course Phy 201 Wm䠵샵Lassignment #031
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11:33:48 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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11:33:51 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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11:47:39 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 = mr^2 = 1.25kg*(.33m)^2 = .14kgm^2
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11:47:41 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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11:48:38 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 --> my answer is right,
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11:49:41 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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11:55:44 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 --> N/A
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11:55:47 ** 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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