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Rotational Dynamics Physics 110 Laboratory mass ring rotational 10-spoke pulley with disk photogate head base mass In this experiment we will investigate several aspects of rotational dynamics by examining rotational energy considerations. We will determine the value of the moment of inertia of a solid ring in two different ways and compare it with the value that we calculate using a theoretically derived expression. The purpose is to become more familiar with rotational dynamics and to examine how the rotational and translational variables are related to each other. Apparatus Asketch of this apparatus is shown above: A hanging mass, m, is connected to the rotational apparatus by a string that passes over a pulley. The pulley has a sensor in it to detect when a spoke passes by it, and the sensor is connected to the computer, allowing us to record the speed, v, of the falling mass (i.e. the tangential speed of the rim of the pulley) as a function of time. The other end of the string is wound around a spool of radius rsp. that is coaxial with the disk of the rotational apparatus. As the mass falls, it causes the disk to spin. An additional ring of radius R can be placed on the disk in a groove so that it also is coaxial with the apparatus. Theory We assume that the pulley has negligible mass compared to the rest of the apparatus, and that the string moves along the pulley without slipping. When the mass is dropped from rest, conservation of energy says that at any instant mg∆h=1mv2+1Iω2, 2 2 where m is the attached mass, I is the moment of inertia of the apparatus, v is the speed of the mass, ω is the rotation rate of the apparatus, and ∆h is the distance the mass has traveled from its starting point (the axes are defined so that ∆h is positive). The tangential component of velocity of the spindle is equal to the velocity of the mass, so v = rsp.ω, where rsp. is the radius of the spindle. Therefore mg∆h=1mv2+1I v 2 2 2 r sp. 1 or I = mr2 2g∆h −1 (1) sp. v2 If, on the other hand, I solve for v, v2 = 2g∆h . 1+I/mr2 sp. 2 But we know v = at and ∆h = (1/2)at for constant acceleration, so v = at = g t. (2) 1+I/mr2 sp. Procedure 1. Make sure that the top of the pulley is at a height so that the string winding around the spool is horizontal and the pulley is tangent to the spindle. 2. Maintain tension in the string and rotate the disk manually to wind the string around the spool. Again, take care that the string remains horizontal throughout the winding process. 3. Configure your Capstone software by selecting “photogate with pulley” as the input, and make an empty chart for velocity versus time. 4. Hang your mass and start taking data using Capstone until the mass is just about to hit the floor. 5. Fit a line to the velocity versus time data and record the acceleration in Table 1. Print a repre- sentative graph with the fit, properly labeling the axes, and hand it in stapled to this packet. 6. Now, place the additional mass ring on the disk and, for the same mass, repeat steps 4 and 5 above to determine the new acceleration value. Record this also in your Table 1. 7. Repeat steps 4-6 for all the other hanging masses. 8. Use a pair of vernier calipers to measure the diameter, 2rsp., of the spool. Calculate rsp. and record it. 9. Use Eqn. 2 to determine the moment of inertia of the system, once with and once without the ring, for each mass. Subtract these values and calculate the rotational inertia of the ring for each trial of a mass. 10. Measure the mass, M, of the ring, and its radius, R, and calculate its rotational inertia using I = MR2. Record these values in the given space on the next page. 11. Choose one mass, release the system, and determine its speed, v, when it has fallen a height ∆h of your choosing. Do this once with, and a second time without, the ring resting on the disk. Record these values in Table 2. 12. Use Eqn. 1 to determine the rotational inertia of the ring from the difference in appropriate moment of inertia values in Table 2. 2 13. Do the moments of inertia calculated from the formula I = MR2 and measured by the apparatus agree within experimental uncertainty? 14. Why does the pulley need to be located so that the string does not make an angle with the spindle at the point it leaves it? 3 Table 1 2 2 2 2 2 m(kg) a (m/s ) I (kg · m ) a (m/s ) I (kg · m ) I (kg · m ) disk only disk only disk + ring disk + ring ring 0.050 0.100 0.200 0.300 0.500 Youshould have five results for the moment of inertia of the ring listed in Table 1. Calculate the average value of the moment of inertia of the ring. Radius of spindle r = ± sp. Mass of ring M= ± Radius of ring R= ± Iring = ± Table 2 2 2 m(kg) h (m) v (m/s) I (kg · m ) v (m/s) I (kg · m ) disk only disk only disk only disk + ring disk + ring 4
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