Sensors: Loggers: Rotary Motion Any EASYSENSE Physics Logging time: 10 seconds Teacher s notes 35 Conservation of angular momentum (1) Introduction The use of the disc accessories allows the Rotary Motion sensor to be used to illustrate the conservation of angular momentum. By fixing a disc to the pulley wheel of the sensor and spinning it, the angular speed of the disc can be measured directly. By adding a second disk to the fixed disk, as it is spinning, the change in angular speed can be seen and measured directly. If the two discs are of identical size and mass, the angular speed after the addition of the second disc should be half of the angular speed immediately before it was added. When a torque Γ is applied to an object that is free to rotate, there will be a change in the angular speed from ω i to ω f. The time taken for the change to take place is Δt. If the moment of inertia is I, then ( ω 1 ω ) Γ = I Δt When a non rotating disc is placed onto a rotating disc the torque of the rotating disk is equal and opposite to the torque on the non rotating disc so there is no net torque in the system. If there is no change in the angular momentum of the system then angular momentum must be conserved. Angular momentum (L) = I i ω i = I f ω f I i is the initial moment of inertia of the rotating disc; ω i is the initial angular speed. The initial moment of inertia of the a disk can be calculated from 1 mr where m is the mass of the disc and r is the radius of the disc. If the second disc has the same moment of inertia as the first disc then the final moment of inertia is twice the initial moment of inertia of the first disk. If angular momentum is conserved, the angular speed after the addition of the disc will be half the angular speed immediately before the addition of the disc. I 1 ω f = ωi = ωi I f The students are asked to calculate the theoretical moment of inertia and compare this to the value derived from the practical. Information is provided in the analysis section to extend the work to calculate energy and energy storage capabilities of flywheels Apparatus 1. An EASYSENSE logger.. Rotary Motion sensor set to Angular Velocity (rads) range. 3. Angular Motion disc set (Rotary Motion accessory). 4. Tripod 5. Retort stand, boss and clamps T35-1 (V)
Set up of the software Use the setup file 35 Conservation of angular momentum. Recording method Recording time Intersample time Graph 10 seconds 100 millisecond Notes The Rotary Motion sensor will need to be set to Angular velocity (rads) for this experiment. The metal discs are in a pair, one disc has a pair of locating holes which fit over the two locating pins on the pulley of the sensor. Friction pads on the disc ensure there is no slipping between the discs when they are spun as a pair. The friction pads should be between the two discs when being used. A spindle on the lower disc helps the upper disc to locate correctly when it is rotating. Locating pin to hold angular momentum disc Locating pin to hold angular momentum disc Assembling the disc accessory set to the Rotary Motion sensor. The discs supplied are of nearly equal mass and diameter. For precision in results the diameter of the discs and the mass of the discs should be found. Spinning the discs by hand should give enough rotation for the experiment; the experiment compares the rotational velocity before and after the addition of the second disc. The discs should be spun in a clockwise direction to produce positive speed / velocity values. Applying a small drop of light machine oil (or silicon oil or methylated spirit) to the Rotary Motion sensor s bearing will give longer and smoother running. You should be aiming to get a spin (maximum recorded time) to stop of 15 seconds or longer. Results and analysis The analysis consists of three stages. 1. Calculation of the moment of inertia of the discs. Calculation of the angular momentum of the discs 3. Calculation of the rotational kinetic energy of the discs (flywheel) T35 - (V)
Moment of inertia of the discs This is a simple calculation; care is required in collecting the data and making sure the correct units are used. Watch out for students who calculate the moment for each disc, they should calculate for the disc attached to the Rotary Motion sensor and then for the combined fixed and dropped disc. Use I = ½mr m = the mass of the disc in kg r = the radius of the disc in m Units are kg m Angular momentum If the students have calculated the moment of inertia correctly, the calculation of angular momentum should not be problem. Use L= Iω L = angular momentum I = moment of inertia ω = angular velocity (rad/s) units are kg m s -1 Angular Kinetic Energy Use K.E. = ½Iω 1. I = moment of inertia. ω = angular velocity (rad/s) Sample results Mass of fixed disc = 0.15083 kg Mass of dropped disc = 0.13747 kg Mass of combined disc = 0.883 kg Angular velocity with fixed disc =.77 rev/s Angular velocity with combined disc = 1.44 rev/s Diameter (r ) = 0.149 m I = ½mr L= Iω 1. I for the fixed disc = 0.075415 x 0.149 = 0.0114 kg m. I for the combined disc = 0.144 x 0.149 = 0.01456 kg m 1. L for the fixed disc = 0.0114 x.77 = 0.0311 kg m /s. L for the combined disc = 0.01456 x 1.44 = 0.0308 kg m /s Kinetic energy For the fixed disc = 0.0056 x 7.67 = 0.0431 J For the combined disc = 0.01555 x.0736 = 0.03 J T35-3 (V)
Flywheels A flywheel is a cheap simple device for storing energy. To optimise the energy storage of a flywheel it needs to spin as fast as possible, this is because the kinetic energy only increases linearly with a mass increase but as the square of the rotation speed. In the students questions they are being asked to identify this pattern. Rapidly rotating devices are subject to large centripetal forces that can rip them apart. This can limit the energy of a device unless the tensile strength of the flywheel material is matched to the forces it will experience. Centripetal force in the rotating mass can be calculated using F = mrω m = mass (kg) r = radius of the rotating mass (m) ω = angular velocity (rad/s) The calculation shows that if dense materials are used they will store more energy, but as the angular velocity increases the force increase will lead to failure at a lower speed than with low density materials. Storing energy on the flywheel, sample calculation A flywheel can be used as an energy store. A car could be adapted to store the energy generated by braking or when the engine is running but the car is not moving. How fast would the disc need to spin to store the energy? If we assume the car has 50 cm (0.5 m) flywheel disc fitted how fast would it need to spin to hold the energy produced by using 10 kg of petrol (approx specific gravity of petrol is 740 kg per m 3 ) (13.5 litres). 10 kg of petrol = 504 MJ Assume the petrol engine has an energy conversion of 15% efficiency (is this realistic?). Energy available is therefore (75.6 MJ). Flywheels have an energy conversion efficiency of about 80%. The fly wheel needs to store 75.6/0.8 = 94.5 MJ. Kinetic energy on a rotating disc is = ½Iω I for a disc = ½mr One revolution of the disc = π radians. Energy stored will equal, ½ I ω = ½ x ½mr x (πω) = π m r ω If ω is measured in revolutions per second, then the stored energy of a flywheel is approximately 10mr x ω (rev/s) Substituting the values of m = 140 kg and r = 50 cm = 0.5 m Energy stored (94.5MJ) = (10 x 140 x 0.5 ) ω 94500000/350 = ω 70000 = ω 70000 = ω 519.615 = ω ω = 519.615 rev/s T35-4 (V)
The disc would need to spin at 500 rev/s or 31176 revs/min Energy storage is 94.5 MJ/140 kg = 0.675 MJ kg which exceeds the energy storage density of steel. A flywheel to store this amount of energy would therefore have to be constructed out of some composite material. There is a similar problem with hard disk storage; to get the best data retrieval from the disc it needs to spin fast, if it spins too fast then the energy stored in the spinning disc can become a critical factor in its design. Inertial constants for different shapes m = mass r = radius k = a constant I =kmr Values of k when the shape is; 1. A wheel or annulus (ring) k=1. A solid disk of uniform thickness k=1/ 3. A solid sphere k = /5 4. Spherical shell (e.g. football or basket ball) k = /3 5. Thin rectangular rod k = 1/ T35-5 (V)