WEBVTT physics/ap-physics-1-2/fullerton
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Hello everyone and welcome back to Educator.com.
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In this mini-lesson, we are going to do one page of the worksheet on circular motion from APlusPhysics.com.
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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.
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Number 1 -- The diagram shows the top view of a 65 kg student at Point (A) on an amusement park ride.
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The ride spins the student in a horizontal circle of radius 2.5 m at a constant speed of 8.6 m/s.
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The floor is lowered and the student remains against the wall without falling to the floor.
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Which vector best represents the direction of the centripetal acceleration of the student at point (A)?
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Centripetal acceleration means center-seeking, so it is toward the center of the circle, for any object moving in a circle.
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At point (A), which direction would that be? That is going to be answer Number 1.
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Number 2 -- Same problem, but now we are to find the magnitude of the centripetal force.
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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.
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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.
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Number 3 -- The magnitude of the centripetal force acting on an object traveling in a horizontal circular path will decrease if what...?
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Let us write our formula for centripetal force, which is mv²/r.
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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.
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If the mass of the object is increased -- if mass goes up, force goes up -- that cannot be it.
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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.
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Number 4 -- Centripetal force acts on a car going around a curve.
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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...?
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Let us write our equation again, Fc = mv²/r.
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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.
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Number 5 -- A car travels at constant speed around a section of horizontal circular track.
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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).
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Again, centripetal or center-seeking must be toward the center of the circular path or that direction.
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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...?
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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.
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Number 7 -- A ball attached to a string is moved at constant speed in a horizontal circular path.
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A target is located near the path of the ball as shown in the diagram.
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At which point along the ball's path, should the string be released if the ball is to hit the target?
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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).
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Our correct answer there is Number 2.
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One last problem -- Which unit is equivalent to meters per second?
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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.
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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.
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That completes page 1 of the worksheet on circular motion.
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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.
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Thanks everyone and make it a great day!