G8-Lab 7

Lab #9: Centripetal Force - p. 1 Physics 210 – Centripetal Force Objective: In this lab, we will explore whether the centripetal force required to keep an object in uniform circular motion agrees with theoretical predictions. Equipment: · PVC pipe · String 25’ (tan color) · Rubber block · 100g and 200g masses · Paper clip · Spring Scale · Caliper (optional) · Computer Theory: We learned that the size of the centripetal force required to keep an object (mass m) in a circular motion (of radius r) at a constant speed (v) is given by the equation below. (The direction of the force must be towards the center of the circle.) 𝐹 𝑐𝑝 = 𝑚𝑎 𝑐𝑝 = 𝑚 𝑣 2 𝑟 (1) The theoretical proof can be a bit abstract, so we will confirm this experimentally in lab today, using a simple pendulum and a twirling mass. Part I – Simple Pendulum When a pendulum swings, the bob moves in a circular path. This means that the centripetal force is involved. 1) Build a pendulum by attaching a 100 g mass (the bob) at one end of the tan string. Do not cut the string (you will not need to for the entire lab). Pick a length for the pendulum and attach the spring scale there. If you wish to be more precise, you can also attach a caliper, as shown (Figure 2). The center of the circular path will be the top of the spring scale. 2) First measure the resting tension (Figure 1) and record it in the Results Sheet. 3) Pull the bob back to about 30° and let it swing freely (Figure 2). Measure the minimum and maximum values of the tension during the motion and record them in the Results Sheet. Figure 1

Lab #9: Centripetal Force - p. 2 Analysis: 4) Draw the free body diagram of the bob while at rest. Show the weight of the bob, the tension on the bob, and the centripetal force on the bob. What can you say about the speed of the bob in this configuration? The speed is zero. The bob is at rest. 5) Draw the free body diagram of the bob at the bottom of the swing. Show the weight of the bob, the tension on the bob, and the centripetal force on the bob. What can you say about the speed of the bob in this configuration? The motion of the bob is at its maximum speed . 6) Draw the free body diagram of the bob at the top of the swing. Show the weight of the bob, the tension on the bob, and the centripetal force on the bob. What can you say about the speed of the bob in this configuration? The motion of the bob is at its minimum speed . 7) Do the values you found for the resting, minimum and maximum tensions agree with your observations in questions 4, 5 and 6? In the experiment, I found the minimum tension is measured at the top and the maximum tension is measured at the bottom. The value of the resting tension is in the middle. These values agree with my observations in questions 4, 5 and 6. Because centripetal force= mv^2/r, centripetal force increases as speed increases. Figure 2

Lab #6: Centripetal Force – p. 4 in circular motion (equation (1)) 𝐹 𝑐𝑝 = 𝑚𝑎 𝑐𝑝 = 𝑚 𝑣 2 𝑟 (1) where m is the mass of the rubber block. This force is supplied by the tension in the string. Since the hanging mass is not accelerating, the tension is equal to the force of gravity on the hanging mass Mg , where M is the mass of the hanging mass. Therefore, 𝑀𝑔 cos 𝜃 = 𝑚 𝑣 2 𝑟 , (4) where θ is the angle between the horizontal and the string. Since r = L cos θ , where L is the length of the string, you can show that 𝑇 2 = 4𝜋 2 𝑚? ?𝑔 (5) 8) Derive equation 5. Show your work. 9) From equation 5, we can see that T vs. L or T vs. M would not yield a linear graph. Instead we will study T 2 vs. L/M . Find the expected slope and fill out the Results Sheet accordingly. 10) Add the following columns to your spreadsheet: L (m) M (kg) Time for 20 cycles (s) Time per cycle T (s) T 2 (s 2 ) L/M (m/kg) 0.50 0.100 11) Use Excel to plot T 2 versus L/M . Apply the linest function to find the slope and calculate the corresponding 95% confidence interval. Enter the values

Lab #6: Centripetal Force – p. 5 in Results Sheet. 12) Does the theoretical slope fall within range of your 95% confidence interval? theoretical slope = T^2* M/L = 4 π ^2 * mL/(Mg) * M/L = 4 π^2 * m /g = 4 π^2 * 0.0245/9.8 = 0.09870 The most probable slope is 0.1314. And for this graph, the lower and upper bounds of the 95% confidence interval are 0.9574 and 0.9865. lower bound = 0.1314254172 - 0.005188625493 *2 = 0.1210481662 upper bound = 0.1314254172 + 0.005188625493 *2 = 0.1418026682 So the theoretical slope does not fall within range of our 95% confidence interval. Interpret/discuss: 95% confidence interval to T2 versus L/M means that there is a 95% probability that the interval is between 0.1210 and 0.1418, but the theoretical value is 0.09870. So the theoretical value does not fall within range of our 95% confidence interval, accurately, about 0.03 smaller. Results Sheet PART I – SIMPLE PENDULUM: Include units! Resting Tension [Group average] 0.9025 N Minimum Tension [Group average] 0.7575 N Maximum Tension [Group average] 1.15 N PART II – TWIRLING MASS Rubber Block Mass [Group average] ____ 24.5 ___ g Expected Slope (using variables) 0.09870 kg s^2/m