Kyle Nixon - Fall22 Centripetal Force Lab Online EDITED 8.16.22(1)

1 Centripetal Force Lab Online Theory Let’s consider a mass moving in a circle, with constant radius ࠵? with a constant tangential speed |࠵?| . This means that at point ࠵? ! , and at time ࠵? ! , the mass has velocity ࠵? ! , but later on at point ࠵? ! and time ࠵? " , the mass has velocity ࠵? " . Now while the direction of the velocity vector clearly changes as the mass moves from the first point to the second, the magnitude of the velocity vector (the speed) stays the same! |࠵? ! | = |࠵? " | = |࠵?| Even though the magnitudes of the velocity vectors are the same the acceleration they are experiencing can’t be zero because the velocity vectors are changing direction. ∆࠵? = ࠵? " − ࠵? ! ≠ 0 It turns out that ∆࠵? points directly radially inwards to the center of the circle, and that the difference in orientation of the two velocity vectors is identical to ∆࠵? the angular displacement of the mass as it moved from ࠵? ! to ࠵? ". If we consider this motion over an extremely small time interval ∆࠵? = ࠵? " − ࠵? ! → 0 ࠵? , then something interesting happens with the geometry of the situation; that being, we get to similar triangles. Where ∆࠵? is the arc length (linear displacement) traveled by the mass during the time interval ∆࠵?. From the rules of simple geometry we know that the ratios of corresponding sides of similar triangles are equal to each other. Which allows us to write the following equation. ∆࠵? ࠵? = ∆࠵? ࠵?

2 Solving this for ∆࠵? gives us, ∆࠵? = ∆࠵? ∙ ࠵? ࠵? If we now divide both sides by the time interval we get, ∆࠵? ∆࠵? = ∆࠵? ∙ ࠵? ∆࠵? ∙ ࠵? Finally, invoking the definitions of acceleration and velocity we obtain ࠵? $ = ࠵? " ࠵? ࠵? ࠵? is the centripetal acceleration the mass is experiencing as it moves in a circle of radius ࠵?. Centripetal means ‘center seeking’. The centripetal acceleration always points towards the center of the curvature of the path the mass is traveling on. From Newton’s Second Law we know that all forces can be written as ࠵? = ࠵?࠵? . That means that the centripetal force acting on the mass causing it move in a circle can be written as: ࠵? $ = ࠵?࠵? $ = ࠵? ࠵? " ࠵? Just like the centripetal acceleration, the centripetal force always points to the center of the curvature of the circular path the mass is traveling on. Any force can act as a centripetal force, be it gravity, tension, friction or a combination thereof. The summation of the forces on the mass, acting in the radial direction, collectively is the centripetal force causing the circular motion. One method to determine the centripetal force acting on a mass is to utilize free body diagrams and force summation equations.

4 Setup 1. Go to the following website: https://www.walter-fendt.de/html5/phen/looping_en.htm 2. You should now see the following: Procedure 1. In the Green area on the right side of your screen make the following settings. a. Radius, R = 0.500 m b. Initial Height, y 1 = 1.260 m c. Gravitational Acceleration, g = 9.81 m/s 2 d. Mass, m = 1.000 Kg 2. Record these values in the Table in this worksheet. 3. Near the bottom left of the Green area, make sure only the ‘weight, contact force’ is selected. a. Contact force is just another name for the Normal force. 4. Near the bottom of the Yellow area please note that is where the value of the Contact Force can be found, as well as the velocity value. 5. In the Green area, click on the start button to start the experiment. a. Once you click on the start button it will turn into a pause button. 6. Right when the mass reaches the bottom of the loop, click on the pause button so that the arrow representing the Contact force points straight up, or at least nearly straight up. a. You will most likely have to do this multiple times before you get one that is close enough. b. Take a screenshot of your screen with the Contact force arrow pointing straight up, or nearly straight up, and make sure to turn it in with this completed worksheet. c. Record the value of the Contact force, N b , in the Table. d. Record the velocity value, v b , in the Table. e. Clicking on the Pause button will have turned it back into the Start button. 7. Click on the Start button to restart the experiment. 8. Once the mass gets to the very top of the loop, click on the Pause button again to pause the experiment.

5 a. You will most likely have to do this multiple times before you get one that is close enough. b. Take a screenshot of your screen with the mass right at or near the very top of the loop, and make sure to turn it in with this completed worksheet. If is perfectly fine if this screenshot is from a different run than the screenshot with the mass at the bottom of the loop is. c. Record the value of the Contact force, N T , in the Table. d. Record the velocity value, v T , in the Table. e. Clicking on the Pause button will have turned it back into the Start button. f. Once the mass has cleared the loop you can click on Pause or Reset to end the experiment.

7 3. Calculate the percent error between the values of question 1 and question 2, using the value from question 1 as the theoretical value. (5 points) % error = (| exp. – theo.|/theo.) x 100 % error = (|49.442 – 49.382|/49.382) x 100 = 0.122% 4. Draw the free body diagram for the mass at the top of the loop, and the write the force summation equation from it. Then use it to calculate the centripetal force of the mass at the top of the loop. (15 points) Fc = NT - mg à 0.223 + (1.00 x 9.81) = 10.033N 5. Calculate the centripetal force for the mass at the top of the loop using the formula mv T 2 /R (10 points) Fc’ = mvT^2/R = (1.00 x (2.260g)^2)/0.500 = 10.215 6. Calculate the percent error between the values of question 4 and question 5, using the value from question 4 as the theoretical value. (5 points) % error = (|10.215 – 10.033|)/10.033) x100 = 1.814% 7. Turn in the two screenshots. (20 points)

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