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Lab G: The ballistic pendulum and firing the cannon horizontally Please watch the following very short video: https://www.youtube.com/watch?v=h_rnqo7XB2U (Notice, it’s like he’s writing backwards!!) Note: In the above video m1 = mball = 0.0664 kg. The pendulum arm has the following properties: m2 = marm = 0.241 kg. The length of the arm is Larm = 0.29 meters. The lab is NOT a formal report! So NO NEED for abstract, theory, etc sections. There are two parts. In BOTH parts you will need to SHOW the equations, and solve for the initial speed. See next page. This is due by WEDNESDAY of Week 10, so 2/18. You should upload your work to Canvas. A single page is fine – if you need more space that’s fine too. m1 = mball = 0.0664 kg. The pendulum arm has the following properties: m2 = marm = 0.241 kg. The length of the arm is Larm = 0.29 meters. Part A: The Ballistic pendulum. The physics is discussed on pg. 226 Giancoli (and in the video above), but basically there’s three moments in time: 1 – the ball is JUST about to strike the pendulum (just before collision) 2- the ball has JUST finished striking the pendulum (just after collision) 3- The ball-pendulum system reaches its maximum height/angle (momentarily stops, before rotating back down) • Between 1 and 2 we have conservation of momentum. But energy is NOT conserved between 1 and 2 (perfectly Inelastic collision, so they stick together). • Between 2 and 3 we have conservation of total energy E. Lastly, from simple geometry the vertical height h relates to Larm and the max angle ϴ by: h = Larm∙[1-cos(ϴ) ] (Giancoli, pg. 195) THUS from knowing the mass and the max angle ϴ the pendulum achieves, you can work backwards to find the speed that the ball leaves the cannon. Measured value: ϴ = 24° ANALYSIS: Show the equations, solve for the speed v. Report the speed v as vA, since you found it using the results from experiment A. Part B: Firing the cannon horizontally. The cannon was fired HORIZONTALLY from an initial height yo = 1.09 meters. It landed on the ground, at a horizontal distance xf. THUS from just knowing that (a) it was fired horizontally, (b) the initial height yo, and (c) the point where it strikes the ground xf, you can work backwards to find the speed that the ball leaves the cannon. Measured value: xf = 1.5 meters ANALYSIS: Show the equations, solve for the speed v. Report the speed v as vB, since you found it using the results from experiment B. If all went well, the two values should be relatively close, but not exact, as the pendulum’s mass was NOT all localized at the bottom of the arm. Reminder: Give the units of the speed (meters/second) when you report the two values. Lab D – The spring constant “K” of a spring, and simple harmonic motion Abstract The purpose of this assignment is to study the estimate the value of the spring constant “k” of a string. The estimate is obtained after performing and analysing the results of two different experiments, involving said string. Theory and Experimental Setup A spring is an elastic object which can store mechanical energy. If a string is stretched or compressed, it will react to this deformation by applying a force. It can be observed that, under certain conditions, this force is proportional to the displacement of the spring. This behaviour is summed up by Hooke’s Law: 𝐹 = −𝑘𝑥 𝑁 The term 𝑘 in Hooke’s Law is called spring constant and is measured in [𝑚]. When the constant 𝑘 is “large”, the spring is “strong” or “rigid”; when the constant 𝑘 is “small” the spring is “weak” and “easy” to yield. To obtain an estimate of the spring constant 𝑘 it is possible to proceed in two simple, but different ways. The first experiment to measure the spring constant consists of applying a known and constant force to the spring and measuring the displacement of the spring. The constant force is exerted by a known mass subject to the gravitational acceleration. It is easily possible to repeat the experiment and change the applied mass to obtain multiple measurements thus reducing the measurement error after an appropriate analysis of the results. In the second experiment, the spring constant is measured indirectly by measuring the period of oscillation of the harmonic motion produced by letting a mass oscillate up and down. The oscillation period is a function of the mass attached to the spring (the spring mass is not considered) and of the spring coefficient. On the other hand, it can be demonstrated that the period of the oscillation is (with a good approximation) independent on the amplitude of the oscillation, so it is not needed to make the mass oscillate with a known amplitude to determine the spring constant. From Hook’s Law and the definition of velocity and position it can be written: 𝐹 = −𝑘𝑥 = 𝑚𝑎𝑦 = 𝑚 𝑑𝑣𝑦 𝑑2 𝑦 =𝑚 2 𝑑𝑡 𝑑𝑡 The solution to this differential equation is the following: 𝑦 = 𝑦𝑚𝑎𝑥 cos 𝜔𝑡 Where 𝑦𝑚𝑎𝑥 is the maximum distance y from equilibrium (or the amplitude of the oscillations) and 𝜔 is the angular velocity: 𝜔= 2𝜋 𝑇 It can be easily proved that the given sine wave is a solution to the differential equation by substitution. It can also be demonstrated the following: 𝑘 𝜔=√ 𝑚 → 𝑇 = 2𝜋√ 𝑚 𝑘 Where T is the natural period of oscillation. Finally, by rewriting the last equation: 𝑚= 𝑘 2 𝑇 4𝜋 2 This equation will be used to estimate the value of the spring coefficient. Results Analysis of Experiment 1 (a) – (b) Graph plot with regression line and regression equation: Spring Force vs Spring Displacement 1.6 1.4 Force [N] 1.2 F = 3.5905 x – 0.6963 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Position [m] From the linear trendline it is determined that the spring coefficient estimated value is: 𝑘 = 3.59 𝑁 𝑚 (c) – (d) 𝑁 Using the regression tool of Excel it can be obtained the slope standard error: 𝜎 = 0.020 𝑚 The computed 95% confidence interval for the spring coefficient is: 𝑘1 = 3.59 ± 0.047 𝑁 𝑚 → 𝑘 ∈ [3.544; 3.637] 𝑁 𝑚 Analysis of experiment 2 (a) – (b) Graph plot with regression line and regression equation: Mass [kg] Mass vs Squared Period 0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 m = 0.0917 T2 – 0.0545 0 0.2 0.4 0.6 0.8 1 Squared Period 1.2 1.4 1.6 1.8 [s2] The slope of the equation is 0.917. From it the spring coefficient can be calculated using the following formula: 𝑘 = slope ⋅ 4𝜋 2 = 0.917 ⋅ 4𝜋 2 = 3.62 𝑁 𝑚 (c) – (d) The slope standard error is 0.000912. The slope 95% confidence interval is the following: slope = 0.0917 ± 0.0021 → slope ∈ [0.0896; 0.0938] Thus, knowing that the spring coefficient uncertainty is proportional to the slope uncertainty, with a factor equal to 4𝜋 2 , it follows: 𝑘2 = 3.620 ± 0.083 𝑁 𝑚 → k ∈ [3.537; 3.703] 𝑁 𝑚 (e) It is very easy to determine that the value estimated with the first experiment does fall into the uncertainty interval obtained with the second experiment (which has a bigger uncertainty interval). This strongly suggests that the two computed values: 𝑘1 and 𝑘2 do agree with each other. Conclusions The results obtained when estimating the spring coefficient can be considered satisfying. Two completely different methods (based on different physical principles) have been used to estimate the spring coefficient, and the results were consistent with each other. It is interesting to note that the measurement uncertainty obtained with the first experiment is smaller compared to the one obtained with the second experiment. This is probably due to the fact that it is difficult to reach a good trade-off of the number of oscillations to measure and to perform dynamic measurements of time. In fact, using too few periods would greatly increase the measurement errors. On the other hand, it is not recommended to let the oscillations last too much because the period would change due to unmodeled friction.
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