Part A: Centripetal Force . 445kg Mass of bob: 445, Radius of Revolution: 17 cm Time for 25 Revolution Period: 18.15/ 25= ,726 Revolutions: 18+ 13 5 Computed Centripetal Force: J 2 10 , 726 )2 Hanging Mass: 950 g = 0910 kg Directly Measured Centripetal Force: F, 2 m . g = (0. 950) (9, 83) = 9:31 M Computa Imputed Percent Difference: 15.67- 121 X100% = 64.2% COX (3, 67 ) Part B: Variation of Mass 17 cm Mass of bob: Radius of Revolution: (1 cu = all /m Time for 25 Revolution Period: 20, 36/25 = 0 .814 Revolutions: 20, 36 Computed Centripetal Force: Hanging Mass: Directly Measured Centripetal Force: Percent Difference:Part C: Variation of Radius Mass of bob: Radius of Revolution: Time for 25 22.5 em Revolutions: 13. 52 Revolution Period: Computed Centripetal Force: 12 Hanging Mass: 1330 g 230 Directly Measured Centripetal Force: 132 Percent Difference: D . Part C: Variation of Radius Mass of bob: 4 4 50 = 0. 4.Rg Radius of Revolution: 2. 25 m 22, 5cm = Time for 25 Revolutions: 34. 64 Revolution Period: 34 . 64 = 1. 386 25 Computed Centripetal Force: FC = 4 TT- ME _ 4 TT2 0. 4 45) ( 2. 25 ) ( 1 . 386) 2 Hanging Mass: 3 20 g Directly Measured Centripetal Force: Percent Difference: Part E: Swinging Ball Time for 25 Time per Revolutions: 24. 46 Revolution: Height of 19.5 cm Length of 3. 4 cm Height of String: 43 cm Ball: Hand: % Difference: Experimentally Theoretical using Determined Angle: Angle:Uniform Circular Motion Equipment List: Hand Rotational Apparatus String Weights and Hangers Balls with Holes Meter Stick and Ruler Protractor Various Springs for Rotational Apparatus I. Introduction: Newton's First Law states that an object in motion will continue in uniform motion (no change in speed or direction of motion) unless a force acts on it. Thus, in order for an object to change direction of motion, some force must be acting on it. An object in circular motion is constantly changing the direction of its motion. Therefore, a force must act on it. This force is always acting perpendicular to the motion to change the direction of the m motion. Since the force is perpendicular to the motion, no component of the force is parallel to the velocity, and thus the object does not speed up or slow down: it merely changes direction of motion. Such a force, directed towards the center of the circle through which the object moves, is called a centripetal force. II. Theory The centripetal force of an object moving in circular motion can be shown to be equal to F = mv (1) However, it is often difficult to measure the velocity of a body moving in circular motion. The velocity can then be computed by knowing the distance around the circle (the circumference) and the time that it takes to make a complete circle (the period). The velocity is the circumference divided by the period: 20 r V = T (2) where T is the period of the circular motion and r is the radius of the orbit. Plugging this into the previous equation we have T 4x ' mr (3) F. =In our experiment, we will use a device in which a spring is connected between a mass and an axle that serves as the center ofthe circular motion. The mass is a pendulum hob. As the hob moves around in a circle, the spring is ' pulling inward on the mass. providing the I ' m centripetal force. Now. the interesting thing about a spring is that it pulls when it is stretched. and the more that it is stretched, The more that it pulls. So. you can measure how hard the spring is pulling by stretching it the same amount that it was stretched when the pendulum bob was swinging around. This can be accomplished by hanging a weight that pulls on the spring. stretching it. The spring then pulls on one side ofa static pendulum hob. and the hanging weight pulls on the other side. rom'lterweight Add enough weight and the spring stretches to pull back on the bob. "the weight of the hanging mass then equals the spring force. When the hanging weight stretches the spring the same that it was stretched while swinging. then the spring force, which equals the hanging weight. will also be equal to the centripetal force. This makes for an experimental measurement ot'dle centripetal force. as seen in the gure below. +2. = use {4) (intern-night In the case of the manually rotating apparatus, the force provided by the spring is the centripetal force. But, consider the case of a person swinging a ball on a string, as show in this diagram. Here the tension in the string is not equal to the centripetal force. Rather, only a component of the tension equals the centripetal force. Consider a free body diagram of the ball and string. It the string is at an angle O with respect to the horizontal, then the x and y components of the tension will be given by T = T sine T, = T cost (5) The vertical component of the tension will balance gravity, and the horizontal component will provide the centripetal force: Tsind = mg mvz (6) T cost = Now, divide the top equation by the bottom equation find the expression: T sine mg T cost mv- (7) sin d cost Now, substituting the expression for v given in equation 5.2, we get sind rg T'g cost 4x 21-2 (8) But, looking at the figure, we see that r is not simply the length of the string. Instead r is given by r= Loose (9) So, substituting this into Equation 5.8, we get sin d T'g (10) cost 4x 2 Loose and, then canceling the cosine terms, we are left with sind = Tg (11) 4x ' LIII. Experimental Procedure. Part A: Centripetal Force 1} Set up the Centripetal Force Apparatus as shown by your instructor. Make sure that you are using the stronger spring to begin with. 2} Weigh the hob and record its mass. 3} Practice a bit turning the axle to keep the bob at a constant distance from the axle. A pointer is available to help you keep the bob moving at the right distance. Adjust the pointer to the shortest radius. When the apparatus is working properly, then the spring should be horizontal and the vertical support strings should he vertical. lf they are not, adjust the apparatus accordingly. Adjust the counterweight to make sure that the apparatus is not unstable while spinning. 4} Turn the device so that the bob passes over the pointer. Note where the center of the bob is when swinging, and measure the distance from the center or the bob to the center of le axle. 5} Now, you want to measure the period of die revolutions. To do this, you'll measure how long it takes to turn 25 times, and then divide by 25 to nd the period of a single revolution. 6} Use equation 5.3 to compute the centripetal force. T} Next, you want to directly measure the centripetal force. Hang the mass over the pulley until the hob is directly over the marker and then use equation 5.4 to compute the force of the spring at that elongation. Find the percent difference between the two measures. 8 U Part B: Variation with Mass 9} Now, add 100g to die mass of the bob. 10} Repeat the procedure given in Part A for this heavier mass. Part C: Variation of Radius. 1 1} Remove the 100g mass from the bob. 1?.) Reset the pointer to its more distant position. 13} Again, repeat the procedure of Part A for lis different radial distance. Pay particular attention to step 2, where you make sure that the counterwe i ghts are properly positioned and the support string is vertical. Part D: Variation with Spring Force 14} Replace the spring with a weaker spring. 15} Repeat the procedure again for the new spring. Part E: Swinging a Ball 16} Now, we will investigate the case of a person swinging a ball on a string, as derived in the theory section of the lab writeup. Find a small ball that you can tie onto the end ofa string. Cut a length of string long enough to swing the ball, but not so long as to swing the ball into a computer or your lab partner. 1?} Practice swinging the ball so that it remains at a constant height. This may take some practice. Be careful not to hit your lab partner, another student, the instructor, or any piece of laboratory equipment. You may want to place a marker of some sort at this height to make sure that you are keeping the ball at the right height. Once you are able to reliably be able to swing the ball at a fixed height, then start timing how long it takes to complete 25 revolutions. While timing, if you do not have a marker for the height, then one lab partner must measure the height of the ball. After completing 25 revolutions, stop timing, but do not change the height of the hand that was doing the swinging. Measure how high this hand is. Measure how long the string was while swinging. The measurement should not be the entire length of string, though. Measure from the hand to the ball. Use this information to compute the angle that the string was while swinging. 18) Knowing the period and the length of the string, use Equation 11 to compute the theoretical angle for the string. Find the percent difference between this value and the experimentally determined angle. 19) Answer the postlab questions.PostLab Questions: 1) What are the major sources of error in this experiment? 2) How did your results change in Part B compared with Part A. Interpret this finding. Is this what you expect? Explain. 3) How did your results change in Part C compared with Part A. Interpret this finding. Is this what you expect? Explain. 4) Why did we not measure the mass of the ball for Part E? 5) The Earth has a mass of 5.97x10"*kg and orbits the Sun in one year at an average distance of 1.496x10*km. What is the centripetal force, in Newtons, that keeps the Earth in its orbit around the Sun? 6) Explain your results for Part D.n 2} 3 a. 4} 5} 6} 7} 'What does the term \"period\" mean in circular motion? For a person swinging a ball on a string, is there any speed at which the string would be horizontal? If so, what is that speed? Explain hem.-r hanging a mass on the apparatus is able to directly measure the centripetal force, even if the apparatus is not rotating at that time. How does centripetal force vary with mass'? How does centripetal force vary with radius of revolution? How does centripetal force vary with period of revolutions? In the centripetal force apparatus, why is it important for the spring to be horizontal and the vertical string to be, in fact, vertical