#$&* course phy 201 Aug 1 4:20pm 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: `a** tau stands for torque and I stands for the moment of inertia. These quantities are analogous to force and mass. Just as F = m a, we have tau = I * alpha; i.e., torque = moment of inertia * angular acceleration. If we know the moment of inertia and the torque we can find the angular acceleration. If we multiply angular acceleration by time interval we get change in angular velocity. We add the change in angular velocity to the initial angular velocity to get the final angular velocity. In this case initial angular velocity is zero so final angular velocity is equal to the change in angular velocity. If we average initial velocity with final velocity then, if angular accel is constant, we get average angular velocity. In this case angular accel is constant and init vel is zero, so ave angular vel is half of final angular vel. When we multiply the average angular velocity by the time interval we get the angular displacement, i.e., the angle through which the object moves. ** STUDENT COMMENT: I believe I am slowly understanding this.. it is hard to grasp INSTRUCTOR RESPONSE: This is completely analogous to the reasoning we used for motion along a straight line. Angular velocity is rate of change of angular position with respect to clock time.Angular acceleration is rate of change of angular velocity with respect to clock time.So the reasoning for velocities and accelerations is identical to that used before. Only the symbols (theta for angular position, omega for angular velocity, alpha for angular acceleration) are different. Torque is different than force, and moment of inertia is different from mass. However if we replace force with torque (tau), and mass with moment of inertia (I), then: Newton's Second Law F = m a becomes tau = I * alpha `dW = F `ds becomes `dW = tau `dTheta andKE = 1/2 m v^2 becomes KE = 1/2 I omega^2. It's important to also understand why this works, but these are the relationships. If you understand the reasoning and equations of uniformly accelerated motion, as well as F = m a, `dW = F `ds, and KE = 1/2 m v^2, then you need only adapt this understanding to the rotational situation. Not easy, but manageable with reasonable effort. The symbols are a stumbling block for many students, so keep reminding yourself of what each symbol you use means. It just takes a little getting used to. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf we know the initial angular velocity of a rotating object, and if we know its angular velocity after a given time, then if we also know the net constant torque accelerating the object, how would we find its constant moment of inertia? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Change in angular velocity= (final angular velocity - initial angular velocity) Angular acceleration= (change in angular velocity)/(change in time) Torque=(moment of I)*(angular acceleration) Therefore: Moment of I=Torque/angular acceleration confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m1r1^2 + m2r2^2 + m3r3^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Moment of inertia of a hoop is M R^2. We would get a total of M1 R1^2 + M2 R2^2 + M3 R3^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!