For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

### Circular Motion

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Question 1 0:21
- Question 2 1:01
- Question 3 1:50
- Question 4 2:33
- Question 5 3:10
- Question 6 3:31
- Question 7 3:56
- Question 8 4:33

### AP Physics 1 & 2 Exam Online Course

### Transcription: Circular Motion

*Hello everyone and welcome back to Educator.com. *0000

*In this mini-lesson, we are going to do one page of the worksheet on circular motion from APlusPhysics.com. *0002

*You can find the link to it down below, so take a minute and go through at least the first page and see if you can solve them and then come back and we will check how you did with your problems. *0008

*Number 1 -- The diagram shows the top view of a 65 kg student at Point (A) on an amusement park ride.*0022

*The ride spins the student in a horizontal circle of radius 2.5 m at a constant speed of 8.6 m/s. *0028

*The floor is lowered and the student remains against the wall without falling to the floor. *0036

*Which vector best represents the direction of the centripetal acceleration of the student at point (A)? *0040

*Centripetal acceleration means center-seeking, so it is toward the center of the circle, for any object moving in a circle. *0046

*At point (A), which direction would that be? That is going to be answer Number 1. *0054

*Number 2 -- Same problem, but now we are to find the magnitude of the centripetal force. *0062

*The centripetal force is mv ^{2}/r, so the mass of our student is 65 kg, our velocity or speed is 8.6 m/s^{2} divided by our radius of 2.5 m. *0068

*When I put all of that together, we have 65 × 8.6 ^{2}/2.5 to give us a force of about 1922 N or roughly 1.9 × 10^{3} N, so the answer is Number 2. *0087

*Number 3 -- The magnitude of the centripetal force acting on an object traveling in a horizontal circular path will decrease if what...?*0110

*Let us write our formula for centripetal force, which is mv ^{2}/r. *0118

*If the radius of the path is increased -- well if radius gets bigger, centripetal force will go down, so that should be a correct answer. *0123

*If the mass of the object is increased -- if mass goes up, force goes up -- that cannot be it. *0132

*If the direction of the motion of the object is reversed -- no, that cannot be it and if the speed of the object is increased -- well it is speed squared, so centripetal force is going to go up significantly, so our best answer there must be Number 1. *0137

*Number 4 -- Centripetal force acts on a car going around a curve. *0153

*If the speed of the car were twice as great, the magnitude of the centripetal force necessary to keep the car moving in the path would be...?*0157

*Let us write our equation again, Fc = mv ^{2}/r. *0165

*If the speed of the car were twice as great, the magnitude of the centripetal force would be -- in this case, if we double the velocity because that is squared, we are going to multiply the force by 4 because of that squared relationship. *0169

*Number 5 -- A car travels at constant speed around a section of horizontal circular track. *0190

*On the diagram below, draw an arrow at Point (P) to represent the direction of the centripetal acceleration of the car when it is at point (P). *0195

*Again, centripetal or center-seeking must be toward the center of the circular path or that direction. *0203

*Number 6 -- A child is riding on a merry-go-round. As the speed of the merry-go-round is doubled, the magnitude of the centripetal force acting on the child...? *0211

*Again, Fc = mv ^{2}/r. If we double the speed because it is squared, we are multiplying the entire thing by 4, so we get 4 times or we quadruple the centripetal force. *0220

*Number 7 -- A ball attached to a string is moved at constant speed in a horizontal circular path. *0236

*A target is located near the path of the ball as shown in the diagram.*0242

*At which point along the ball's path, should the string be released if the ball is to hit the target? *0246

*Once that string is released, there is no longer a centripetal force and the ball is going to travel in a straight line, so it looks like you were here at (B), the line that is tangent to the circle that would have the ball at (B) going in a straight line, would hit the target if it were released at (B). *0251

*Our correct answer there is Number 2. *0269

*One last problem -- Which unit is equivalent to meters per second? *0272

*We have meters per second and our choices are hertz times seconds -- well remember a hertz is equal to 1/second, so answer 1, a hertz times second would be 1/second × 1 second, which is 1, so no. *0278

*How about 2, a hertz meter -- that is 1/second times a meter or meters per second -- that must be our correct answer, Number 2. *0292

*That completes page 1 of the worksheet on circular motion. *0303

*If this went swimmingly, went great -- Excellent -- keep on going, but if you had trouble with it, now would be a great time to go review the lesson on circular motion. *0307

*Thanks everyone and make it a great day!*0315

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