#$&* course Phy 121 032. `query 32*********************************************
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Given Solution: `a** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhat part or parts of the system experience(s) an increase in rotational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Anything that’s rotating. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The wheel, the bolts, the axle, and anything else that's rotating about an axis experiences an increase in rotational KE. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhat part or parts of the system experience(s) an increasing translational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass that’s descending. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Only the descending mass experiences an increase in translational KE. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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 are faster so they do. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .5MR^2=moment of inertia for styrofoam and the wheel’s inertia=Styrofoam inertia+radius*angV confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE=1/2Iomega^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It is the rate at which the velocity (which is itself the rate at which the object moves, or turns) changes. We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. • Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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: KE=.5(3.75e-2)(860^2)=1400J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE of a rotating object is KE = .5 I omega^2, where I is the moment of inertia and omega the angular velocity. Since I is given in standard units of kg m^2, the angular velocity should be expressed in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm = 860 rad/sec, approx.. The initial KE is 0, and from the given information the final KE is KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 / sec^2 = 14000 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "