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22:27:20 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?......!!!!!!!!...................................
RESPONSE --> `tau = I `omega , where I is the moment of inertia, `tau is the torque and `omega is the angular acceleration.
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22:27:33 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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22:29:09 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 --> Newton's Second Law in rotational form says that `tau = I `omega , where I is the moment of inertia, `tau is the torque and `omega is the angular acceleration.
We would find the difference in angular velocities and divide that by time to get the angular acceleration. We would then divide the torque by the angular acceleration to get the moment of inertia..................................................
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22:29:18 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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22:30:39 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 --> We would multiply its mass by the radius squared.
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22:30:58 ** 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. We would add each one together to get the total for the three hoops.
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22:31:51 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 --> We would find it the same way as last question. Multiply the mass by the radius squared for each mass. Then add all three together.
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22:31:56 ** 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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22:35:36 problem 8.16 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
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RESPONSE --> question 8.16 asks for the average angular acceleration of a car that slows down from 4500rpm to 1200 rpm in 2.5 s. It also asks the # of revs completed during that time.
The average angular acceleration is equal to 528 rev/s2. The total number of revs completed during that time is 1140..................................................
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22:35:42 ** alpha = v^2 / r so v = `sqrt( alpha * r ) = `sqrt( 100,000 * 9.8 m/s^2 * .07 m) = `sqrt( 69,000 m^2 / s^2 ) = 260 m/s approx.
Circumference of the circle is 2 `pi r = 2 `pi * .07 m = .43 m. 260 m/s / ( .43 m / rev) = 600 rev / sec. 600 rev / sec * ( 60 sec / min) = 36000 rev / min or 36000 rpm. All calculations are approximate. **......!!!!!!!!...................................
RESPONSE --> ok
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22:39:22 gen problem 8.26 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?......!!!!!!!!...................................
RESPONSE --> I had problems with 8.26 in the book. It is the problem about the wrench and I was unsure of how to approach it because it asks to estimate the force applied at the open socket of the wrench.
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22:39:30 ** 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.. **......!!!!!!!!...................................
RESPONSE --> ok
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22:39:51 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> did not have this ?
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22:39:58 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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22:40:05 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 --> gen
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22:40:09 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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