course Phy 201 U߾ڡݎTyassignment #030
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14:00:17 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 --> Since torque=moment of inertia *angular acceleration it is the same as F=m*a. We would use this equation to find the angular accel. Now we multiply `dt by the angular accel. to get the change in angular velocity. Since it starts from rest then the final angular velocity will be equal to the final angular velocity. Now we can take the intial and final angular velocity to find the avg angular velcoiy and multiply that by the `dt to get the angle through which it rotated.
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14:01:04 ** 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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14:04:29 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 --> First we take the intial and final angular velocity and find the change in velocity. Then we take the change in angular velcoity and divide by the `dt to get the angular acceleration. Then we use the tau=I*alpha equation to find I. I=tau/alpha
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14:05:37 ** 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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14:16:27 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 --> Since the moment of inertia of a hoop is MR^2 and we know the mass and radius then the I=M1 R1^2+M2 R2^2+M3 R3^2
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14:18:47 ** 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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14:21:10 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 --> It would be the same equation as in the last qu.estion We know the mass and the distance from the axis which would be equal to the R. So we know MR^2
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14:22:03 ** 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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14:28:01 Principles of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds.
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RESPONSE --> Since we know omegaF and omegaO we can find `d`omega which would be 0rpm-6500rpm=-6500rpm. Since we know time we can find alpha which is -6500rpm/3s=-2200rpm/s
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14:33:26 The change in angular velocity from 6500 rpm to rest is -6500 rpm. This change occurs in 3.0 sec, so the average rate of change of angular velocity with respect to clock time is ave rate = change in angular velocity / change in clock time = -6500 rpm / (3.0 sec) = -2200 rpm / sec. This reasoning should be very clear from the definition of average rate of change. Symbolically the angular velocity changes from omega_0 = 6500 rpm to omega_f = 0, so the change in velocity is `dOmega = omega_f - omega_0 = 0 - 6500 rpm = -6500 rpm. This change occurs in time interval `dt = 3.0 sec. The average rate of change of angular velocity with respect to clock time is therefore ave rate = change in angular vel / change in clock time = `dOmega / `dt = (omega_f - omega_0) / `dt = (0 - 6500 rpm) / (3 sec) = -2200 rpm / sec. The unit rpm / sec is a perfectly valid unit for rate of change of angular velocity, however it is not the standard unit. The standard unit for angular velocity is the radian / second, and to put the answer into standard units we must express the change in angular velocity in radians / second. Since 1 revolution corresponds to an angular displacement of 2 pi radians, and since 60 seconds = 1 minute, it follows that 1 rpm = 1 revolution / minute = 2 pi radians / 60 second = pi/30 rad / sec. Thus our conversion factor between rpm and rad/sec is (pi/30 rad / sec) / (rpm) and our 2200 rpm / sec becomes angular acceleration = 2200 rpm / sec * (pi/30 rad / sec) / rpm = (2200 pi / 30) rad / sec^2 = 73 pi rad / sec^2, or about 210 rad / sec^2.
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RESPONSE --> ok
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14:46:01 Principles of Physics and General College Physics problem 8.16. Automobile engine slows from 4500 rpm to 1200 rpm in 2.5 sec. Assuming constant angular acceleration, what is the angular acceleration and how how many revolutions does the engine make in this time?
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RESPONSE --> First we find the `domega=4500rpm-1200rpm=3300rpm. Now we find alpha=3300rpm/2.5s=1320rpm/s=1320*pi/30 rad/s^2=138 rad/s^2. Know we find the omegsAve=(4500rpm+1200rpm)/2=2850rpm. This multiplied by the time gives us our total rpm during this time interval which is 2850rpm*2.5s=7125revolutions.
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14:47:26 The change in angular velocity is -3300 rpm, which occurs in 2.5 sec. So the angular acceleration is angular accel = rate of change of angular vel with respect to clock time = -3300 rpm / (2.5 sec) = 1300 rpm / sec. Converting to radians / sec this is about angular accel = -1300 rpm / sec ( pi / 30 rad/sec) / rpm = 43 pi rad/sec^2, approx.. Since angular acceleration is assumed constant, a graph of angular velocity vs. clock time will be linear so that the average angular velocity with be the average of the initial and final angular velocities: ave angular velocity = (4500 rpm + 1200 rpm) / 2 = 2750 rpm, so that the angular displacement is angular displacement = ave angular velocity * time interval = 2750 rpm * 2.5 sec = 6900 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (4500 rpm + 1200 rpm) / 2 * (2.5 sec) = 6900 revolutions and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (4500 rpm - 1200 rpm) / (2.5 sec) = 1300 rpm / sec, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 6900 revolutions can also be expressed in radians as 6900 rev = 6900 rev (2 pi rad / rev) = 13800 pi rad, or about 42,000 radians.
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RESPONSE --> ok
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15:09:26 gen Problem 8.23: A 55 N force is applied to the side furthest from the hinges, on a door 74 cm wide. The force is applied at an angle of 45 degrees from the face of the door. Give your solution:
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RESPONSE --> First we start out finding the torque applied perpindicular to the door would be 55N*.74m=40.7N m. Since we are pushing at 45deg then our foce perpindicular would be 40.7N m*sin(45deg)=29N.So now the torque=29N*.74m=21.5N m
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15:10:10 ** ** If the force is exerted perpendicular to the face of the door, then the torque on the door is 55 N * .74 m = 40.7 m N. The rest of the given solution here is for a force applied at an angle of 60 degrees. You can easily adapt it to the question in the current edition, where the angle is 45 degrees:The torque on the door is 45 N * .84 m = 37.8 m N. If the force is at 60 deg to the face of the door then since the moment arm is along the fact of the door, the force component perpendicular to the moment arm is Fperp = 37.8 m N * sin(60 deg) = 32.7 N and the torque is torque = Fperp * moment arm = 32.7 N * .84 m = 27.5 m N. STUDENT COMMENT: Looks like I should have used the sin of the angle instead of the cosine. I was a little confused at which one to use. I had trouble visualizing the x and y coordinates in this situation.INSTRUCTOR RESPONSE: You can let either axis correspond to the plane of the door, but since the given angle is with the door and angles are measured from the x axis the natural choice would be to let the x axis be in the plane of the door. The force is therefore at 60 degrees to the x axis. We want the force component perpendicular to the door. The y direction is perpendicular to the door. So we use the sine of the 60 degree angle.
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RESPONSE --> ok
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15:25:30 gen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
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RESPONSE --> First we can find the circum. of the circle which is 2 pi .07m=.44m. Since aCent=v^2/r then v=sqrt(aCent*r) we get v=sqrt(100,000g's*.07m),v=sqrt(100,000*9.8m/s^2*.07m)=262m/s. Since .44m is one revolution then .4m=.44m/revolution so now we can find rpm. 262m/s/.44m/rev=595r/s.To convert this to rpm we have 595rev/s*60s/min=35700r/min
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15:29:15 ** 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. **
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RESPONSE --> ok
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15:33:57 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 big wheel is 2/25 the size of the small wheel then the alpha of the big wheel will be equal to .08*7.2rad/s^2=.58rad/s^2
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15:35:00 ** 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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15:47:34 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> First of all we know alpha which = .58rad/s^2.If we convert 65rpm to rad/s we get 65*pi/30 rad/s=6.8rad/s.Now we can find the change in time= 6.8rad/s/.58rad/s^2=11.7s
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15:48:07 ** 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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15:48:13 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 --> ok
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15:48:18 ** 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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