#$&* course PHY 201 030. `query 30 was submitted 15 Apr 2011 around 10:51 PM. 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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: If we know the initial and final angular velocity, we can find the change in these velocities. (d omega = omega_f - omega_0) Also with the rate of change in angular velocity and time, one can find the angular acceleration. The constant moment of inertia can be found using this formula and knowing the known torque: tau = I * alpha I = tau / alpha 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: 3 ********************************************* 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: The moment of inertia depends on the radius of the circle it’s spinning in. The r represents the radius at which all the mass of the hoop is concentrated. I = m * r^2 The moment of inertia of a concentric configuration of 3 uniform hoops is: I = m1* r1^2 + m2 * r2^2 + m3 * r3^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Moment of inertia of a hoop is M R^2. We would get a total of M1 R1^2 + M2 R2^2 + M3 R3^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow do we find the moment of inertia a rigid beam of negligible mass to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The moment of inertia a rigid beam of negligible mass to which 3 masses are attached is: I = m1 r1^2 + m2 r2^2 + m3 r3^2, similar to the hoop rotating around its center like a ferris wheel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* Question: `qPrinciples of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To go from 6500 rmp to rest in 3 sec = -6500 rmp So the avg rate of change of angular velocity with respect to clock time is: -6500 rpm / 3 s = -2200 rpm / s The angular acceleration of blender blades slowing to rest from 6500 rmp: dOmega = omega_f - omega_0 dOmega = 0 - 6500 rpm dOmega = -6500 rpm ave rate = dOmega / dt ave rate = (omega_f - omega_0) / `dt ave rate = (0 - 6500 rpm) / (3 sec) ave rate = -2200 rpm / s, must be changed to (rad/s) Since 1 revolution corresponds to an angular displacement of 2 pi rad and 60 s = 1 min: 1 rpm = 1 rev / min = 2 pi rad / 60 s = pi/30 rad / s angular acceleration = 2200 rpm / s * (pi / 30 rad / s) / rpm angular acceleration = (2200 pi / 30) rad / s^2 angular acceleration = 73 pi rad / s^2 = 229 rad / s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qPrinciples 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 many revolutions does the engine make in this time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To go from 4500 rpm to 1200 rpm in 2.5 sec = -3300 rpm, so the rate of change is: -3300 rpm / 2.5s = -1320 rpm / s The angular acceleration and how many revs it will take the engine in this time is: angular acceleration = -1320 rpm / s ( pi / 30 rad/s) / rpm angular acceleration = 44 pi rad/s^2 The ave ang vel with be the average of the initial and final angular velocities: ave ang vel = (4500 rpm + 1200 rpm) / 2 ave ang vel = 2850 rpm or 47.5 rev / s The angular ds will be: angular ds = ave ang vel * time = 47.5 rev/s * 2.5 s = 119 revs The uniformly accelerated motion formula will give: dTheta = (omega_0 + omega_f) / 2 * dt dTheta = (75 rev / s + 20 rev / s) / 2 * (2.5 s) dTheta = 119 revs omega_f = omega_0 + alpha * dt alpha = (omega_f - omega_0) / dt alpha = (75 rev/s - 20 rev/s ) / (2.5 s) alpha = 22 rev / s^2, converted as: 43 pi rad/s^2 or 130 rad/s^2 The angular ds of 120 revs is: 120 revs = 120 rev * (2 pi rad / rev) 120 revs = 240 pi rad, or about 754 rads confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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, or 47.5 rev / sec. so that the angular displacement is angular displacement = ave angular velocity * time interval = 47.5 rev/s * 2.5 sec = 120 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (75 rev / s + 20 rev / s) / 2 * (2.5 sec) = 120 revolutions, approx.. and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (75 rev/s - 20 rev/s ) / (2.5 sec) = 22 rev / sec^2, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 120 revolutions can also be expressed in radians as 120 rev = 120 rev (2 pi rad / rev) = 240 pi rad, or about 750 radians. STUDENT COMMENT I didn’t know I was supposed to express my answer in radians. INSTRUCTOR RESPONSE Revolutions and radians both express rotation and it's easy to convert one to the other. However in situations that involve the trigonometry you want your angles to be in radians, as you will if you want to relate motion along the arc to the angular motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qgen 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = 55 N W = 74 cm or 0.74 m Angle = 45 deg The torque on the door would be 55 N * 0.74 m = 40.7 m*N Since the force at a 45 degree angle with the door and not perpendicular, the components of this force parallel and perpendicular to the door will be: 55 N * cos (45 deg) = 39 N 55 N * sin (45 deg) = 39 N When the force is exerted perpendicular to the door, we can use the following equation to calculate the torque: torque = perpendicular component of force * moment arm torque = 55 N * sin (45 deg) * 0.74 m torque = 30 m*N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** ** If a 55 N force is exerted perpendicular to the face of the door at a point 74 cm from the hinges, the torque on the door would be 55 N * .74 m = 40.7 m N. However the force is not exerted perpendicular to the door, but at a 45 degree angle with the door. The components of this force parallel and perpendicular to the door are therefore 55 N * cos(45 deg) = 30 N and 55 N * sin(45 deg) = 30 N, approx.. The component parallel to the door face pulls on the hinges but doesn't tend to make the door swing one way or the other; this component does not contribute to the torque. The component perpendicular to the door face is the one that tends to induce rotation about the hinges, so the torque is exerted by this component. The torque is torque = perpendicular component of force * moment arm = 55 N * sin(45 deg) * .74 meters = 30 m * N, approx.. 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 are referring to the problem from the previous edition of the text, in which the force made a 60 degree angle with the door. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qgen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given: 7 cm = 0.07 m experiences 100,000 g’s The centripetal acceleration of the object’s circular motion is given by the following equation: alpha = v^2 / r v^2 = alpha * r v = sqrt( alpha * r ) v = sqrt( 100,000 * 9.8 m/s^2 * .07 m) v = sqrt( 68,600 m^2 / s^2 ) v = 262 m/s Circumference of the circle is 2 pi r = 2 pi * 0.07 m = 0.43 m 262 m/s / (0.43 m / rev) = 609 rev / s 609 rev / s * (60 s / min) = 36540 rev / min = 36540 rpm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the angular acceleration of the larger wheel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: small wheel = rad 2 cm = accel at 7.2 rad/s^2 large wheel = 25 cm wheel The angular ds which are equal to distance along the rim divided by radii will be in inverse proportion to the radii. The angular velocities and angular accelerations will also be in inverse proportion to radii. The angular acceleration for the second wheel will therefore be: 2/25 that of the first = 2/25 * 7.2 rad/s^2 = 0.58 rad/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.. ** STUDENT QUESTION I really struggled with these questions. I’ve studied your answers but am still not certain. Why would it be an inverse relationship between the 2 wheels? INSTRUCTOR RESPONSE If the rims of two wheels, one with twice the diameter as the other, are traveling at identical speeds, then since the circumference of the larger is doulbe that of the smaller, the smaller wheel rotates through two revolutions while the larger rotates through only one. The reason is that when a wheel travels through a revolution, its rim moves a distance equal to the circumference. When the first wheel rotates through a revolution its rim travels a distance equal to its circumference, so the rim of the smaller wheel travels the same distance, which is twice its circumference, to that it travels through two revolutions. The larger wheel is 2 times the diameter of the smaller, but it travels through 1/2 as many revolutions. The wheel with lesser radius travels through more revolutions. So lesser radius implies greater angular velocity. In this case the angular velocity is inversely proportional to the radius. If the radii of the two wheels are r1 and r2, then the circumference of the second is r2 / r1 times that of the first (the actual ratio is 2 pi r2 / (2 pi r1), but that reduces to r2 / r1). If the second wheel travels through a revolution, the second travels through r2 / r1 times as many revolutions. So the first wheel travels through r2 / r1 times the angle in a give time interval. It follows that omega1 = r2 / r1 * omega 2, so that omega1 / omega2 = r2 / r1, an inverse proportion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow long does it take the larger wheel to reach 65 rpm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The time it will take the larger wheel to reach 65 rpm is: 65 rpm = 65 * 2 pi rad / min 65 rpm = 65 * 2 pi rad / (60 s) 65 rpm = 6.8 rad / s The angular acceleration for the second wheel was 0.58 rad/s^2 so: dt = (change in ang vel) / (ang accel) dt = 6.8 rad / s / ( 0.58 rad / s^2) dt = 12 s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: 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): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!