A-Level Physics Circular Motion Guide

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 A-level physics. Okay, let's get started and not lose any time. The first thing that we need to be looking at is the radian as a measure of an angle. Now remember if we wanted to convert from degrees to radians, what we need to do is multiply. by pi and divide by 180 degrees and if we're going the other way around if we're going from radians to degrees we need to multiply by 180 degrees and we need to divide by pi a couple of useful ones just to remember so one full oscillation one full circle is 360 degrees which is equal to 2 pi radians. Additionally 180 degrees is half of that, that's pi, and 90 degrees is half of that, that is pi over 2 radians. Okay, next thing we need to remember are the time period which is 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 circular motion. The time period is 1 over the frequency and the frequency is 1 over the time period. Now additionally we can 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 velocity is given by the circumference, which is 2 pi r divided by the time period t. Okay, next one, omega, our angular velocity. Now remember our angular velocity tells us what angular displacement we are covering in a given time period. So it's the angular equivalent of 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, linear velocity and angular velocity. are related by this equation that v is equal to omega times r so once again the angular velocity omega that tells us how much angular displacement let's say delta theta we cover 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 over t, if our circle is of radius r, we come back to our 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 per second for our units for our angular velocity the conversion factor if we're going from rpm to radians per second is multiplying by 2 pi and then divided by 60 for instance if I had something which was spinning at 200 revolutions per minute so rpm so that means that it does 200 complete cycles in one minute so the actual distance the actual angular displacement that that is going to cover is going to be 200 full circles which is 200 times 2 pi and the amount of seconds in one minute is 60 so because it's per minute we per 60 seconds and 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 centripetal forces next now. What is a centripetal force? So that's a net force which causes the object to move in a circular path and 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 common examples of those forces are for example the tension if you have something which is being rotating a circle it could also be the gravitational force it could be the frictional force, etc. The formula for it is given by mv squared divided by r. Now from that equation we can entail that there is a centripetal acceleration which is towards the center of rotation and that centripetal acceleration is given by the formula of v squared divided by r. Hey, next one. Now, in circular motion, the linear velocity changes, however, the linear speed is constant. How can this be? Remember, you can 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 centripetal 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 fx cos theta and cos of 90 degrees 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 centripetal force and one for the acceleration as the following f is going to equal m Omega r squared divided by r, which is going to give us m omega squared r squared divided by r. We can cancel those out, which is going to give us m omega squared times r. Similarly, for the acceleration, this is exactly the same equation, just without the mass in front of it. This is going to equal to omega squared times 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 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 of the motion constant. We would normally try and do that with a marker. We could have... that we have 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 in which, for example, you are keeping the mass constant, but you are varying the radius r. So be prepared to be quite flexible in how we're going to approach this problem. We can measure the mass m with a top-pamp balance, and then we could calculate the weight, which actually provides the centripetal force, and that weight is providing the centripetal 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 at least 10, I would say, full 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 over t and then we could calculate v squared for different values of m. In terms of our analysis we're going to plot a graph of the force which is just mg against v squared. Now If the formula for centripetal 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 that f is equal to mv squared divided by r now if we have f on the y-axis we have v squared on the x-axis 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 squared is on the x-axis c is zero our gradient is the mass of the bung divided by the radius. So we could say that our gradient, let's call that grad, is equal to m over r. And depending what is known in this question, for instance, if we know the radius and we wanted to find out the mass of the bunk, the m mass of the bunk will be equal to our gradient, let's call it grad, times 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 degrees 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, R 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 that better be equal to the weight mg and r sine theta this component of the normal reaction will actually be providing the centripetal force because that's the net force acting towards the direction of turning so r sine theta is going to equal to mv squared divided by r Very very similar in a conical pendulum that is at an angle theta. The vertical component 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 centripetal force so T sine theta. is equal to mv squared over r. Circular motion at an angle could be represented in many different situations. For instance, you may have the lift force or some frictional force which is making something turn. So 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 ask us to calculate the speed in this situation. If that's the case, we need to have a look at this system of two equations. The way we would normally do it is just rearrange one of the equations for one of the unknowns, for instance t. We can say that t is going to equal to mg over cos theta. And then what I'm going to do is I'm going to input this equation into this expression over here. So rather than T sine theta, I'm going to write Mg sine theta over cos theta is equal to Mv squared divided by 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 I'm going to get the G tan. of the angle is equal to v squared divided by 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 example, it could be a mass on a string that is being moved in a vertical circle, or it could be a roller coaster loop on a plane doing a vertical loop, etc, etc. Now, 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. In position two, Mg and R are in the opposite directions. Let's apply Newton's second law for each of those directions, each of those positions. So, for position one, Ma 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 acceleration in this case, because we're moving in a circle now, is centripetal. So this is going to equal to mv squared over r is going to equal to mg plus r. Now this means that in position 1, the normal contact force will be equal to mv squared over r minus mg. Now in position 2, we're going to follow exactly the same procedure, and we're going to write... that mv squared over r is going to equal to now we're going to have to take them away and it 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 squared over 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 this is an insufficient revision and now it will be up to you to do every available past paper question on circular motion to really make sure that you understand this topic. Thank you very much for watching.

Now remember if we wanted to convert from degrees to radians, what we need to do is multiply. by pi and divide by 180 degrees and if we're going the other way around if we're going from radians to degrees we need to multiply by 180 degrees and we need to divide by pi a couple of useful ones just to remember so one full oscillation one full circle is 360 degrees which is equal to 2 pi radians. Additionally 180 degrees is half of that, that's pi, and 90 degrees is half of that, that is pi over 2 radians.

Okay, next thing we need to remember are the time period which is 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 circular motion. The time period is 1 over the frequency and the frequency is 1 over the time period. Now additionally we can 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 velocity is given by the circumference, which is 2 pi r divided by the time period t. Okay, next one, omega, our angular velocity.

Now remember our angular velocity tells us what angular displacement we are covering in a given time period. So it's the angular equivalent of 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, linear velocity and angular velocity.

are related by this equation that v is equal to omega times r so once again the angular velocity omega that tells us how much angular displacement let's say delta theta we cover 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 over t, if our circle is of radius r, we come back to our 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 per second for our units for our angular velocity the conversion factor if we're going from rpm to radians per second is multiplying by 2 pi and then divided by 60 for instance if I had something which was spinning at 200 revolutions per minute so rpm so that means that it does 200 complete cycles in one minute so the actual distance the actual angular displacement that that is going to cover is going to be 200 full circles which is 200 times 2 pi and the amount of seconds in one minute is 60 so because it's per minute we per 60 seconds and 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 centripetal forces next now. What is a centripetal force? So that's a net force which causes the object to move in a circular path and 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 common examples of those forces are for example the tension if you have something which is being rotating a circle it could also be the gravitational force it could be the frictional force, etc.

The formula for it is given by mv squared divided by r. Now from that equation we can entail that there is a centripetal acceleration which is towards the center of rotation and that centripetal acceleration is given by the formula of v squared divided by r. Hey, next one.

Now, in circular motion, the linear velocity changes, however, the linear speed is constant. How can this be? Remember, you can 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 centripetal 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 fx cos theta and cos of 90 degrees 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 centripetal force and one for the acceleration as the following f is going to equal m Omega r squared divided by r, which is going to give us m omega squared r squared divided by r. We can cancel those out, which is going to give us m omega squared times r.

Similarly, for the acceleration, this is exactly the same equation, just without the mass in front of it. This is going to equal to omega squared times 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 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 of the motion constant.

We would normally try and do that with a marker. We could have... that we have 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 in which, for example, you are keeping the mass constant, but you are varying the radius r. So be prepared to be quite flexible in how we're going to approach this problem.

We can measure the mass m with a top-pamp balance, and then we could calculate the weight, which actually provides the centripetal force, and that weight is providing the centripetal 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 at least 10, I would say, full 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 over t and then we could calculate v squared for different values of m. In terms of our analysis we're going to plot a graph of the force which is just mg against v squared.

Now If the formula for centripetal 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 that f is equal to mv squared divided by r now if we have f on the y-axis we have v squared on the x-axis 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 squared is on the x-axis c is zero our gradient is the mass of the bung divided by the radius.

So we could say that our gradient, let's call that grad, is equal to m over r. And depending what is known in this question, for instance, if we know the radius and we wanted to find out the mass of the bunk, the m mass of the bunk will be equal to our gradient, let's call it grad, times 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 degrees 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, R 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 that better be equal to the weight mg and r sine theta this component of the normal reaction will actually be providing the centripetal force because that's the net force acting towards the direction of turning so r sine theta is going to equal to mv squared divided by r Very very similar in a conical pendulum that is at an angle theta.

The vertical component 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 centripetal force so T sine theta. is equal to mv squared over r.

Circular motion at an angle could be represented in many different situations. For instance, you may have the lift force or some frictional force which is making something turn. So 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 ask us to calculate the speed in this situation.

If that's the case, we need to have a look at this system of two equations. The way we would normally do it is just rearrange one of the equations for one of the unknowns, for instance t. We can say that t is going to equal to mg over cos theta.

And then what I'm going to do is I'm going to input this equation into this expression over here. So rather than T sine theta, I'm going to write Mg sine theta over cos theta is equal to Mv squared divided by 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 I'm going to get the G tan. of the angle is equal to v squared divided by 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 example, it could be a mass on a string that is being moved in a vertical circle, or it could be a roller coaster loop on a plane doing a vertical loop, etc, etc. Now, 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. In position two, Mg and R are in the opposite directions. Let's apply Newton's second law for each of those directions, each of those positions.

So, for position one, Ma 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 acceleration in this case, because we're moving in a circle now, is centripetal. So this is going to equal to mv squared over r is going to equal to mg plus r. Now this means that in position 1, the normal contact force will be equal to mv squared over r minus mg.

Now in position 2, we're going to follow exactly the same procedure, and we're going to write... that mv squared over r is going to equal to now we're going to have to take them away and it 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 squared over 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 this is an insufficient revision and now it will be up to you to do every available past paper question on circular motion to really make sure that you understand this topic. Thank you very much for watching.

Transcript for:A-Level Physics Circular Motion Guide

Transcript for:
A-Level Physics Circular Motion Guide