#$&* course PHY 121 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat part or parts of the system experience(s) an increase in rotational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The wheel, the bolts, and the axle and the parts that rotate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The wheel, the bolts, the axle, and anything else that's rotating about an axis experiences an increase in rotational KE. ** STUDENT QUESTION I’m not quite sure what translation kinetic energy is, but it seems like it means that kinetic energy is somehow moved from place to place??? I think maybe the string moves the energy (if you can call it that) from the wheel to the paper clip. INSTRUCTOR RESPONSE Translational KE is 1/2 m v^2, where v is the velocity of the object itself as it moves from one point to another (more technically, v is the velocity of the object's center of mass). This is contrasted with rotational KE, in which an object can stay in one place and rotate about some axis. An object can also move from point to point while rotating, so it can have both rotational and translational KE. For example the ball rolling down the grooved track moves from point to point, but it also rotates as it moves. In the current example, the wheel, axel and the bolts embedded in it rotate about the axle, but the wheel doesn't go anywhere. So there is no translational KE in the wheel, just rotational. The only thing with translational KE is the descending mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat part or parts of the system experience(s) an increasing translational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The descending mass. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Only the descending mass experiences an increase in translational KE. ** STUDENT COMMENT: i must have overlooked the definition of translational KE, i didn't know what it was INSTRUCTOR RESPONSE: translational motion is motion from one point to another; in rotational motion about an axis each point keeps following the same circle, repeating the same points and otherwise never going anywhere, relative to that axis. The axis of course could be moving from one place to another--i.e., undergoing translational motion--so an object can undergo rotational motion as well as an independent translational motion. A simple example is a ball rolling down an incline. It moves from one end of the incline to the other, but as it rolls it also rotates about an axis through its center. The axis is horizontal and perpendicular to the incline. Since the center of the ball is moving, the axis of rotation is also moving. The axis is translating from one point to another as the ball rotates about it. If you drive straight down the road, the back wheels of your car rotate about an axis which runs straight through the center of your car's real axel. The axel has translational motion (i.e., it moves down the road) and the wheels have the same translational motion. Both the wheels and axel also rotate about a common central axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qDoes any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ones on the outside of the wheel, according to angular velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 ** STUDENT COMMENT: Oh... I need to think of it in terms of angular velocity INSTRUCTOR RESPONSE: Think in terms of angular velocity as well as velocity. At any instant all masses on the wheel have the same angular velocity, but the masses further from the center have greater velocity (and therefore greater KE) than those closer to the center. STUDENT COMMENT: i had the right idea here, but had it backward, i thought the closer to the fixed point the greater the velocity INSTRUCTOR RESPONSE: That would be the case for a satellite orbiting a planet. However in this case the entire wheel is rotating at a single angular velocity, so closer points don't move as fast as distinct points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m r^2 for the bolts. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. ** STUDENT COMMENT: I = .5mr^2 for the disk I = .5mr^2 for each of the bolts INSTRUCTOR RESPONSE: The moment of inertia of a particle of mass m at distance r from the axis of rotation is m r^2. A particle has all its mass concentrated at one specific location. A hoop consists of a collection of particles, all at the same distance from the axis of rotation. If we add up the m r^2 contributions from all the particles in the hoop, we get M R^2, where M is the mass and R the radius of the hoop. Thus the moment of inertia of the hoop is M R^2. The disk consists of a collection of particle spread out at many different distances from the axis. If we 'cut up' the disk into individual particles, we find that the sum of the m r^2 contributions of the particles is 1/2 M R^2, where M is the mass of the disk and R its radius. Thus the moment of inertia of the disk is 1/2 M R^2. The mass of a bolt isn't all concentrated at a single distance from the axis, but all the particles that make up the bolt are pretty close to the center of the bolt, so it doesn't differ from a particle by much. Its moment of inertia is pretty close to m r^2, where m is the mass of the bolt and r its distance from the axis. You add the moment of inertia of the disk to the moments of inertia of the bolts, and you end up with the moment of inertia of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You determine the velocity, how fast it falls. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It is the rate at which the velocity (which is itself the rate at which the object moves, or turns) changes. We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. • Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question:Principles of Physics and General College Physics problem 8.43: Energy to bring centrifuge motor with moment of inertia 3.75 * 10^-2 kg m^2 to 8250 rpm from rest. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2= 14000 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE of a rotating object is KE = .5 I omega^2, where I is the moment of inertia and omega the angular velocity. Since I is given in standard units of kg m^2, the angular velocity should be expressed in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm = 860 rad/sec, approx.. The initial KE is 0, and from the given information the final KE is KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 / sec^2 = 14000 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!