course phy201 ́gՕzZJ~dWݥassignment #032
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12:56:57 Query experiment to be viewed. What part or parts of the system experiences a potential energy decrease? What part or parts of the system experience(s) a kinetic energy increase?
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RESPONSE --> The weight loses PE and gains KE as it descends. The wheel and axle gain KE as they rotate.
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12:57:04 ** 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 --> ok
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12:57:46 What part or parts of the system experience(s) an increase in angular kinetic energy?
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RESPONSE --> The parts of the system that rotate: the wheel and axel.
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12:58:04 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **
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RESPONSE --> ok
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12:58:30 What part or parts of the system experience(s) an increasing translational kinetic energy?
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RESPONSE --> None of the rotating components have translational motion, only the descending weight.
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12:58:35 ** Only the descending mass experiences an increase in translational KE. **
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RESPONSE --> ok
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12:59:14 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.
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RESPONSE --> Bolts farther away from the center have greater velocity and thus gain greater KE.
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12:59:18 ** 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 --> ok
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13:48:11 What is the moment of inertia of the Styrofoam wheel and its bolts?
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RESPONSE --> Each bolt is mr^2 added together plus .5mr^2 for the wheel itself.
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13:48:14 ** 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 --> ok
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13:48:55 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?
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RESPONSE --> .5mv^2 of the falling mass must equal the angular KE of the wheel because the two parts move as one system
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13:50:18 ** 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 --> I see that I should have used the velocity to find the angular velocity before calculating the angular KE
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13:53:52 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.
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RESPONSE --> 14000J b/c 8250*2`pi/60=864rad/s. .5*.0375*864=14000J
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13:53:58 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 --> ok
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14:29:50 Query 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 Erath due to its rotation about the Sun? What is the angular kinetic energy of the Earth due to its rotation about its axis?
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RESPONSE --> KE of Earth around Sun: .5*6E24*((2*`pi*1.5E11)/(3600*24*365))^2=2.68E33J. KE of Earth about its axis: .5*((2/5)*6E24*(6.4E6)^2)((2`pi/(24*3600))^2=2.60E29J
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14:29:53 ** 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 --> ok
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14:42:41 Query problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal mass, length equal diameter, dropped concentric with disk. Resulting angular velocity?
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RESPONSE --> (.5mr^2)+((1/12)mr^2)=(7/12)mr^2. (7/12)/.5=1.17. 1.17*2.4=2.8rev/s
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14:44:37 ** 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 --> Your answer makes sense because I expect that the angular velocity would decrease rather than increase, but in order to get your result, I have to calculate I for the disk as 2/5mr^2, which is supposed to be I for a sphere. .5mr^2 is supposed to be I for a solid disk, unless there's something I'm not getting here.
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15:04:35 Univ. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger.
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RESPONSE --> 1) a=((.025*1.5*9.8)/((.5*.8*.025^2)+(.5*1.6*.05^2)+(1.5*.025^2))*.025=2.88m/s/s 2) a=((.05*1.5*9.8)/((.5*.8*.025^2)+(.5*1.6*.05^2)+(1.5*.05^2))*.05=2.88m/s/s=6.13m/s/s
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15:04:40 ** 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. **
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RESPONSE --> ok
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