#$&* course phy 201 Aug 1 4:20pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat part or parts of the system experience(s) an increase in rotational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The wheel experienced the increase in angular kinetic energy. More specifically, the outer bolts experienced the greatest increase. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Since the wheel was not rolling on a surface, there was no translational kinetic energy. 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 outer bolts gain more KE because of their higher value of moment of inertia. Or a common sense form, the outer bolts simply moved faster than the inner bolts and, having the same mass as the inner bolts, they had more KE as a result of this increased 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: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can determine the velocity of the outer portion of the wheel, which is equal to the velocity of the falling mass. We can then divide by the radius to find an angular velocity, which we can use with the moment of inertia to find the KE. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It is the rate at which the velocity (which is itself the rate at which the object moves, or turns) changes. We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. • Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can determine the velocity of the outer portion of the wheel, which is equal to the velocity of the falling mass. We can then divide by the radius to find an angular velocity, which we can use with the moment of inertia to find the KE. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 (if necessary): ------------------------------------------------ Self-critique rating: #*&!