course phy201 assignment #032?€?o????????Physics I
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18:02:20 ** 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). **
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RESPONSE --> ** 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). **
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18:02:50 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **
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RESPONSE --> The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE.
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18:03:23 ** Only the descending mass experiences an increase in translational KE. **
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RESPONSE --> Only the descending mass experiences an increase in translational KE.
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18:04:20 ** 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 **
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RESPONSE --> The bolt on the outside would be traveling at a greater velocity and would carry more KE
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18:05:11 ** 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. **
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RESPONSE --> The moment of inertia for the center of its mass=its radias times angular velocity.
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18:05:44 ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: 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. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **
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RESPONSE --> ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: 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. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **
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18:06:10 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.
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RESPONSE --> 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.
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18:06:45 ** 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 is I=2/5 M r^2=6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37kg 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. **
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RESPONSE --> 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 is I=2/5 M r^2=6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37kg 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. **
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18:07:07 ** 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.
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RESPONSE --> ** 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.
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