V-axisX-axisAFigure 1. An object (yellow point) travels from point A to point B along the arc of a circle that iscentered 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 forceacting on the object. For uniform circular motion, it turns out that this force must always bepointing towards the center of the circle that the object is travelling around and we say that thereis a centripetal force present.You can use Newton's second law and a little bit of geometry to understand why an objectundergoing uniform circular motion must be experiencing a force that is pointing towards thecenter of the circle the object is travelling on. If the acceleration vector pointed above or below theplane of the circle, the velocity vector would also end up pointing out of the plane that the circleis on. Since the velocity vector always stays in the plane of the circle, it must be that theacceleration is also in the plane of the circle. Next recall that if a force acts along the direction ofvelocity it either increases or decreases the magnitude of velocity. Since the speed is constant inuniform 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 thecenter of the circle or radially out from the circle's center.To sce which direction the acceleration is pointing, consider Figure 1 which shows anobject of mass m, moving along a circle of radius r centered about the red point O. The objectstarts at point A and travels with constant speed v. After a small time interval Ar it reaches pointB 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 themagnitude of velocity is still the same but the direction has changed. By carefully examining thefigure, you can see that the new y component of velocity at point B is vcos(AO) and the new xcomponent is - v sin(A). Note the minus sign. The change in velocity is thusAv = 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:72Sprsng 2020VAsin(A0) = A0-(4)Plugging this into Equation (2) gives:v'AAV - -v(A0)(5)The acceleration is then the change in velocity divided by the change in time:(6)ArUsing 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 itpasses through point A) so the direction of the force is towards O i.e. towards the center of thecircle. If we had chosen any other point on the circle to do this analysis, we would have found thatthe force was also always pointing to the center of the circle. It is important to note that this forcehas to be provided to the object. Otherwise the objecct would not change its direction and wouldcontinue to move along its initial direction. If you take the magnitude of the vectors on both sideof 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, thetension force in the string attached to the carriage (which the force sensor measures) is still the centripetal force.True or False?O TRUEO FALSE

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Answered: Image /qna-images/answer/5f83d054-3b6f-43df-b9c9-08a76e960a28.jpg