205_sec 26_Lab 7_Nath, Sankeshwarkar, Incabo, Shah, Balamurugan

Lab 7: Rotational Dynamics Saniyah Sankeshwarkar, Susmita Nath, John Incabo, Aesha Shah, Padmavati Balamurugan I. Gently drop a non-rotating rod into a circular disc (no external torques) Parts 1-3: - Point just before the collision: 0.04 kg.m/s 2 - Point just after the collision: -0.0075 kg.m/s 2 Graphs: - Total energies (initial, i; final, f; ratio, f/i) - initial: - i rod = 0.012 J - i disc = 0.011 J - f = 0.0007 J - final/initials - f/i for rod = 0.0007/0.012 = 0.0583 - f/i for disc = 0.0007/0.011 = 0.0636 - Total angular momenta (i, f, f/i): - initial: - i rod = 0.03 kg m 2 /s - i disc = 0.01 kg m 2 /s - final: f = 0.04+ 0.0075 = 0.0475 kg m 2 /s - final/initials - f/i for rod = 0.0475/0.03 = 1.583 kg m 2 /s - f/i for disc = 0.0475/0.01 = 4.75 kg m 2 /s

Part 4: Graphs: - Total energies (initial, i; final, f; ratio, f/i): - initial: - I rod = 0.0350 J - i disc = 0.0316 J - f = 0.0203 J - final/initials - f/i for rod = 0.0203/0.0350 = 0.58 J - f/i for disc = 0.0203/0.0316 = 0.642 J - Total angular momenta (i, f, f/i): - initial: - i rod = 0.010031 kg m 2 /s - i disc = 0.0091099 kg m 2 /s - final: f = 0.010357 N - final/initials - f/i for rod = 0.010357/0.010031 = 1.032 kg m 2 /s - f/i for disc = 0.010357/0.0091099 = 1.137 kg m 2 /s II. Use a small mass to rotate the circular disc (external torque) Part 1: No credit: Why is the external torque only approximately mgR pulley ? Its angular momentum is conserved but kinetic energy is not conserved and it changes.

d) The change in theta (angle) and omega (angular velocity) from the time the drive mass is released to when it hits the table ➔ Δθ = h/rpulley = 0.14m/0.025 = 560 rads - = (T)/(I ɑ disk ) = (30.228)/(1.621*10 -4 ) = 186477.483 ➔ ɷ f 2 = 0 + (2)( )(Δθ) = 14451.81 rads/second ɑ Part 4: Label the points where the mass was released and when it hits the table. Enter numerical values, and ratios of experimental to theoretical, for Angular Momentum and Energy (disc) for when the drive mass hits the table . Theoretical calculations: - Angular Momentum = L f = (I disk )( ɷ f ) = (1.621*10 -4 )(14451.81) = 2.343 kg m 2 /s - Energy disc = (½)(I disk ) ( ɷ f 2 ) = (½)( 1.621e-4)( 14451.81)(14451.81) = 16927.68 J Experimental: - Energy disc when mass hits table: 0.0580 J - Angular Momentum: 0.0122 kg m 2 /s

Part 5: Give analysis on how close (or far) these values are using Percent Difference Angular acceleration: 30.228 kg*m^2 Experimental Value: 0.0122 kg*m^2 Percent difference: 199.839% difference QUESTIONS 1. What if the rod were already rotating at the same angular velocity at the disc before you dropped it onto it— would the disc’s angular velocity change after dropping? Why or why not? If the rod was already rotating at the same angular velocity as the disc before that we dropped it onto it, as opposed to dropping a non-rotating rod, the disc’s angular velocity would change after dropping the rod onto it. This is due to the fact that in any type of system for this scenario, it should be realized that the angular momentum of the system as a whole must be conserved. In this specific scenario, the system would be the disk that is rotating as well as the rod that was dropped onto the disk (that has the same angular velocity as the disc). For our experiment, where a non-rotating rod was dropped onto the disc, our work showed that the initial angular momentum for the non-rotating rod for the system was 0. However, setting the rod to have the same angular velocity as the disc would (in terms of the angular momentum equation for the system) only affect the initial angular momentum of the rod, but not the disc. This is because the initial angular momentum of the rod would actually have a value as opposed to 0. As a result, it can be agreed upon when we say that this change would end up changing the disc’s angular velocity after dropping the rod. Since the left side of the angular momentum equation (initial values) changed values, the right side of the equation would have to change alongside it. 2. If you were a bug (instead of the falling mass) hanging from the string wrapped around the same pulley you used in the lab, and you wanted to descend to the lab table as softly as possible (smallest possible acceleration) so as not to squash yourself. You have your choice of discs– the usual one, or one with half the mass but twice the radius – which one would you choose? (Remember – you are only swapping the disc (aluminum) part of the setup). In order for the ladybug to have the smallest possible acceleration when attempting to go down to the lab table, the disc that should be chosen in order to support this scenario would be the disc that is the one that was used originally within the experiment for this lab. This is due to the fact that the acceleration imposed on the object (which is connected to a disc pulley) is affected by the mass of the disc. It should be realized that the radius of the disc used within the pulley is independent of the acceleration of the object attached to the pulley system. As a result, in order to reach the scenario with the ladybug descending with the smallest possible acceleration, the disc that has the most mass would be the best, since we know that the larger the mass of the disc is, the smaller the acceleration is for the object in question within the system. That way, the ladybug would be able to land on the lab table as softly as possible.