course Phy 121
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18:14:15 `q001. Note that this assignment contains 4 questions. If an object rotates through an angle of 20 degrees in five seconds, then at what rate is angle changing?
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RESPONSE --> 20 degrees / (5 seconds) = 4 deg / sec
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18:15:19 The change of 20 degrees in 5 seconds implies a rate of change of 20 degrees / (5 seconds) = 4 deg / sec. We call this the angular velocity of the object, and we designate angular velocity by the symbol `omega.
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RESPONSE --> ok
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18:16:54 `q002. What is the average angular velocity of an object which rotates through an angle of 10 `pi radians in 2 seconds?
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RESPONSE --> `omega = `d`theta / `dt 10 `pi radians / 2 seconds = 5 `pi rad/s
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18:18:58 The average angular velocity is equal to the angular displacement divided by the time required for that displacement, in this case giving us `omega = `d`theta / `dt = 10 `pi radians / 2 seconds = 5 `pi rad/s.
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RESPONSE --> ok
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18:20:20 `q003. If an object begins with an angular velocity of 3 radians / sec and ends up 10 seconds later within angular velocity of 8 radians / sec, and if the angular velocity changes at a constant rate, then what is the average angular velocity of the object? In this case through how many radians this the object rotate and at what average rate does the angular velocity change?
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RESPONSE --> (8 rad/s + 3 rad/s) / 2 = 5.5 rad/s 5.5 rad/s * 10 s = 55 rad in 10 sec (8 rad/s - 3 rad/s) = 5 rad/s ( 5 rad / sec ) / (10 sec) = .5 rad/s^2
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18:23:26 Starting at 3 rad/s and ending up at 8 rad/s, the average angular velocity would be expected to be greater than the minimum 3 rad/s and less than the maximum 8 rad/s. If the angular velocity changes at a constant rate, we would in fact expect the average angular velocity to lie halfway between 3 rad/s and 8 rad/s, at the average value (8 rad/s + 3 rad/s) / 2 = 5.5 rad/s. Moving at this average angular velocity for 10 sec the object would rotate through 5.5 rad/s * 10 s = 55 rad in 10 sec. The change in the angular velocity during this 10 seconds is (8 rad/s - 3 rad/s) = 5 rad/s; this change takes place in 10 seconds so that the average rate at which the angular velocity changes must be ( 5 rad / sec ) / (10 sec) = .5 rad/s^2. This is called the average angular acceleration. Angular acceleration is designated by the symbol `alpha. Since the angular velocity in this example changes at a constant rate, the angular acceleration is constant and we therefore say that `alpha = `d `omega / `dt. Again in this case `d`omega is the 5 rad/sec change in the angular velocity.
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RESPONSE --> ok
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18:24:11 `q004. If an object starts out with angular velocity 14 rad/s and accelerates at a rate of 4 rad/s^2 for 5 seconds, then at what rate is the object rotating after the 5 seconds? Through how many radians will the object rotate during this time?
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RESPONSE --> (4 rad/s^2) (5s) = 20 rad/s 14 rad/s + 20 rad/s = 34 rad/s at the end (14 rad/s + 34 rad/s) / 2 = 24 rad/s `d`theta = (24 rad/s) ( 5 sec) = 120 rad
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18:25:15 Changing angular velocity at the rate of 4 rad/s^2 for 5 sec the angular velocity will change by (4 rad/s^2) (5s) = 20 rad/s. Since the angular velocity was already 14 rad/s at the beginning of this time period, it will be 14 rad/s + 20 rad/s = 34 rad/s at the end of the time period. The uniform rate of change of angular velocity implies that the average angular velocity is (14 rad/s + 34 rad/s) / 2 = 24 rad/s. An average angular velocity of 24 radians/second, in 5 seconds the object will rotate through an angle `d`theta = (24 rad/s) ( 5 sec) = 120 rad.
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RESPONSE --> ok
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assignment #030 [뺗܃ܝ Physics I 04-12-2006
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18:29:26 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 --> tau = I * alpha or torque = moment of inertia * angular acceleration If the moment of inertia and torque is known then you can find change in angular acceleration. Multiply angular acceleration by time interval to get change in agular velocity. Add change in angular velocity to initial angular velocity to get final angular velocity. multiply the average angular velocity by the time interval to get the angular displacement
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18:29:35 ** 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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18:31:26 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 --> From initial and final angular velocity find change in angular velocity (`d`omega = `omegaf - `omega0). from this and the given time interval find Angular acceleration = change in angular vel / change in clock time Then from the known torque and angular acceleration find moment of intertia tau = I * alpha so I = tau / alpha
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18:31:59 ** 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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18:32:19 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 --> Moment of inertia of a hoop is M R^2. So 3 would = M1 R1^2 + M2 R2^2 + M3 R3^2
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18:33:52 ** 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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18:34:33 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 --> Moment of inertia of a mass r at distance r is m r^2. Total of m1 r1^2 + m2 r2^2 + m3 r3^2.
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18:34:57 ** 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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