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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

Transcription: Circular Motion

Hello everyone and welcome back to

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

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 mv2/r, so the mass of our student is 65 kg, our velocity or speed is 8.6 m/s2 divided by our radius of 2.5 m.0068

When I put all of that together, we have 65 × 8.62/2.5 to give us a force of about 1922 N or roughly 1.9 × 103 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 mv2/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 = mv2/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 = mv2/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