course Phy 121
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19:37:27 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 --> The proportionality constant between the inferred average velocities of the weight and average angular velocities of the disk can then be related to the radius of the axle.
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19:37:34 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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19:38:51 If the falling weight accelerates uniformly, does it follows that the rotating disk has a uniform angular acceleration?
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RESPONSE --> It seems the acceleration of the weight is approximately uniform. It can then be inferred that the acceleration of the disk is approximately uniform, and that the average velocity of the weight is therefore equal to the average of its initial and file velocities, with a similar relationship for angular velocities of the disk.
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19:38:56 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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19:42:34 gen Problem 8.30 55 N force on the end of a door 74 cm wide.
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RESPONSE --> This does not match up with any of the homework problems I did for chapter 8 ... actually the 8.30 is a ""potter shaping a bowl problem"" -- I hope I am not missing something here.
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19:42:47 ** The torque on the door is 55 N * .74 m = 40.7 m N. If the force is at 60 deg to the face of the door then since the moment arm is along the fact of the door, the force component perpendicular to the moment arm is Fperp = 40.7 m N * sin(60 deg) = 35 N (approx) and the torque is torque = Fperp * moment arm = 35 N * .74 m = 27 m N (approx). {}{}STUDENT COMMENT: Looks like I should have used the sin of the angle instead of the cosine. I was a little confused at which one to use. I had trouble visualizing the x and y coordinates in this situation.INSTRUCTOR RESPONSE: You can let either axis correspond to the plane of the door, but since the given angle is with the door and angles are measured from the x axis the natural choice would be to let the x axis be in the plane of the door. The force is therefore at 60 degrees to the x axis. We want the force component perpendicular to the door. The y direction is perpendicular to the door. So we use the sine of the 60 degree angle.
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RESPONSE --> ok
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19:42:50 ...!!!!!!!!...................................
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RESPONSE --> ok
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19:51:40 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.
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RESPONSE --> Assuming that 1 rad = .31 m; then 7 m/s would equal 22 rad / s^2; Thus; I am going to say that tau = 3.6 * .31^2 * 22 = 7.6 m *N
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19:52:19 ** 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 --> Sorry ... I forgot to calculate the force.. missed that one
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19:52:23 Univ. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?
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RESPONSE --> Not Required
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19:52:26 ** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s. The angular acceleration is therefore alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx.. The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx.. To achieve the necessary angular acceleration we have tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N. The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque tauFrictAx = -96 N * .52 m = -50 m N. The frictional torque of the wheel is in the direction opposite motion and is therefore tauFrict = -6.5 m N. The net torque is the sum of the torques exerted by the crank and friction: tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N. The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore F = tau / x = 67 m N / (.5 m) = 134 N. If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx. Starting at 120 rpm = 12.6 rad/s the time to come to rest will be `dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **
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RESPONSE --> ok
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