Today we're going to be having a revision lesson in which we're going to go through all of the major concepts in circular motion in Alevel physics. Okay, let's get started and not not lose any time. The first thing that we need to be looking at is uh the radian as a measure of an angle. Now remember if we wanted to convert uh from degrees to radians what we need to do is multiply by pi and divide by 180°. And if we're going the other way around if we're going from radians to degrees, we need to multiply by 180° and we need to divide by pi. a couple of uh useful ones uh just to remember. So one full oscillation, one full circle is 360° which is equal to 2 pi radians. Um additionally 180° is half of that that's pi and uh 90° is half of that that is pi / 2 radians. Okay, next thing uh we need to remember are the time period which is um the time it takes to complete one full circle. Essentially the time it takes to complete one full orbit for an object which is moving in a uh circular motion. The time period is one over the frequency and the frequency is one over the time period. Now, additionally, we can uh find the time period or the velocity of an object that is moving in a circle. And let's say that the radius of this circle is r. Now, if that's the case, the uh velocity is given by the circumference, which is 2 pi r / the time period t. Okay, next one. Omega our angular velocity. Now remember our angular velocity tells us how many uh how what angular displacement we are covering in a given time period. So it's the angular equivalent of uh normal linear velocity. And the equations that we can use to find them are that omega is 2 pi over the time period. And because the time period is 1 over the frequency, we can also say that omega is equal to 2 pi f. Additionally, the linear velocity and angular velocity are related by this equation that v is equal to omega * r. So once again the um angular velocity omega that tells us how much angular displacement let's say delta theta we cover in um in in a second of rotation whereas the linear velocity tells us the tangential velocity at a given point and what that magnitude is. They're related by v is equal to omega r. If we substitute one of those formulas for omega, for example, 2 pi r of t, if our circle is of radius r, we come back to our uh normal equation that the linear velocity is equal to the circumference divided by the time period. We will also need to remember how to convert from RPM, which is revolutions per minute to radians/ second for our units for our angular velocity. The conversion factor if we're going from RPM to radians/s is multiplying by 2 pi and then dividing by 60. For instance, if I had something which was spinning at 200 revolutions per minute. So, RPM. So, uh that means that it does 200 complete cycles in 1 minute. So, the actual distance the actual angular displacement that uh that it's going to cover is going to be 200 full circles which is 200 * 2 pi. And the uh amount of seconds in 1 minute is 60. So because it's per minute be per 60 seconds and uh if we put that into a calculator we get about 20.9 call that 21 radians per second. Okay. So let's have a look at centripedal forces next. Now what is a centripedal force? So that's a net force which causes the object to move in a circular path. And the that force has two really really important features. Number one, it is always always directed towards the center. And number two, it is perpendicular to the linear velocity of that object. A uh common examples of those forces are for example the um uh tension. If you have something which is uh being rotated in a circle, it could also be the gravitational force. It could be the frictional force uh etc. The formula for it is given by mv v² / r. Now um from that equation we can entail that there is a centripedal acceleration which is towards the center of rotation and that central acceleration is given by the formula of v ^ 2 / r. Okay. Next one. Now in circular motion the linear velocity changes. However the linear speed is constant. How can this be? Remember you can uh have acceleration at constant speed because velocity is a vector quantity and only the magnitude i.e. the speed remains the same. However the direction is constantly changing. So the centropedo acceleration changes the direction of motion only. Now this next question is really really really important. That's why I've put this in red. Why does the speed of an object in circular motion not change? Now the reason for that is because the net force is directed perpendicular to the direction of motion i.e. the velocity. Now this means there is no work done by this force and hence there is no change in the speed. Remember work done in general is equal to the force times the displacement times the angle between them. So it's f(x) cos theta and cos of 90° is equal to zero which means that the work done will be zero if the force f and the displacement x are perpendicular to each other. Now because v is equal to omega r we can rewrite the following two equations one for the centripal force and one for the acceleration as the following. F is going to equal M omega R 2 / R which is going to give us M omega^ 2 R 2 / R. We can cancel those out which is going to give us m omega^ 2 * r. Similarly for the acceleration uh this is exactly the same equation just without the mass in front of it. This is going to equal to omega^ 2 * r. Let's have a look at an experiment to try and investigate circular motion. So we have the following setup over here in which we have a bunk which we're going to uh spin that's connected on a piece of string that goes through a glass cylinder and it reaches a mass on the other end. I'm going to try and keep the radius uh of um of the uh of motion constant. Uh we would normally try and do that with a uh with a marker. Uh we could have uh that that we've placed on this end of the string just where it meets the glass cylinder. And we're going to be varying the mass m. There are many different variations of this experiment. So something could come up in a question which for example you are uh keeping the mass constant but you are varying the radius r. So be prepare be prepared to be quite flexible in how we're going to approach this problem. uh we can measure the mass m with a top hand balance and then we could calculate the weight uh which actually provides the centripedal force and um that weight is um providing the centrial force once again which is equal to just mg. We're going to measure the radius R with a ruler. And then we're going to be timing 10 uh at least 10 I would say uh four oscillations, 10 full time periods with a stop clock. And we're doing that to reduce the percentage uncertainty, improve the accuracy in our experiment. We're going to calculate t by dividing by 10. After that, we could calculate our velocity using v is equal to 2 pi r / t. And then we could calculate v^ squ for different values of m. In terms of our analysis, we're going to plot a graph of the force which is just mg uh against v squared. Now if the um if the formula for centripedal force is true, the graph will be a straight line through the origin. Additionally, we could use this graph to determine the mass of the bung should we wish to remember the f is equal to mv^ 2 / r. Now, if we have f on the y ais, we have v ^ 2 on the xaxis, what's left for our gradient and our intercept is zero. So, let's just write that over here. So, f is on the y axis, v ^2 is on the x axis, c is zero. Our gradient is the mass of the bung divided by the radius. So uh we could say that our gradient let's call that grad is equal to m / r. And depending what is known in this question for instance if we know the radius and we wanted to find out the uh mass of the bunk the m mass of the bunk will be equal to our gradient. It's called grad time r. Okay. Now let's have a look at circular motion at an angle. So if we have something like a car which is turning to the left, the normal reaction r is always at 90° to the uh to direct to the angle of the slope. So that means that this angle here is going to be equal to this angle here. This is because of similar triangles by the way. And we can split the normal reaction r into two components r cos theta and r sin theta. Now in this setup, our cos theta is going to be balancing out the weight because that's the vertical component acting upwards. So if the car is not moving up or down, this better be equal to the weight mg. And r sin theta, this component of the normal reaction will actually be providing the centropedal force because that's the net force acting towards the direction of of turning. So r sin theta is going to equal to mv^ 2 / r. Very very similar in a conical pendulum. Uh that is um at an angle theta the vertical component is is the adjacent which is equal to t cos theta and that is equal to mg. Additionally, the horizontal component is the one component which is acting towards the center of rotation that actually provides the centripal force. So, T sin theta is equal to MV² / R. Circular motion at an angle could be represented in many different situations. For instance, you may have the lift force or um or some frictional force which is making something turn. So uh please be aware that tension and the normal reaction are not the only examples of forces in which this type of problem could be applied to. A typical problem would uh ask us to calculate the uh the the speed in this situation. If that's the case, we need to have a look at this system of two equations. Uh the way we would normally do it is just rearrange one of the equations for one of the unknowns. For for instance t um we can say that t is going to equal to mg uh over cos theta. And then what I'm going to do is I'm going to input this equation into the into this expression over here. So rather than t sin theta I'm going to write mg sin theta over cos theta is equal to mv^ 2 / r. Notice that my other unknown which is the mass in this case is going to cancel out and sine over cos is equal to tan. So uh I'm going to get that g tan of the angle is equal to v ^ 2 / r. In other words, v is going to equal to the square root of gr tan theta. Okay. And finally, let's have a look at vertical circular motion. So, this is just an example of some washing which is spinning in the drum of a washing machine. But this could be a bunch of other situations as well. For it could, for example, it could be a mass on a string that is being uh moved in a vertical circle or it could be a roller coaster in a loop on a plane uh doing a vertical loop etc etc. Now um let's imagine two positions one and two. Now notice that in position one mg and the normal reaction R are in the same direction. Position two MG and R are in the opposite directions. Let's apply Newton's second law for each of those directions uh each of those positions. So for position one um m a is going to equal the sum of the two forces. Now they're in the same direction so I'm going to add them. So I'm going to say this is equal to mg plus r. Now remember my um acceleration in this case because we're moving in a circle now is centripedal. So this is going to equal to mv ^2 / r is going to equal to mg + r. Now this means that in position one the normal contact force will be equal to mv ^2 / r minus mg. Now in position two, we're going to follow exactly the same procedure and we're going to write that mv^ 2 / r is going to uh equal to now we're going to have to take them away and will be r minus mg because we can see that they are in the opposite direction. So just rearranging it for r once again we're going to get that r will be equal to mv ^2 / r plus mg. Now notice that R will be the greatest in position two because we are adding those two quantities. Okay folks, now this was most of the A level specification on circular motion. Thank you very much for watching. If you have enjoyed this video, please consider subscribing and please do remember that uh this is insufficient revision and now it will be up to you to do every available pass paper question on circular motion to really make sure that you understand this topic. Thank you very much for watching.