course phy201 ·?????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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course phy201 ·?????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?
......!!!!!!!!...................................
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
.................................................
......!!!!!!!!...................................
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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