerequisites of Uniform Circular Motion above, but it's position, given by $x_{p}$ and $y_{p}$ are the same. >Given that $\frac{dy_{p}}{dt}$ is equal to the vertical velocity component $v_y$ and $\frac{dx{p}}{dt}$ is equal to the horizontal velocity component $v_x$, we can rewrite the expression as: >$\overrightarrow{a} = -(\frac{v^2}{r})\cos{\theta} \hat{i} + (\frac{v^2}{r}\sin{\theta}) \hat{j} = a_{x} + a_{y}$ >The magnitude of the centripetal acceleration can therefore be found as: >$ a = \sqrt{(a^2)_{x} + (a^2)_{y}} = \frac{v^2}{r}\sqrt{ \cos {\theta}^2 \sin {\theta}^2 } $ >Hence $a = \frac{v^2}{r}$ given the trig identity $sin {\theta}^2 + cos {\theta}^2 = 1$ Orbits, Cars turning etc can be generalised through the use of Uniform Circular Motion. We can substitute this expression for centripetal acceleration into Newton's Second Law to allow us to find the force on an object whilst it travels in a circle: >[!Formula:] >$F = m\frac{v^2}{r}$ One final thing - let's take a closer look at a body undergoing uniform circular motion. ![[Pasted image 20230930143407.png]] This cow isn't going to want to stay on the path (the circle) - it wants to go run off into the prairie with its brethren, taking it on a forwards route wherever it goes. Therefore, there is a *tensional force* required to keep it in check - given by the arrow below. This exerts a force on the rope - creating our centripetal force (the normal force of it all!). ## Periods of Revolution The time taken for an object to complete one orbit in Uniform Circular motion can also be given as follows: $T = \frac{2\pi r}{v}$ **Where:** - $v$ is the speed of the object - $r$ is the radius of the orbit. Pretty self explanatory! Try to get this down somewhere so you can apply it in [[Rotational Mechanics - Minus the Angular Momentum|rotational motion]]. # Example Problems >[!Example]- Example Problem: Module 6-3 Problem 43 - Unbanked Turns >What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29 km/h and the $\mu_s$ between tires and track is 0.32? > >**Solution:** >Note the force body diagram %%put the actual diagram here%% >When dealing with frictions, we're going to want to just take the normal forces - those being the centripetal force and 'friction'. These are going to want to be equal as *only the frictional force can make the car turn* - why do you think drivers slam the brakes to do this? >$r_{min} = \frac{v^2}{\mu g}$ >However, $v$ is in the units of km/h. We need to convert this to m/s before we can plug our value into our equation: >$29 \times \frac{1000}{3600} = 8.05 m/s$ >$r_{min} = \frac{8.05^2}{2.9}$ >$r_{min} = 20.66m \approx 21m$ >[!Example]- Example Problem: Module 6-3 Problem 42 - Unbanked Turns 2 >Suppose the coefficient of static friction between the road and the tires on a car is 0.60 and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of 30.5 m radius? > >**Solution:** >Let's break down the language first. "On the verge" corresponds to us having to find the maximum static frictional force - which means we can relate the forces as shown: >$\mu_{s}F_{N} = \frac{mv^2}{r}$ >$0.6mg = m\frac{v^2}{r}$ >This is an expression for the total resultant force on the system. "On the verge" means there will not be any change in the object's motion. Let's begin to substitute our values now by cancelling out the $m$: >$5.886 = \frac{v^2}{30.5}$ >This gives us a maximum value of $13.4$ m/s. Pretty fast if you ask me! >[!Example]- Example Problem: Module 6-3 Problem 46 - Banked Turns >A police officer in hot pursuit drives her car through a circular turn of radius 300 m with a constant speed of 80.0 km/h. Her mass is 55.0 kg. What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.) > >**Solution a):** >Ignoring the car, we get a value for $m$ as 55, $v$ as 22.2 after converting to m/s and $r$ as 0.3. We can just substitute these values into our equation for force $F = m \frac{v^2}{r}$ to get: >$F = 55\left( \frac{22.2^2}{0.3} \right) = 90354\dots \approx 90400 N$ >**Solution b)** >%%normal force is going to be on an angle - resolve and solve!%%Now for the tricky part. We can imagine the officer while turning as sitting on a seat at an angle which we will call $\theta$. This can be visualised using a triangle as follows: > >![[UCM problem46.png]] > >Due to the shorter radius of curve the centripetal acceleration on the car is high. This creates an imbalance in the force felt by both sides of the car, leading to the tilt. We can visualise the %%%% >[!Example]- Example Problem: Module 6-3 Problem 48 >A roller-coaster car at an amusement park has a mass of 1200 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 18 m, assume that its speed is not changing. At the top of the hill, what are the >(a) magnitude $F_N$ and >(b) direction (up or down) of the normal force on the car from the track if the car’s speed is v = 11 m/s? >**What are:** >(c) the $F_N$ normal force >(d) the direction if $v = 14 m/s$? > >**Solution:** >*a)* Just plug our values into $F = m \frac{v^2}{r}$! >$F = 1200 \frac{11^2}{18} = 8100 N$ >The gravitational normal incident on the object is given by: >$F_{G} = 1200 \times 9.81 = 11772N$ > >*b)* We can note that centripetal force is actually the normal force of the object. This means the normal force will actually be pointing upwards, as the magnitude of the upwards normal is much greater than the downwards value. > >*c)* Rinse and repeat with different velocity values: >$F = 1200 \frac{14^2}{18} = 13100 N$ > >*d)* Find the weight of the object! This is given by: > >$F_{G} = 1200 \times 9.81 = 11172 N$ >Since the magnitude of the centripetal force is going to exceed this value, the rocket will point downwards, as the centripetal normal force and the weight of the object point downwards! >[!Example]- Example Problem: Module 6-3 Problem 53 >An old streetcar rounds a flat corner of radius 9.1 m, at 16 km/h. What angle with the vertical will be made by the loosely hanging hand straps? > >**Solution:** >Oh boy! This is going to need a force body diagram - so let's get cracking. >![[Pasted image 20230930174806.png]] >Our goal here is to find theta - by any means. Since the tension force is equal to the centripetal force, let's make the identity: >$T\sin \theta = m \frac{v^2}{r}$ >The weight of the object is just given by $mg$. Let's revise our diagram - now including the centripetal normal force: > >![[Pasted image 20230930174959.png]] >This gives us the expression: >$T\cos \theta = mg$ >Now divide the two identities with each other. This should give you: >$\tan \theta = \frac{v^2}{rg}$ >Substitute to get our answer! > >$\theta = \arctan \frac{v^2}{rg} = 12.5 ^\circ$ >[!Example]- Example Problem: Module 4-5 Problem 67 >A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone during the circular motion? > >**Solution:** >We're going to have to combine the concepts of both circular and projectile motion. We can take the position of the apex to be at height $h = 2$. >Remember that we can take the time for the maximum height to occur as a function given below: >$t_{max} = \frac{v\sin\theta}{g}$ >We can therefore use the first projectile motion SUVAT derivation to get a value for the parabolic angle: >$x_{t} = x_{0} + v_{x,0}t$ >Finally, we can find the velocity of the object by >$10 = v(\frac{v\sin\theta}{g})$ >%%find the vertical velocity based on the height it launches from using suvat. this gives you a time frame. since the horizontal velocity in projectile motion is interpreted as constant, we can divide whatever time value we get from the vertical value by the horizontal. after this should be smooth sailing from here - the magnitude of the horizontal (tangential vel) is equal to the velocity within UCM (as the magnitude of the tangential stays the same! the direction does not only) %% # What's next? Naturally, the next step for you would be to go to learn all about Rotational Motion, which you can do here: [[Rotational Mechanics - Minus the Angular Momentum]] You can navigate back to the home page here: [[Mechanics - A Contents Page]] To get back to the main physics page, here: [[The (Incomplete) Physics Almanac]] Happy navigating!