College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
V-axis X-axis A Figure 1. An object (yellow point) travels from point A to point B along the arc of a circle that is centered at point O (red point). velocity of an object, either due to a change in speed or direction, means the object is accelerating. According to Newton's second law, an object will accelerate only if there is a non-zero total force acting on the object. For uniform circular motion, it turns out that this force must always be pointing towards the center of the circle that the object is travelling around and we say that there is a centripetal force present. You can use Newton's second law and a little bit of geometry to understand why an object undergoing uniform circular motion must be experiencing a force that is pointing towards the center of the circle the object is travelling on. If the acceleration vector pointed above or below the plane of the circle, the velocity vector would also end up pointing out of the plane that the circle is on. Since the velocity vector always stays in the plane of the circle, it must be that the acceleration is also in the plane of the circle. Next recall that if a force acts along the direction of velocity it either increases or decreases the magnitude of velocity. Since the speed is constant in uniform circular motion, the force must be pointing perpendicular to the object's velocity vector. This only leaves two possible directions for the acceleration vector, either pointing towards the center of the circle or radially out from the circle's center. To sce which direction the acceleration is pointing, consider Figure 1 which shows an object of mass m, moving along a circle of radius r centered about the red point O. The object starts at point A and travels with constant speed v. After a small time interval Ar it reaches point B covering an arç of length As = vAt. The small angle covered in this time interval is: As (1) At point A, the velocity points in the positive y direction, as shown. At point B the magnitude of velocity is still the same but the direction has changed. By carefully examining the figure, you can see that the new y component of velocity at point B is vcos(AO) and the new x component is - v sin(A). Note the minus sign. The change in velocity is thus Av = v,-v, =-v(l- cos(A0))y-vsin(AO)i (2) For very small time intervals, the angle A0 will be very small and you can write: cos(A6) = 1 (3) and: 72 Sprsng 2020 VA sin(A0) = A0- (4) Plugging this into Equation (2) gives: v'A AV - -v(A0) (5) The acceleration is then the change in velocity divided by the change in time: (6) Ar Using Newton's Law F=ma we then find: mv (7) Note that this is the force on the object when it is at point A (or an arbitrarily small time after it passes through point A) so the direction of the force is towards O i.e. towards the center of the circle. If we had chosen any other point on the circle to do this analysis, we would have found that the force was also always pointing to the center of the circle. It is important to note that this force has to be provided to the object. Otherwise the objecct would not change its direction and would continue to move along its initial direction. If you take the magnitude of the vectors on both side of Eguation onc finds.that the masnitude of the foree on the ohiect is:
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Transcribed Image Text:V-axis X-axis A Figure 1. An object (yellow point) travels from point A to point B along the arc of a circle that is centered at point O (red point). velocity of an object, either due to a change in speed or direction, means the object is accelerating. According to Newton's second law, an object will accelerate only if there is a non-zero total force acting on the object. For uniform circular motion, it turns out that this force must always be pointing towards the center of the circle that the object is travelling around and we say that there is a centripetal force present. You can use Newton's second law and a little bit of geometry to understand why an object undergoing uniform circular motion must be experiencing a force that is pointing towards the center of the circle the object is travelling on. If the acceleration vector pointed above or below the plane of the circle, the velocity vector would also end up pointing out of the plane that the circle is on. Since the velocity vector always stays in the plane of the circle, it must be that the acceleration is also in the plane of the circle. Next recall that if a force acts along the direction of velocity it either increases or decreases the magnitude of velocity. Since the speed is constant in uniform circular motion, the force must be pointing perpendicular to the object's velocity vector. This only leaves two possible directions for the acceleration vector, either pointing towards the center of the circle or radially out from the circle's center. To sce which direction the acceleration is pointing, consider Figure 1 which shows an object of mass m, moving along a circle of radius r centered about the red point O. The object starts at point A and travels with constant speed v. After a small time interval Ar it reaches point B covering an arç of length As = vAt. The small angle covered in this time interval is: As (1) At point A, the velocity points in the positive y direction, as shown. At point B the magnitude of velocity is still the same but the direction has changed. By carefully examining the figure, you can see that the new y component of velocity at point B is vcos(AO) and the new x component is - v sin(A). Note the minus sign. The change in velocity is thus Av = v,-v, =-v(l- cos(A0))y-vsin(AO)i (2) For very small time intervals, the angle A0 will be very small and you can write: cos(A6) = 1 (3) and: 72 Sprsng 2020 VA sin(A0) = A0- (4) Plugging this into Equation (2) gives: v'A AV - -v(A0) (5) The acceleration is then the change in velocity divided by the change in time: (6) Ar Using Newton's Law F=ma we then find: mv (7) Note that this is the force on the object when it is at point A (or an arbitrarily small time after it passes through point A) so the direction of the force is towards O i.e. towards the center of the circle. If we had chosen any other point on the circle to do this analysis, we would have found that the force was also always pointing to the center of the circle. It is important to note that this force has to be provided to the object. Otherwise the objecct would not change its direction and would continue to move along its initial direction. If you take the magnitude of the vectors on both side of Eguation onc finds.that the masnitude of the foree on the ohiect is:
This prelab question is about the experimental setup shown in the Lab Manual as Figure 1 for Experiment 7. Even though the carriage is slowing down, the tension force in the string attached to the carriage (which the force sensor measures) is still the centripetal force. True or False? O TRUE O FALSE
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Transcribed Image Text:This prelab question is about the experimental setup shown in the Lab Manual as Figure 1 for Experiment 7. Even though the carriage is slowing down, the tension force in the string attached to the carriage (which the force sensor measures) is still the centripetal force. True or False? O TRUE O FALSE
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