--- ## Uniform circular motion - "Uniform circular motion" is describes the motion of an object that is **traveling in a circle *around* some central point with a ==*constant,*== *non-zero* velocity**. In order to maintain this velocity the object must also experience some constant **non zero acceleration towards the center of revolution since the velocity vector is always changing direction**. *(Recall acceleration is needed to change either the magnitude of a velocity vector OR the direction).* - When an object is in uniform circular motion its velocity vector will **always be tangent to the circle and its acceleration vector will always points towards the middle of the circle.** Since the acceleration vector always points towards the center of revolution, it should follow that the net force on that object must also point in that direction *(if the object is to continue traveling at a uniform speed around the circle)*. - In relation to circular motion, the variable$f$ denotes frequency - or the *number of revolutions per second,* measured in hertz. The variable $T$ is used to denote the period - or the time the object takes to travel around **the circle *once*. - The variable $\omega$ is used to denote the *angular velocity* *( the change in the object's angle over time)* which is discussed more in the *rotational motion* section of these notes. - The variable $r$ is used to denote the radius - or the distance of the object from its **center of revolution**. $\newline$ $\large\text{- Uniform circular motion equations -}$ $\newline$ $\text{\small\color{grey}Frequency and period: }\;\;\;\;\huge\boxed{\normalsize{\color{LightCyan}f =\frac{1}{T}}\;\;\;\;\;\;\; T=\frac{1}{f}}$ $\text{\small\color{grey}Velocity: }\;\;\;\;\huge\boxed{\normalsize\large v=\frac{2\pi\,r}{T}\;\;\;\;\;\;\;{\color{LightCyan}v=2\pi fr}\;\;\;\;\;\;\;{\color{LightCyan}v=\omega r}}$ $\text{\small\color{grey}Acceleration: }\;\;\;\;\huge\boxed{\normalsize\large {\color{LightCyan}a=\frac{v^2}{r}}\;\;\;\;\;\;\;a=(2\pi f)^2r\;\;\;\;\;\;\;{\color{LightCyan}a=\omega^2r}}$ $\newline$ >[!quote] >##### *Circular motion visual* >![[2D motioncircularmotionstuff.png|850]] #### *Forces involved in UCM (uniform circular motion)* - Objects in uniform circular motion must experience a constant **centripetal force** which is always directed towards **the center of revolution**. This force is what causes the the object's inward acceleration, *which ill remind you, is needed to change the **direction of the object's velocity vector** (as it goes around the circle).* - The centripetal force is different from the **centrifugal force**, which is an *apparent force (not a real force)* that appears to **pull** spinning objects away from the center of revolution. This apparent force is due to the object's **inertia** which wants to resist the constant change in direction that is caused by the centripetal force. This causes the object to **feel as if it is being "pressed up" against the boundary of revolution** . Since the centrifugal force is an *apparent force* it should never appear on free body diagrams or be used during calculations. - *Note: the centripetal force should also never appear on FBD as it is always due to other forces like the normal force or friction force.* $\text{Uniform circular motion force equations}$ $\Huge\boxed{\;\color{LightCyan}\large\vec{F}_\textit{net}=m\vec{a}=\frac{mv^2}{r}=m\omega^2 r\;}$ $\newline$ - The apparent weight of object **changes depending on where it is when moving in uniform circular motion.** When an object travels through a valley, or dip, the centripetal acceleration points upwards, in the opposite direction of the weight force. ==When this happens the object's apparent weight **is greater than its true weight** because the normal force of the object is greater than it's weight [[Forces|force]] *(Recall that the "centripetal force" is due to the normal force in cases like these)*. Likewise an object traveling **over a hump** has a **lower apparent weight** because the centripetal force is pointing in the same direction as the object's weight force *(here the weight force IS the centripetal force as well)*== *(meaning the weight force must be greater than the normal force to account for the downwards net force!)*. - Its important to note that the "centripetal force" is not some outside invisible force acting on an object. The centripetal force is - for most objects, the result of a changing normal force. *(the normal force changes due to inertia)*. $\text{\small Apparent weight at bottom of loop: }\;\huge\boxed{\normalsize w_{\text{app}}=n=w+\frac{mv^2}{r}}$ $\text{\small Apparent weight at top of loop: }\;\huge\boxed{\normalsize w_{\text{app}}=n=\frac{mv^2}{r}-w}$ $\newline$ - The **critical speed** of an object undergoing uniform circular motion is the minimum velocity that object needs to complete a full "loop" *(like a person biking through a loop that goes up-side down)*. At this speed the normal force **pushing against the object traveling around the loop is equal to 0**. ==When this velocity is reached, the weight force of the object alone provides just enough centripetal acceleration for the object to complete the up-side down part of the loop without falling off.== *If the velocity was any lower, the normal force would be negative which is impossible.* - The **normal force must be greater than 0 for the object to remain in a circular motion**. $\newline$ $\text{Critical velocity: }\;\;\;v_c=\sqrt{\frac{rw}{m}}=\sqrt{\frac{rwg}{m}}=\sqrt{gr}$ $\newline$ >[!quote] >##### *Circular motion and forces* >![[Forcescircularmotion.png|850]] --- ## Rotational motion - When two forces of differing magnitude and or direction are applied at two different locations **on a single object, that object will experience rotational motion**. The difference between *rotational motion* and *circular motion* is that objects undergoing rotational motion, ==**rotate about a point that lies somewhere *within themselves* whereas objects undergoing circular motion rotate about an *external* point.** This *internal point of rotation* is known as a **hinge or pivot**.== - **Angular velocity is a measure of the rate of change of the angle/orientation** of an object around its point of revolution. Angular velocity differs from regular velocity because it **it is not dependent on the radius of revolution** (*objects that are further from the point of revolution must **have a greater velocity because the circumference of the circle they are traversing increases***). - Angular velocity and acceleration are measured using **radians**. Therefore, they are denoted using the unit/term: *radians per second* $rad/s$ - In the equations below $s$ represents arc length, $f$ denotes the frequency of rotation and $\alpha$ denotes the angular acceleration. $\text{Angular velocity: }\;\huge\boxed{\;\normalsize\large{\color{LightCyan}\omega=\frac{\Delta\theta}{\Delta t}}\;\;\;\;w=2\pi f\;\;\;\;\;\omega=\frac{s}{rt}\;\;\;\;\;{\color{LightCyan}\omega=\frac{v}{r}}\;}$ $\text{Angular acceleration: }\;\huge\boxed{\;\normalsize\large{\color{LightCyan}\alpha=\frac{\Delta\omega}{\Delta t}}=\frac{w_f-w_i}{t_f-t_i}\;\;\;\;\;\alpha=\frac{\tau}{mr^2}\;}$ $\text{Kinematic equations:}\;\;\huge\boxed{\;\normalsize\large{\color{LightCyan}\alpha=\frac{a_t}{r}\;\;\;\;v=\omega r}\;\;\;\;\alpha=\frac{F}{mr}\;}$ $\newline$ - The symbol $\phi$ denotes the **angular displacement** of an object. The angular displacement is the difference between an object's starting and ending angle/orientation. Remember that **counterclockwise displacement is denoted using a positive angle and clockwise displacement is denoted as a negative angle**. >[!tip] >##### *Rolling motion* >- If a perfectly circular object is rolling and not "slipping" we can use the following equation to find the **linear velocity at any point along it's edge** *(where $\small R$) is the radius of the object*. > - **Note that this equation is extremely useful for ==converting linear velocity to angular velocity, so long as you know the radius of revolution==**. > >$\Huge\boxed{\normalsize\large\; v =\omega R\;}$ >$\Huge\boxed{\normalsize\; a_{obj}=\alpha R\;\;\;\;\;\;\;\; v_{obj}=\omega R\;}$ >$\newline$ >- Recall that the velocity of a point on a wheel **changes depending on where the point is**. At the bottom of the wheel for example, the point has an instantaneous velocity *of zero*. The chart below provides information about this. > - To calculate the velocity of a point at the leftmost or rightmost point of a tire, you must **use the Pythagorean theorem to add together the speed of the tire/car as a whole plus the downwards, *y* speed of the point**. If you are instead measuring the velocity of a point that is at some angle in between the axis, you must use trigonometric functions to calculate the x and y velocities independently, making sure to add the vehicle speed to the x velocity. *(Then as always use the Pythagorean theorem to solve.)* > >--- >![[rollling motion.png|850]] #### *Torque* - The ability of a force to cause a rotation depends on three factors: **the magnitude of the force, its distance $r$ from the pivot *(the axis about which the object can rotate)* and the angle at which the force is applied** *(relative to the object)*. The strength of this "rotational force" **is known as torque.** - Recall that a ==**torque causes an angular acceleration.**== - To calculate the torque experienced by an object **we have to first identify the radial line**. This is a line which starts at the object's pivot and passes through the point at which the force is applied. The **magnitude, of the component of this force which is *perpendicular* to the radial line determines how much *torque* that force is applying.** - If there are multiple forces acting on an object the ***net torque* will be the sum of all their torque vectors $F_{\perp}$ combined.** - In the following equation $\tau$ denotes the torque, $r$ denotes the distance the force is from the pivot and $F_{\perp}$ denotes the magnitude of the component of the force which is *perpendicular* to the radial line. - The symbol $\phi$ denotes the angle between the *radial line* and the force vector. - Torque is measured in units of "newton-meter:" $\small N\cdot m$ $\newline$ $\text{Torque: } \;\;\Huge\boxed{\normalsize\color{LightCyan}\large\;\tau=rF_{\perp}=rF\sin(\phi)\;}\;\;{\color{grey}\small\text{and}}\;\;\Huge\boxed{\large\tau=I\alpha}$ $\newline$ - Another way we can think of torque *visually* is by using something known as **the moment arm**. This technique involves drawing an imaginary, infinitely long line **that is perfectly aligned with the force vector**. The **shortest possible line that passes through both the "force vector line" and the pivot point is known as the moment arm** *(this line will always be perpendicular to the force vector line)*. The length of this line, multiplied by the force is equal to the torque done. - The length of the moment arm is denoted as $r_{\perp}$ $\newline$ $\text{Torque equation using moment arm: }\;\;\Huge\boxed{\large\color{LightCyan}\;\tau = r_{\perp}F\;}$ $\newline$ - This equation for torque establishes a very **fundamental relationship between distance and force**. If a force gets closer to an object's pivot **its magnitude must also increase in order for the applied torque to remain constant**. *This is why its much easier to tighten a bolt with a long rod, the further away you are from the pivot/bolt the more torque you can impart on the object!* - This relationship is often referred to as **mechanical advantage or leverage**. >[!quote] >##### *Torque visual* >![[torque image.png|850]] --- ## *Gravitational torque and center of mass* - *"Gravitational torque"* is the term used to describe the **torque force that gravity exerts on object's at their center of mass**. *(An object only experiences a gravitational torque if they can rotate)*. - Objects which are free to rotate will always **come to rest with their center of gravity below the pivot point**, except when their center of gravity is **directly above the pivot**. This is the only location where the gravitational torque will be 0 and the object can *balance* perfectly *(this is what happens when you "balance" and object)* - When ==**the pivot is at the center of gravity the torque due to gravity is 0**==. - When using the center of gravity formula to find the center of gravity between **multiple different points**, evaluate the center of gravity on the x axis first, and then evaluate it on the y axis, *plugging in each mass's corresponding x or y value*. This technique allows you to find the center of gravity on a 2D plane. *You can do the same for x y and z on a 3D plane also)*. - **==The combined center of gravity of two objects always lies on a linear line that connects their individual centers of gravity.==** - To find the center of gravity between two objects *(or more)* we can also create an equation where **the sum of their torques *(due to gravity)* is equal to 0**. We can then solve for the **pivot's position *(also the center of gravity's position)*** by re-arranging the equation in the following manner: $\newline$ $\text{\small\color{grey}Net torque is 0 at center of gravity: }\;\;\tau_{net}=0=\tau_1+\tau_2=(x_{cg}-x_1)m_1g-(x_2-x_{cg})m_2g$ $\newline$ $\therefore\text{Center of gravity: }\;\Huge\boxed{\large \,x_{cg} =\frac{x_1m_1+x_2m_2}{m_1+m_2}}$ $\text{Gravitational torque: }\;\Huge\boxed{\large\;\color{LightCyan}\tau_g=mg\,r\;}$ $\newline$ >[!check] >##### *Where does the weight force act?* >- Always remember the **weight force ALWAYS acts on an object AT its center of gravity.** --- ## *Moment of inertia* - The moment of inertia of an object is a quantity that represents **the "rotational inertia" of that object**. In other words - how much it resists being spun up, and spun down. The moment of inertia depends on the **distribution of the object's mass.** Objects that have their mass concentrated far away from their pivot are harder to rotate then objects that have their mass concentrated close to their pivot. *This is why pulling your arms in while spinning causes you to spin faster.* - An object's moment of inertia **depends on its mass and *how* that mass is distributed.** - The moment of inertia is denoted using the variable: $I$. The unit used for **the moment of inertia is kg per m squared $\small kg/m^2$. - Calculating the moment of inertia is difficult since its hard to know/calculate *how* an object's mass is distributed. Therefore we often generalize the object as **a primitive shape with evenly distributed mass** which allows us to use simple, pre-determined equations. >[!quote] >##### *Moment of inertia equations* >![[Rotational and Circular motion.png|850]] - Since the **moment of inertia** is the rotational motion's version of mass, and **torque** is its version of force we can use those two variables to re-write **Newton's third law for rotation**: $\newline$ $\large\text{Newton's third law for rotation: }\;\;\Huge\boxed{\large\color{LightCyan} \tau=I\alpha}\;\;\color{grey}\longrightarrow\normalsize\;\;F_{\perp}=\frac{I\alpha}{r}$ $\newline$ --- ## Non uniform circular motion - If an object is revolving around some point with a **non-constant velocity that object is said to be moving with non uniform circular motion**. The ==**tangential acceleration of the object represents the rate of change of its velocity** *around the circle*.== This is different from the *centripetal acceleration* which is directed towards the center of the circle and is representative of the object's changing direction. - The centripetal and tangential acceleration are directly proportional. If the tangential acceleration of an object increases the tangential velocity will also increase *(the object starts going faster around the circle)* which causes the centripetal force and thus, acceleration to also increase. $\text{Tangential acceleration: }\;\;\;\large a_t=\frac{\Delta v}{\Delta t}=\frac{\Delta(\omega \,r)}{\Delta t}=\Huge\boxed{\;\color{lightcyan}\large\alpha\,r=a_t\;}\;\;\color{grey}\small\therefore\;\text{recall: }\alpha =\frac{F}{mr}$ $\text{Using Newton's third law: }\;\Huge\boxed{\color{lightcyan}\large\;a_t=\frac{F}{m}\;}$ $\newline$ >[!bug] >##### *You cant use tangential acceleration in kinematic equations!* >- If you want to find the amount of distance that say *a point on a circle has traveled if that circle is accelerating and rotating* you must find the arc length it has traveled. >- To do this first you must find the total angle that the point/object has traveled around the center of revolution. You can do this by using **angular velocity and angular acceleration in the regular displacement kinematics equation:** > >$\text{Angle displacement equation: }\;\large\Delta\theta=\theta_i+\omega\,t+\frac{1}{2}\omega\, t^2$ >- After you have found the total number of angles traversed by the point or object *(also known as the angle displacement)* using the above equation we can **plug that and the radius into the *arclength* equation to find the total distance traversed by the point/object**. > - *Note: this arclength equation only works with radians which shouldn't be a problem since $\small\omega$ is always measured in radians.* > >$\text{Length traveled: }\;\large l=\theta\,r$ >[!bug] >##### *How to determine if a rotating object is speeding up or slowing down:* >- If the angular acceleration (the acceleration directed towards the center of rotation) and **tangential acceleration** have the same sign the object will accelerate. *This means that the angular acceleration points towards the center of rotation and the tangential acceleration is directed counter clockwise.* >- If the angular acceleration and angular velocity have different signs the object will decelerate. --- ## Static equilibrium and rotation - In order for an object **==to be in static equilibrium the net force AND the net torque must both be equal to zero==**. - When an object is in static equilibrium the net torque **at all locations is equal to zero no matter ==where the pivot is located==**. - As a result if you are trying to determine weather or not an object is in static equilibrium its often advantageous to place the pivot at the source of the **most complex or unclear force**. Choosing to place this imaginary pivot at that location makes it easier to determine if the object is in rotational equilibrium because you wont have to consider whatever complex/unclear force acts at that position. *Recall that if a force is acting AT a pivot point its distance $r$ is zero and therefore it has no effect on torque.* $\newline$ $\Huge\boxed{\large\text{When in static equilibrium: }\tau_{net}=F_{net}=0}$ $\newline$ --- ## Calculator <iframe src="https://amesweb.info/Physics/Circular-Motion-Calculator.aspx"width="900"height="3000"></iframe>