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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the rating:, before you look at the given solution.
030. `query 30
Question: `qintroductory 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?
Your solution:
Confidence rating::
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):
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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?
Your solution:
Confidence rating::
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. **
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Question: `qHow do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
Your solution:
Confidence rating::
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. **
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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?
Your solution:
Confidence rating::
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. **
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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.
Your solution:
Confidence rating::
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.
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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 how many revolutions does the engine make in this time?
Your solution:
Confidence rating::
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.
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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.
Give your solution:
Your solution:
Confidence rating::
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.
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Question: `qgen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
Your solution:
Confidence rating::
Given Solution:
`a** a_cent = v^2 / r so v = `sqrt( a_cent * 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.
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Question: `qWhat is the angular acceleration of the larger wheel?
Your solution:
Confidence rating::
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.. **
inverse relationship between the 2 wheels?
INSTRUCTOR RESPONSE
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.
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Question: `qHow long does it take the larger wheel to reach 65 rpm?
Your solution:
Confidence rating::
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. **
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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?
Your solution:
Confidence rating::
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.
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