course Phy 121
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23:27:04 introductory set 8. If we know the constant moment of inertia of a rotating object and the constant net torque on the object, then how do we determine the angle through which it will rotate, starting from rest, in a given time interval?
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RESPONSE --> using tau = I* alpha; we can find alpha or the angular acceleration. Knowing the acceleration; angular acceleration * time = 'dv or the change in velocity over the time interval. Once I know the 'dv, I know that I have the final velocity (if there were no intial velocity). Thus the avg velocity over the time would be change in velocity / time period... Thus vavg * time = 'ds
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23:27:11 ** 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. **
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RESPONSE --> ok
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23:33:16 If 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?
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RESPONSE --> the final velocity - initial velocity will give me the change in velocity...From the equation 'dv = a*'dt.. I can then say that a = 'dv / 'dt; Thus using this for the original equation where tau = I * alpha The moment of inertia = torque / alpha (since now we have the torque, and alpha which is the angular acceleration.
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23:33:23 ** 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. **
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RESPONSE --> ok
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23:37:06 How do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
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RESPONSE --> I would assume that we would total up the moment of inertia for the 3 hoops which would be 3*M R^2 (assuming that the three hoops are identical)... If mass of the three are different then each of the compnent would need to be added seperatels MR^2 + MR^2 + MR^2
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23:37:11 ** 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. **
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RESPONSE --> ok
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23:39:32 How do we find the moment of inertia a light beam to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation?
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RESPONSE --> I am thinking that this problems follows the same principal of the hoops; althougt the light beam thing is throwing me off a bit. the moment of inertia = MR^2 where each of these component would be added together.
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23:39:37 ** 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. **
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RESPONSE --> ok
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23:44:09 gen problem 8.20 small wheel rad 2 cm in contact with 25 cm wheel, no slipping, small wheel accel at 7.2 rad/s^2. What is the angular acceleration of the larger wheel?
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RESPONSE --> Since the smaller whell is .08 the size of the larger wheel then I will assume that the the accleration of the larger whell would be .08 of the smaller...since the smaller wheel will move much faster than the larger one... In this case is would be: .08 * 7.2 = .576 rad/s/s
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23:44:17 ** Since both wheels travel the same distances at the rim, angular displacements (which are equal to distance along the rim divided by radii) will be in inverse proportion to the radii. It follows that angular velocities and angular accelerations will also be in inverse proportion to radii. The angular acceleration of the second wheel will therefore be 2/25 that of the first, or 2/25 * 7.2 rad/s^2 = .58 rad/s^2 approx.. **
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RESPONSE --> ok
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23:59:19 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> We need to reach a speed of 65 rpm ... knowing that the circumerence to complete on revelotion would be = 2*'pi*rad so 65 * 2 * 3.14 / time (60 seconds) = equals 6.8 rad/sec Knowing from the previous problem that the angular acceleration is .576 rad /s/s; Then using the formula 'dt = 'dv / a 'dt = 6.8 / .576 = 11.8 seconds It would take the object 11.8 seconds to reach a speed of 65 rpm
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23:59:24 ** 65 rpm is 65 * 2 `pi rad / min = 65 * 2 `pi rad / (60 sec) = 6.8 rad / sec, approx. At about .6 rad/s/s we get `dt = (change in ang vel) / (ang accel) = 6.8 rad / s / ( .6 rad / s^2) = 11 sec or so. **
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RESPONSE --> ok
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23:59:30 Univ. 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?
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RESPONSE --> Not Required
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23:59:33 ** 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. At a distance of .208 m from the axis of rotation the velocity will be .208 m * 230 pi rad / sec = 150 m/s, approx.. The angular acceleration at the .208 m distance is aCent = v^2 / r = (150 m/s)^2 / (.208 m) = 108,000 m/s^2, approx.. The electrostatic force of attraction between sawdust and blade is nowhere near sufficient to provide this much acceleration. **
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RESPONSE --> ok
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