#$&* course Phy 231 7/20 8 032. `query 32*********************************************
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Given Solution: `a** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhat part or parts of the system experience(s) an increase in rotational kinetic energy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The wheel, the things attached to the wheel, like the bolts, and the axle that held the wheel all experienced an increase in rotational kinetic energy. 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 weight that falls from the table has increasing translational kinetic energy because it is moving from one place to another. 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 bolts that are attached to the outer edge of the wheel have greater kinetic energy because they have a greater velocity. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Moment of inertia for the wheel = (1/2)(m r^2) Moment of inertia for each bolt = (m r^2) *r is the distance from the center* Add all of the moments of inertia up and you get the moment of inertia for the whole system. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The moment of inertia for the center of its mass=its radius times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You find the velocity of the falling weight at a point in meters / second. Then you divide that by the radius of the wheel and get the angular velocity in radians/second. Finally, you multiply (1/2) by the moment of inertia by the square of the angular velocity to get the kinetic energy. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The question was how we use measurements of the motion of the descending mass to find the angular KE: By observing position vs. clock time we can estimate velocities, and determine the velocity of the descending mass at any point. The string is wound around the rim of the wheel. So the rim of the wheel moves at the same speed as the string, which is descending at the same speed as the mass. So if our measurements give us the speed of the descending mass, we know the speed of the wheel. If we divide the velocity of the rim of the wheel by its radius we get the angular velocity of the wheel. Recall that angular velocity is designated by the symbol omega. • Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to 1/2 I omega^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 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. Your solution: 8250 rpm (2pi / 1rev)(1min/60s) = 864 rad/s Final KE = (1/2)(3.75*10^-2)(864 rad/s)^2 = 13997 J 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 Question: `qQuery gen problem 8.58 Estimate KE of Earth around Sun (6*10^24 kg, 6400 km rad, 1.5 * 10^8 km orb rad) and about its axis. What is the angular kinetic energy of the Earth due to its rotation about the Sun? What is the angular kinetic energy of the Earth due to its rotation about its axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Around the Sun: Circumference: 2 pi 1.5*10^8km = 9.42 * 10^ 8 km Velocity = 9.42*10^8km / 365 days = 2582130km/day (day/24hrs)(1hr/3600s)(1000m/1km) = 29886 m/s Angular velocity = 29.886 km/s / 1.5*10^8km = 1.99*10^-7 rad/s KE = (1/2)(6*10^24)(29886 m/s)^2 = 2.68 * 10^33 J Rotation on the axis: Moment of inertia = (2/5)(6*10^25kg)(6400000m)^2 = 9.83*10^37 kg m^2 Angular velocity = 1rev/24hr (1hr/3600s)(2pi / 1 rev) = .00175 rad/s Angular KE = (1/2)(9.83*10^37)(.00175 rad/s)^2 = 1.51 * 10^32 J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The circumference of the orbit is 2pi*r = 9.42*10^8 km. We divide the circumference by the time required to move through that distance to get the speed of Earth in its orbit about the Sun: 9.42 * 10^8 km / (365days * 24 hrs / day * 3600 s / hr) =29.87 km/s or 29870 m/s. Dividing the speed by the radius we obtain the angular velocity: omega = (29.87 km/s)/ (1.5*10^8 km) = 1.99*10^-7 rad/s. From this we get the angular KE: KE = 1/2 mv^2 = 1/2 * 6*10^24 kg * (29870 m/s)^2 = 2.676*10^33 J. Alternatively, and more elegantly, we can directly find the angular velocity, dividing the 2 pi radian angular displacement of a complete orbit by the time required for the orbit. We get omega = 2 pi rad / (365days * 24 hrs / day * 3600 s / hr) = 1.99 * 10^-7 rad/s. The moment of inertia of Earth in its orbit is M R^2 = 6 * 10^24 kg * (1.5 * 10^11 m)^2 = 1.35 * 10^47 kg m^2. The angular KE of the orbit is therefore KE = .5 * I * omega^2 = .5 * (1.35 * 10^47 kg m^2) * (1.99 * 10^-7 rad/s)^2 = 2.7 * 10^33 J. The two solutions agree, up to roundoff errors. The angular KE of earth about its axis is found from its angular velocity about its axis and its moment of inertia about its axis. The moment of inertia of the Earth as it spins on its axis is I=2/5 M r^2= 2/5 * 6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37 kg m^2. The angular velocity of the Earth about its axis is 1 revolution / 24 hr = 2 pi rad / (24 hr * 3600 s / hr) = 7.2 * 10^-5 rad/s, very approximately. So the angular KE of Earth about its axis is about KE = .5 I omega^2 = .5 * 9.8 * 10^37 kg m^2 * (7.2 * 10^-5 rad/s)^2 = 2.5 * 10^29 Joules. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not exactly sure what I did wrong in the second part with the angulae KE of the Earth about the axis…
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Given Solution: `a** The moment of inertia of the disk is I = 2/5 M R^2; the moment of inertia of the rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the disk so L = R. The masses are equal, so we find that the moments of inertia can be expressed as 2/5 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 2/5 M R^2 + 1/12 M R^2 = 29/60 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (29/60 M R^2) / (2/5 M R^2) = 29/60 / (2/5) = 29/60 * 5/2 = 145 / 120 = 29 / 24. Since angular momentum I * omega is conserved an increase in moment of inertia I results in a proportional decrease in angular velocity omega so we end up with final angular velocity = 24 / 29 * initial angular velocity = 24 / 29 * 2.4 rev / sec = 2 rev/sec, approximately. Self-critique (if necessary): I thought the moment of inertia for a disk was 1/2 m r^2?????????? And 2/5 was for a sphere????? Or am I missing something?????????
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Given Solution: `a** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + 1.5 kg * (.025 m)^2= .0032 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx. The resulting angular acceleration is alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx. The moment of inertia of the system, with block suspended from the larger disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx. The resulting angular acceleration is alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "