WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about uniform circular motion.
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Our objectives include calculating the speed of an object traveling in a circular path or portion of the circular path.
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Calculating the period and frequency for objects moving in circles at constant speed.
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Explaining the acceleration of an object moving in a circle at constant speed.
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Solving problems involving calculations of centripetal acceleration.
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Defining centripetal force and recognizing that is not a special kind of force
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but that is provided by forces such as tension, gravity, and friction.
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Solving problems involving calculations of centripetal force.
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Uniform circular motion, objects travel a circular path at constant speed, when they do that we call that uniform circular motion.
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The distance around the circle is its circumference.
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If there is a circle and we call the distance from the center to the edge of the radius, then circumference is 2π r,
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the distance around the outside.
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If instead, you define a diameter through the center point across a circle as your diameter then circumference is just π × D.
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The average speed formula that we talked about from our kinematic section still applies here.
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Average speed is distance traveled / time but for something moving in a circle once around the circle,
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it would be 2 π r the circumference ÷ the time it takes to go once around the circle.
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If it is once around the circle, you can say that is 2 π r ÷ period, where the period is the time for one complete cycle or revolution.
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Frequency is the number of revolutions or cycles which occur each second.
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It gets the symbol f and the units are 1/s, which are also known as a Hertz abbreviated capital Hz.
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Frequency is the number of cycles per seconds or the number of revolutions per second.
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Corresponding to frequency, we have the period that is the time it takes for one complete revolution or cycle.
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Its symbol is T and the units are seconds.
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T period time for one cycle is the time for one revolution.
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Frequency in period are closely related.
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The frequency is 1/ the period and the period is 1/ the frequency.
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1 When you know one, you can easily find the other.
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A couple of examples here starting off by talking about a race car.
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The combined mass of a race car and its driver is 600 kg.
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Traveling at constant speed the car completes one lap around a circular track of radius 160m in 36 s, calculate the speed of the car.
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We will start with their circle.
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A radius of 160 m, average velocity is distance travel ÷ time.
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Its distance traveled is once around so that is a circumference ÷ time or 2 π × the radius 160 m/36 s which is 27.9 m/s .
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A very simple example but probably worth getting a good solid foundation in the basics before we move further.
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Taking a look at a toy train, a 500g toy train completes 10 laps of its circular track in 00:01:40.
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If the diameter of the track is 1m, find the trains period T and its frequency f.
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Period is going to be, it takes it 100s or 1min and 40s to do 10 laps.
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Period being the time for 1 lap or 1 revolution is just going to be 10 s.
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Once we know that, the frequency becomes 1/ the period or 1/10s which could be 0.1 Hz.
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Let us take a look at a more detailed example but still pretty straightforward.
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Allen makes 38 complete revolutions on the playground around about 30s.
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If the radius of the roundabout is 1 m, determine the period of the motion.
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The frequency of the motion, the speed at which Allen revolves, and how sick he is when he is out.
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A, period of the motion, the period it takes him 30s to do 38 revolutions, the period is 0.789s.
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The frequency is just 1/ the period which would be 1/0.789s or 1.27 Hz and the speed at which Allen revolves.
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Average speed is distance travel ÷ time.
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The distance he travels is 2 π × the radius 1m, he does that 38 times.
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To do that 38 times, it takes him 30s so the speed would be 7.96 m/s.
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A little more on uniform circular motion.
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Is an object undergoing uniform circular motion accelerating?
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We draw our circle for something moving at constant speed and is it accelerating?
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It is kind of a trick question.
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If we draw the velocity along the path, notice that it is changing direction depending on where it is at.
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And because velocity is a vector, it has direction and acceleration is change in velocity.
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It has a vector, because you have a change in direction, you have a change in velocity.
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Therefore, yes you are accelerating.
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Yes, it is accelerating due to that change in direction even though the speed is staying the same.
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How do you determine the direction of that centripetal acceleration?
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Here, we have a look at an object moving counterclockwise around the circle.
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At this point, it is instantaneous velocity is up and a little bit later its velocity is in that direction.
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If we want to find its acceleration, that will be the change in velocity ÷ time which will be final velocity - initial velocity ÷ time.
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Velocity final - velocity initial that is the same as final velocity + negative initial velocity.
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Let us see if we can draw it to see the direction of the change in velocity.
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Our final velocity, this vector here in green I’m going to try and reproduce here.
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It is sliding over but something like that.
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If our initial velocity is up, negative initial velocity must be down from that point.
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Something like that, where this is VF this is – VI.
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To add those vectors, we draw a line from the starting point of the first to the ending point of our last.
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That is going to be the direction of our acceleration vector.
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If we look here by these, if we draw that same vector here, we are pointing toward the center of the circle.
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That is why I refer to it as a centripetal acceleration.
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Centripetal acceleration, oftentimes abbreviated ac because centripetal means center seeking.
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It always points towards the center of the circle, even though it is moving in a constant speed the object going
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in uniform circular motion is constantly accelerating towards the center of the circle.
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How do you find the magnitude of acceleration?
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The magnitude is straightforward.
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That is the square of the speed ÷ the radius of the circle, AC =V² /r.
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You got to know that one.
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Let us do another example here.
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Miranda drives a car clockwise around a circular track of radius 30m.
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It is a circle that you get the idea 30m.
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She completes 10 laps around the track in 2 minutes.
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Find Miranda’s total distance traveled in the average speed and centripetal acceleration.
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Distance traveled is going to be the distance around the circle × 10 laps or 2 π × the radius 30m × 10 laps which will be about 1885m.
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To find her average speed, average speed is distance over time.
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That will be 1885m and it took her 2 minutes or 120s to do that, which is 15.7 m/s.
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Centripetal acceleration, that is going to be the square of speed ÷ the radius which will be 15.7m/s² ÷ 30m or 8.22 m/s².
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Let us take a look now at what causes the centripetal acceleration.
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We have acceleration, we must have a force.
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We are going to call that a centripetal force.
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If an object is traveling in a circle, it is accelerating towards the center of the circle, we have to find that.
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For an object to accelerate there must be a net force.
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This is what we call a centripetal force, it is a center seeking force.
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Centripetal points towards the center in order to cause a center seeking acceleration.
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When we talked about net force in the x direction, caused an acceleration
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in the x direction or net force in the y direction cause an acceleration in the y direction,
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we can also note now the net force towards the center of a circle causes an acceleration towards the center of the circle.
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A part to note here, as centripetal force is not a new force.
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It is just a label placed on an existing force when it is directed to the center of circle.
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Tensions can cause a centripetal force, gravity, even friction, can provide us centripetal force.
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A centripetal force is not any magical new force that just comes into existence because something was in a circle.
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Something has a force towards the center of the circle that causes it to move in a circle.
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We will call that forces centripetal force.
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It is just a more generic label for any force going towards the center of the circle.
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As such, you do not want to label anything FC or F centripetal on a free body diagram.
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Be more specific, write down what it is that is causing that force.
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Calculating centripetal force.
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If net force is mass × acceleration, net force for the center of the circle is mass × acceleration towards the center of circle.
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We know that centripetal acceleration is V² /r.
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We can write net force must be equal to MV² /r, that easy.
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If you want a look at units, the units of force are going to be kilogram m² / s² mass × speed² ÷ distance which is going to be kilogram m/s²,
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which is our definition of a Newton, the unit of force.
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Let us take another example.
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If the car is accelerating, is its speed increasing?
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That depends.
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If you think about that, we could have a car traveling with some velocity to the right and it could be accelerating to the right.
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If that is the case, yes its speed is going to increase.
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On the other hand, we can have a car with some velocity to the right and acceleration to the left.
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Is its speed increasing?
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No, it is decreasing.
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Or we could take a look at a car traveling in a circular path with some centripetal acceleration
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that is accelerating but it is moving at constant speed.
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If the car is accelerating is its speed increasing?
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Perhaps, it could be but not necessarily.
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In the diagram below, the car travels clockwise a constant speed in a horizontal circle.
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At the position shown in the diagram which air indicates the direction of the centripetal acceleration of the cart.
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It is got to be A toward the center of the circle.
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Here we have a ball attached to a spring to a string move at constant speed in a horizontal circular path.
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The target is located near the path of the ball is shown.
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At which point along the balls path should the string be released removing the centripetal force if the ball is to hit the target?
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If you want the ball to hit the target, he would release it at the point when it is instantaneous velocity
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is tension to the circle which would be right at B.
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The moment you get rid of that centripetal force, it no longer has a centripetal acceleration.
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Therefore, it travels in the straight line by Newton’s first law.
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Let us take a look at the bucket swung in a horizontal circle.
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The diagram shows a 5 kg bucket of water swung a horizontal circle of radius 0.7m at the constant speed of 2 m/s.
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What is the magnitude of the centripetal force on the bucket of water?
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The centripetal force, the magnitude of the centripetal force will be MV² /r or 5 kg × its speed 2m /s² ÷ 0.7m which is 28.6 N.
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What happens if we do this in a vertical circle?
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The diagram now shows the same bucket, same radius, constant speed 3 m/s,
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find the magnitude of tension of a string at the top of the circle and at the bottom of the circle.
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Let us draw our free body diagram for the top and we will start with that as our analysis.
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We have a tension pulling down, strings can only pull, and we have the weight MG.
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When we write Newton’s second law, I am going to write Newton’s second law in this centripetal direction for the center of the circle.
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Towards the center the circle is down so we have T + MG and that must equal mass × acceleration or MV² /r.
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Which implies then the tension must be MV² /r - MG which will be 5 kg × 3 m/s² ÷ the radius 0.7m - 5 × 10 or T =14.3 N.
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Let us do the same analysis down here when the bucket is at the bottom of the circle.
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At the bottom, our free body diagram looks a little different.
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Our tension is pointing up, gravity still pulls down, and we will write Newton’s second law equation
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f net C = T is pointing towards the center of the circle so that is positive.
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Gravity MG is pointing away from the center of a circle so that is negative.
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T - MG =MV² /r.
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When we solve for tension, T =MV² /r + MG which is going to be 5 × 3² / 0.7 + 5 × 10.
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Therefore, the tension now in the string is 114.3 N.
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Quite a difference.
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Let us take a look at the demon drop problem.
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Put up this after an amusement park ride at an amusement park that used to visit when I was a kid.
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The diagram shows at the top of you the 65 kg student here at point A on an amusement park ride.
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The ride spins a student in a horizontal circle of radius 2 1/2 m at the constant speed of 8.6 m /s.
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While that is happening, the floor is lowered and the student remains against the wall without falling to the floor.
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There is no floor there but the student remains against the wall.
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First off, draw the direction of the centripetal acceleration of the student on the diagram.
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Centripetal acceleration, center seeking, nice easy towards the center of the circle.
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There we go for that part.
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For part 2, determine the centripetal acceleration of the student and the centripetal force acting on the student.
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Centripetal acceleration is V² /r so that is just going to be 8.6 m/s² ÷ our radius is 2.5m which will be 29.6 m/s².
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If we want the centripetal force, the centripetal force is going to be MAC which is 65kg × centripetal acceleration 29.6 m/s² or 1924 N.
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Some force is pushing the student or pulling the student toward the center of the circle with 1924 N of force.
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Number 3, what force keeps the student from sliding to the floor?
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Let us draw a picture.
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Here is our student against the wall, is it spinning in the direction kind of like that.
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If I draw a free body diagram from the side, we have weight down, normal force from the wall, and we must have force of friction.
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What keeps the student from sliding to the floor?
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It must be the force of friction.
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How do you get a formula for force of friction?
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Remember, force of friction is μ × the normal force.
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You must have a big normal force and that is what is causing the centripetal force is the normal force.
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Force of friction keeps a student from sliding.
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What is the minimum coefficient of friction between the student and the wall required to keep the student from sliding down the wall?
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In this case, in order to not slide down the wall, force of friction and the weight of the student have to be balanced.
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Force of friction must equal the weight of the student.
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We have just said force of friction is μ × the normal force.
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μ × the normal force must equal MG which implies then that μ must equal MG / the normal force.
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We also know for all of this to work that the normal force is providing the centripetal force which must be equal to mac.
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We can write that μ =MG / mac which is just G /centripetal acceleration or 10/ centripetal acceleration was 29.6.
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We find that we have a μ of 0.34, that is the coefficient of friction that we need in order to keep the student from sliding down the wall.
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Hopefully, that is a pretty good refresher of fundamentals on uniform circular motion.
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Thanks so much for joining us at www.educator.com and make it a great day everyone.