course phy201 Please see my response on the last question Thanks! ·?????X?D???w????·assignment #031?S???t???}·?`?·???
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09:27:54 experiment to be viewed. What is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk?
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RESPONSE --> Not exactly sure I understand the question. But the angular velocity of the axle and the velocity of the linear displacement of the string are proportional. The proportional constant would be related to the circumference of the disc.
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09:28:42 GOOD STUDENT RESPONSE The angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking. ** The speed of the falling object is the same as the speed of a point on the rim of the axle. The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r. The disk rotates with the axle so it has the same angular velocity. **
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RESPONSE --> ok
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09:28:56 If the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration?
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RESPONSE --> yes
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09:29:19 GOOD STUDENT RESPONSE yes, because the angle of acceleration is proportional to the velocity of the disk with the radius(which is constant) as the constant of proportionality. And the velocity of the disk will be the same as the velocity of the falling weight which is dependent on the acceleration of the weight. ** If v changes at a uniform rate then since r is uniform, omega = v / r changes at a uniform rate. **
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RESPONSE --> ok
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09:33:10 Principles of Physics and General College Physics Problem 8.28: Moment of inertia of bicycle wheel 66.7 cm diameter, mass 1.25 kg at rim and tire.
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RESPONSE --> I'll choose the formula for a thin hoop which is MR^2 1.25 Kg * (.33m)^2 = .14 Kg*m^2
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09:33:36 The mass of the rim and tire is all located at about the same distance from the axis of rotation, so the rim and tire contribute m * r^2 to the total moment of inertia, where m is the mass and r the distance from the axis of rotation of the rim and tire. The distance r is half the diameter, or 1/2 * 66.7 cm = 33.4 cm = .334 m, and the mass is given as 1.25 kg, so the moment of inertia of rim and tire is I = m r^2 = 1.25 kg * (.334 m)^2 = 1.4 kg m^2.
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RESPONSE --> got it
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09:34:21 Why can the mass of the hub be ignored?
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RESPONSE --> because it is so close to the center that the squared term (R) is negligible
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09:34:42 The radius of the hub is less than 1/5 the radius of the tire; because its moment of inertia is m r^2, where r is its 'average' distance from the axis of rotation, its r^2 will be less than 1/25 as great as for the rim and tire. Even if the mass of the hub is comparable to that of the rim and tire, the 1/25 factor will make its contribution to the moment of inertia pretty much negligible.
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RESPONSE --> ok
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09:54:53 gen Problem 8.38 arm, 3.6 kg ball accel at 7 m/s^2, triceps attachment 2.5 cm below pivot, ball 30 cm above pivot. Give your solution to the problem.
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RESPONSE --> first we calculate the moment of inertia which is the .31m radius times the 3.6 Kg mass of the ball or 1.1 Kg * m^2 next we calculate the angular acceleration which is 7.0 m/s^2 / radius of .31m or 22.6 rad/s^2 Tau (torque) = 1.1 Kg*m^2 * 22.6 rad/s^2 = 25.2 mN To generate this torque at the pivot point, the tricep muscle, located .025m away must apply a force of 25.2 mN / .025m or 1008 N. This is about 250 lbs of force which sounds too high. But a 3.6 Kg ball is very heavy!
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09:58:23 ** The moment of inertia of a 3.6 kg ball at a point 30 cm from the axis of rotation is I = m r^2 = 3.6 kg * (.30 m)^2 = .324 kg m^2. At a 30 cm distance from axis of rotation the 7 m/s^2 acceleration becomes an angular acceleration of alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2. The necessary torque is therefore tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx.. The muscle exerts its force at a point x = 2.5 cm from the axis of rotation and perpendicular to that axis so we have F = tau / x = 7.6 m N / (.025 m) = 304 N. **
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RESPONSE --> I worked the problem correctly but screwed up when I didn't square the radius when calculating moment of inertia.
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{??L?????·?·???|R? assignment #032 ?S???t???}·?`?·??? Physics I 11-21-2007
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13:26:24 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 --> Any loss of altitude (if conserving force is gravity)or loss of tension (in the case of a rubber band or spring) represents a loss of potential energy. When these sources of stored energy lose potential energy it is converted to increased angular velocity which is an increase in the KE of the rotating mass.
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13:26:48 ** 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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13:27:09 What part or parts of the system experience(s) an increase in angular kinetic energy?
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RESPONSE --> the wheel and the axle
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13:31:11 ** 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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13:32:19 What part or parts of the system experience(s) an increasing translational kinetic energy?
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RESPONSE --> as the weight accelerates downward its translational KE increases
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13:32:27 ** Only the descending mass experiences an increase in translational KE. **
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RESPONSE --> ok
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13:33:50 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.
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RESPONSE --> Yes. The ones further from the center of roation move at a higher velocity and thus have a higher KE than those of the same mass but closer to the center and moving more slowly
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13:33:57 ** 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 -->
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13:34:18 ** 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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13:36:15 What is the moment of inertia of the Styrofoam wheel and its bolts?
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RESPONSE --> I don't have all the info in front of me, but I know we calculated it by ignoring the mass of the styrofoam and the center axle/washers and then took the mass of each bolt times the square of its distance from the center and added all these.
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13:36:48 ** 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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13:38:04 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?
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RESPONSE --> Ignoring frictional and other minor losses, angular KE of the wheel should be equal to the PE loss of the falling mass
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13:39:46 ** 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 understand. Wasn't sure exactly what you were looking for but I understand.
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13:43:05 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 --> The 8250 RPM's converts to 863 rad/s angular velocity The energy required will equal 1/2 * I*v^2 or 1/2 * 3.75 10^-2 * 863^2 = 1.4 10^4 J
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13:43:31 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 --> got it
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14:19:43 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 --> The earth has a moment of inertia around its axis of 2/5 MR^2 = 1.5 10^31 Kg*m^2 It rotates around it's axis at an angular velocity of 7.3 10^-5 rad/s taking 1/2 * I * v^2 =3.99 10^22 Joules rotational KE around axis around the sun it has a moment of inertia of 5.98 10^24 * 1.5 10^11 m = 8.9 10^35 Kg*m^2 It's angular velocity is 2pi/3.15 10^7 sec = 1.99 10^-7 rad/s squaring angular velocity and multiplying by moment of inertia and then by .5 gives a KE around the sun of 1.76 10^22 Joules
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14:22:03 ** 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 --> Thought I had this right but I guess not.
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14:41:21 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 --> The disc has a mass = to the rod but because of it's shape the rod has only .65 the disc's moment of inertia. the angular momentum is L = I * w so if I is increased from 1 to 1.65 and momentum is conserved than angular velocity must drop in the same proportion so new angular velocity is 15.1 rad /s / 1.65 = 9.15 rad/s
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14:42:55 ** 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 --> why 2/5 for a uniform disc? Isn't that for a sphere?
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