Lab 5: Conservation of Energy


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1 Lab 5: Conservation of Energy Equipment SWS, 1meter stick, 2meter stick, heavy duty bench clamp, 90cm rod, 40cm rod, 2 double clamps, brass spring, 100g mass, 500g mass with 5cm cardboard square taped to bottom, motion sensor II, photogate sensor, rubber tube (diameter 2.5 cm), paper tube (diameter 2.5 cm), bubble wrap, 30cmlong string, vernier calipers. 1 Purpose The purpose of this experiment is to explore conservation of energy in three mechanical systems: a falling mass, a pendulum, and a mass oscillating on a spring. 2 Theory If a mass m moves from its initial position r i to its final position r f under the influence of a force F, the work done by the force is given by ( ) ( ) 1 1 F dr = r i 2 mv2 f 2 mv2. (1) i Equation 1 is obtained using Newton s 2nd law to replace F by ma and then integrating over the path (i.e. trajectory) of the mass. Equation 1 is a completely general result, which generally depends on the path taken by the particle. However, if the result of doing that integral is independent of the path taken by the particle, the force F is said to be a conservative force. Examples of conservative forces include the gravitational force, the ideal spring force, and the electrical force. The quantity appearing in each of the terms on the right hand side of Eq. 1 is called the kinetic energy K = 1 2 mv2. (2) As noted above, the integral in Eq. 1 is independent of path for a conservative force, so in that case we can break it up into the sum of two integrals, one that goes from the starting point r i to some arbitrary reference point, and another that goes from to r f : r i F dr = = r0 r i F dr + F dr F dr (3) ri F dr, (4) 1
2 where we have reversed the starting and ending points of the first integral in Eq. 3, which reverses its sign, to obtain Eq. 4. From Eqs. 1 and 4 we obtain ( ) 1 ri ( ) 1 rf 2 mv2 F dr = 2 mv2 F dr. (5) i The fact that Eq. 5 holds for any choice of points r i and r f means that the the quantity on each side of the equality is independent of the choice of r i and r f. That is, it is a constant for any choice of position r. We call that constant the total energy E: E 1 2 mv2 r f F dr, (6) where the velocity and position correspond to the state of the mass at any given time. We have already designated the first term on the right hand side of Eq. 6 as the kinetic energy; the second term defines the potential energy U: U(r) r F dr, (7) which is a function only of r (and the choice of the reference position ), since we have assumed that F is a conservative force. Thus, we may rewrite Eq. 6 as E = K + U, (8) which expresses the principle of the conservation of energy for conservative forces. In the next two sections, we find explicit expressions for the potential energy of a linear spring and for uniform gravity. 2.1 Linear spring Let s consider an ideal spring that obeys the force law F = ky, where y is the distance the spring is stretched along its length. According to Eq. 7, the spring s potential energy is given by: U s (y) = F (y ) dy = y 0 ky dy y 0 (9) = 1 2 k ( y 2 y 2 0). (10) If we take the reference potential energy to be zero at y 0 = 0, then U(y) = 1 2 ky Uniform gravitational field We now consider a mass m in a uniform gravitational field, whose force is given by F = mg, where the upward direction is taken to be positive. According to Eq. 7, the potential energy of the m in the gravitational field is given by: U g (y) = F (y ) dy = mg dy (11) y 0 y 0 = mg (y y 0 ). (12) 2
3 If we take the reference potential energy to be zero at y 0 = 0, then U(y) = mgy. Note that the gravitational potential energy depends on only on the vertical distance y the mass moves and not on the horizontal distance. Why? 3 Experiment You will investigate the conservation of energy for three different cases: free fall in a uniform gravitational field, a pendulum, and a linear spring. 3.1 Free fall of a mass When an object falls in a uniform gravitational field, it s total energy should be conserved, provided all other forces are small compared to gravity. In this experiment you investigate the free fall of a small rubber tube and of a paper tube of roughly the same size Description A tube of mass m is held horizontally a distance h above the level beam of a photogate sensor. The tube is dropped so that it falls without rotating and its velocity is measured as it passes through the photogate. The total energy at the time it is dropped is compared to the total energy when the tube passes through the photogate Programming Measure the diameter of the tube with the vernier calipers provided, checking for uniformity. If the diameter of the tube is not uniform, discuss this in your error analysis. Program SWS for the digital photogate sensor. In the sensor setup window enter the diameter of the tube. Open the digits display window and choose velocity. Double click in the display area of the digits display to open the display setup window. Program for 3 decimal digits. Repeat the measurement a number of times to make sure you get consistent results Taking Data Move the photogate so that its plane is horizontal and is opposite the base plate that holds the vertical rod supporting the photogate. Adjust the height of the photogate beam to be about 25 cm above the table and place the photogate over the edge of the table next to the vertical rod in the bench clamp. Drop the rubber tube taking care to make sure it doesn t rotate as it falls and remains horizontal as it passes through the photogate beam. Use the vertical rod next to the photogate as a guide for dropping the tube through the photogate without hitting it. Click Record, and using the meter stick drop the tube onto bubble wrap from a height of h = 10 cm above the beam. Measure height to the middle of the tube. Record the velocity through the photogate. You will want to take some practice runs and a few runs for real. Repeat for h = 20 and 30 cm Analysis Compare your measured velocities with hose predicted by conservation of energy. Is there any friction in this experiment? If so, how would it affect your data? Try dropping the paper tube from a height of 50 cm above the photogate beam and analyze the data as before. Is energy conserved? If not, what happens to the energy? Try different heights. 3
4 3.2 Pendulum In this experiment, the motion of the mass is not solely in the same direction as the conservative force, and thus illustrates some new features of energy conservation Description A 100 g mass is hung from a 30 cm string and used as a pendulum. The weight is pulled to one side and let go from various heights. At the lowest point the mass passes through the beam of a photogate sensor and its velocity is measured. The total energy at the beginning of the measurement is compared to the energy at the bottom of the pendulum s trajectory Programming Measure the diameter of the 100g weight with calipers. Follow the procedures in the previous section to measure the velocity of the mass Taking Data Adjust the photogate on the lab bench so that the legs point up. The photogate beam should be directly below the suspension point of the pendulum. Adjust the height of the suspension point of the pendulum so that the full width of the weight swings through the center of the photogate beam taking care to avoid having grooved portions of the weight pass through the beam (why is this important?). Pull the weight to one side so that the middle of the weight is 10 cm above the level of the beam. Let go and record the velocity of the weight at its lowest point. Repeat for h = 8, 6, and 4 cm. Take care to describe how you measure the change in height of the mass so that you make sure that you are measuring the change in height of the center of mass Analysis How well is energy conserved? Is any deviation from perfect conservation within your experimental uncertainty? Does air resistance appear to play a significant role in this experiment or does it appear to be negligible? Note that the mass has a force exerted on it by the string, but this force does not contribute to the potential energy. Why? 3.3 Gravity and a spring In this experiment you investigate conservation of energy for a case in which there are two conservative force laws, that of a spring and gravity, acting simultaneously Description A 0.5 kg mass m is suspended from a vertical spring. The mass is set into vertical oscillatory motion and its position is measured as a function of time by a motion sensor. The velocity and acceleration are calculated by SWS. The total energy of the mass is compared at different points in the motion. The potential energy of the springmass system is due to both gravitational potential energy and the potential energy of the spring. 4
5 3.3.2 Theory The mass m performs simple harmonic motion with an angular frequency given by ω = k/m, where k is the spring constant. Define the coordinate y to be zero when the spring is unstretched, as it is when no mass is attached. Take the downward direction to be negative. In this case the total energy can be written as E = 1 2 mv ky2 + mgy. (13) Here, y is defined so that it is both the position of the spring and the height of the mass. Show that when the mass is attached so that the spring is stretched but not oscillating (the force of gravity and the spring on the mass exactly balance), the total energy is given by E = 1 mgb, (14) 2 where is y = B is the distance the spring is stretched by the mass (B > 0). In this experiment, you will explore conservation of energy when the mass is oscillating. If energy is conserved, the total E in Eq. 13 should be the same at any point in the motion, assuming no friction Programming Program SWS for the motion sensor, using the default values. Open the graph window, choosing position, velocity, and acceleration for the vertical axes.you will probably find the default sampling rate of 20 Hz to be satisfactory, but feel free to experiment Preliminaries Hang the spring over the edge of the bench. Tape the top of the spring to the rod from which it is suspended so that the spring cannot slip off the rod inadvertently. Adjust the height of the spring so that with the 0.5 kg mass attached the bottom of the mass is cm from the floor. Measure the force constant k for the spring. Place the motion sensor on the floor directly beneath the the mass. Check that the face of the motion sensor is horizontal. There is a switch on the motion sensor which sets the width of the acoustic beam of the sensor. You will probably find that the narrower width works the best but feel free to experiment. The motion sensor should be plugged into the interface so that position increases as the mass goes higher. Set the mass into motion by pulling it down a reasonable distance and letting go. Recall that the motion sensor does not record distances less than 0.15 m from its face. Do not set the mass in motion by lifting it up. If you lift it too far it will crash into the motion sensor. After the motion settles down, click Record and do the following. Simultaneously observe the motion of the mass and the graphs of position, velocity, and acceleration. Is the acceleration maximum or minimum when the velocity is zero? When the velocity is maximum is the acceleration maximum or minimum? Observe the springmass system. Is there any kinetic energy that is not accounted for by 1 2 mv2? If so, this would be worthwhile estimating quantitatively in your error analysis. 5
6 3.3.5 Experiment and Analysis Set the mass into motion, and when the motion settles down click REC, and click STOP after 3 or 4 cycles of the motion. Using the cross hairs on the position graph, determine the distance the mass has traveled from the highest position (H) to the following lowest position (L) on the position graph. This distance will be 2A, where A is taken to be a positive number. Determine the total energy E for the following 4 positions of the mass during the motion: the highest and lowest positions of the mass utilized in finding the amplitude A, and the two positions of maximum velocity following these highest and lowest positions of the mass. Recall that the coordinate y gives the position of the end of the spring, not the position of the mass. The highest position of the mass occurs when y = y 0 + A and the lowest position of the mass when y = y 0 A. Maximum velocity occurs when y = y When the mass is at the highest chosen position and the kinetic energy is zero, the energy E H will be E H = 1 2 k(y 0 + A) 2 + mg(y 0 + A). (15) 2. When the mass has the maximum down velocity following the highest chosen position of the mass the energy E 01 is given E 01 = 1 2 mv ky2 0 + mgy 0. (16) 3. When the mass is at the lowest chosen point and the kinetic energy is zero, the energy is given by E L = 1 2 k(y 0 A) 2 + mg(y 0 A). (17) 4. When the mass has the maximum up velocity following the lowest chosen position of the mass the energy E 02 is given by E 02 = 1 2 mv ky2 0 + mgy 0. (18) Obtain the velocity from the velocity curve of the graph display. Compare the total energy at these 4 points. Is energy conserved? What factors might introduce error into your measurements and calculations? Questions The analysis presented in this section assume no friction. How will friction affect your experimental results? Do you see evidence of friction in the curve of position vs time? Extra: Copy the record or position, velocity, and acceleration over two or three minutes to an Excel file. Save the file as a CSV file and plot the position, velocity, and acceleration as a function of time using Python. From your plots, determine how much energy is lost per cycle. Is the energy lost per cycle the same throughout the experiment? Is the fraction of energy lost per cycle the same throughout the experiment? 6
7 4 Finishing Up Please return the bench to the condition in which you found it. From the File menu click New and then click Don t Save. Thank you. 7
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