course phy 201 10:30 7/22 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `aWhat is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk? GOOD STUDENT RESPONSE &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok ********************************************* Question: `qIf the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes! velocity changes at a uniform rate and radius is always uniform so omega is a uniform rate. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Inertia = 3.6 kg * (.30 m)^2 = .324 kg m^2. acceleration--> alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2. torque --> tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx.. F = tau / x = 7.6 m N / (.025 m) = 304 N Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating:ok"