#$&* course Phy 241 November 19 around 11:45pm. 030. `query 30*********************************************
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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 and KE = 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): OK, it was a good review on why this works, from newton’s second law. ------------------------------------------------ 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Since there’s 3 hoops: M1R1^2 + M2R2^2 + M3R3^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Moment of inertia of a hoop is M R^2. We would get a total of M1 R1^2 + M2 R2^2 + M3 R3^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qHow do we find the moment of inertia a rigid beam of negligible mass to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would be the same as the last question except you would use “r” and “m”: m1r1^2 + m2r2^2 + m3r3^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Moment of inertia of a mass r at distance r is m r^2. We would get a total of m1 r1^2 + m2 r2^2 + m3 r3^2. Note the similarity to the expression for the hoops. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `qUniv. 9.72 (64 in 10th edition). motor 3450 rpm, saw shaft 1/2 diam of motor shaft, blade diam .208 m, block shot off at speed of rim. How fast and what is centrip accel of pt on rim? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Angular velocity of shaft: 3450rpm*2=7900rpm Angular velocity: (7900*2pi)/(60sec)=230 pi rad/sec Diameter is 0.208 so radius is 0.104. Velocity of rim of blade: 0.104m*230pi=75.12m/s Centrip acceleration: 75^2/0.104=54086.54 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The angular velocity of the shaft driving the blade is double that of the motor, or 3450 rpm * 2 = 7900 rpm. Angular velocity is 7900 rpm = 7900 * 2 pi rad / 60 sec = 230 pi rad / sec. The rim of the blade is half the .208 m diameter, or .104 m, from the axis. At a distance of .104 m from the axis of rotation the velocity will be .104 m * 230 pi rad / sec = 75 m/s, approx.. The centripetal acceleration at the .104 m distance is a_cent = v^2 / r = (75 m/s)^2 / (.104 m) = 54 000 m/s^2, approx.. The electrostatic force of attraction between sawdust and blade is nowhere near sufficient to provide this much acceleration. ** STUDENT QUESTION: Since you're multiplying meters * rad/s, you should get rad*m / s. But we end up with just meters/second. How did this happen? INSTRUCTOR RESPONSE: A radian is the angle for which the arc distance is equal to the radius. So when a unit of radius is multiplied by the number of radians, you get units of arc distance. That is, in this context a radian multiplied by a meter is a meter. STUDENT COMMENT I don’t see how some of the numbers were calculated I get different values when I plugged in those numbers. INSTRUCTOR RESPONSE Remember that all my arithmetic is done by mental approximation and isn't guaranteed, though it should usually be closer than it was on this problem. I made a poor approximation of the angular velocity in rad / s, more that 10% low. That was compounded when the quantity was effectively squared, so the final solution was more than 20% low. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: