#$&* course PHy 201 7/17/2012 1:00 Samantha RogersPHY 201
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Given Solution: `aThe angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking. The speed of the falling object is the same as the speed of a point on the rim of the axle. The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r. The disk rotates with the axle so it has the same angular velocity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it does, because the acceleration? angle will be directly proportional to the velocity of the disk with the radius. The avg velocity of the disk will equal the velocity of the falling weight which is dependent on gravity and acceleration. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD 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. ** 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. 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. Why can the mass of the hub be ignored? 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qgen 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3.6kg ball, 7m/s^2, 2.5cm attachment, 30cm above pivot Plug values into I = m r^2, 3.6 * .30m^2 = .32 kgm^2 7m/s^2 / .3 = 23.2 rad/s^2 .32 * 23.2 = 7.42 Nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** STUDENT QUESTION Ok. I see what I was supposed to do. The quantity I got for the tension was just about right though. But I didn’t understand how you determined the angular acceleration ‘alpha by doing a / r. INSTRUCTOR RESPONSE A radian of angular displacement corresponds to a displacement along the arc which is equal to the radius. So any arc distance is equal to the angle, in radians, multiplied by the radius: • `ds = r * `dTheta. It immediately follows that the angular displacement is equal to the arc displacement divided by the radius: • `dTheta = `ds / r. All this follows immediately from the definition of a radian. It's then easy to understand that an angular velocity of a radian/second corresponds to an arc displacement equal to the radius every second. So velocity along the arc is equal to the radius multiplied by the angular velocity. Once you understand this it isn't difficult to see that a rad/s^2 of angular acceleration corresponds to an acceleration equal to 1 radius / sec^2 along the arc. More formally: Since omega_Ave = `dTheta / `dt and `dTheta = `ds / r, omega_Ave = (`ds / r) / `dt = r * `ds / `dt = v_ave / r, where v_ave is average velocity along the arc. So v_Ave = omega_Ave * r. Also since alpha_Ave = `dOmega / `dt, and `dOmega = `dv / r, we have alpha_ave = (`dv / r) / `dt = (`dv / `dt) / r = a_Ave / r, where a_Ave is the acceleration along the arc. So a_Ave = r * alpha_ave. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qUniv. 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n/a confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a ------------------------------------------------ Self-critique Rating: n/a " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!