course Phy 201 ic_class_091111Brief experiment
Class 091111
When you rotate the metal rod back and forth, holding it in the middle, it's easier than if you hold it by the end. The reason is that the mass of the rod is, on the average, about twice as close to your hand when you hold it by the middle, as when you hold it by the end. The further the mass is from the axis of rotation, the more difficult it is to accelerate the system. At this point this is an intuitive statement; we will get more specific as we develop the tools to analyze the system. The main point here is that it's not just the mass of the system that resists our effort to get the thing rotating; it's also a question of how far the masses are from the axis of rotation.
You rotate the rod by applying a torque--a 'twisting force'. If you want to rotate the rod back and forth through a given angle, at a fixed time interval, then you need more torque (more 'twisting force') when you hold it at the end than at the middle.
The property of the system that resists rotational acceleration is called moment of inertia. Moment of inertia takes account of both the mass and how far the mass is from the axis of rotation. The further the mass is from the axis of rotation, the greater the moment of inertia. Here are a couple of definitions:
I = m r^2.
It should be clear, then, that the further the mass is from the axis, the greater its moment of inertia (greater r implies greater r^2).
I = sum( m_i * r_i^2)
To understand the meaning of the _i subscripts, consider a system with three masses at three different distances from the axis. We could represent the masses with the symbols m_1, m_2 and m_3, and the distances from the axis with r_1, r_2 and r_3.
- The moment of inertia of the first mass would be m_1 * r_1^2. This is just the expression m_i * r_i^2, with i = 1.
- the second and third masses would have moments of inertia m_2 * r_2^2 and m_3 * r_3^2. These expressions are of the same form m_i * r_i^2, this time with i = 2, then with i = 3.
- The total moment of inertia would be the sum of the individual moments of inertia
I = m1 * r_1^2 + m_2 * r_2^2 + m_3 * r_3^2.
This expression is of the form I = sum(m_i * r_i^2), with i taking the values 1, 2 and 3.
Example: Assume that the strap rotating on the die has negligible moment of inertia, compared to that of two dominoes, each of mass .02 kg, resting 14 cm from the axis of rotation. What is the moment of inertia of the system?
Solution: The moment of inertia of the first domino is m_1 * r_1^2 = .02 kg * (.14 m)^2 = .0004 kg * m^2. The second domino has the same mass and lies at the same distance from the axis so its moment of inertia is the same: m_2 * r_2^2 = .02 kg * (.14 m)^2 = .0004 kg m^2.
The total moment of inertia of the 2-mass system is therefore
I = sum(m_i * r_i^2) = m_1 * r_1^2 + m_2 * r_2^2 = .0004 kg m^2 + .0004 kg m^2 = .0008 kg m^2.
Example: The strap itself actually has moment of inertia .0002 kg m^2. What is the moment of inertia of the system?
Solution: The moment of inertia of any system is obtained by adding up the moment of inertia of all the different parts of the system. This system consists of the strap, with moment of inertia .0002 kg m^2, and the two dominoes each with moment of inertia .0004 kg m^2.
The total moment of inertia of the system is therefore
I = .0004 kg m^2 + .0004 kg m^2 + .0002 kg m^2 = .001 kg m^2.
Definition of Torque
The torque exerted by a force on a system depends both on the amount of force, and on how much 'leverage' the force has on the system. It takes less force to accelerate the rod with two hands if your hands are further apart; however when they're closer together it's possible to get the system moving faster.
Suppose you drill a hole through one end of the rod, and insert an axel through that end. The axel is mounted in such a way as to keep one end of the rod fixed while allowing the rod to rotate about the axel (i.e., the axel keeps that end from moving, but don't restrict the rotation of the rod about the axel).
Suppose you then exert a force of 30 Newtons perpendicular to the rod, at a point 10 cm (i.e., .1 meter) from the fixed end. Then with that force, at that position, you are exerting a torque of (.1 meter) * (30 Newtons) = 3.0 meter * Newtons, which tends to rotate the system.
If you exert your 30 Newton force at a distance of .2 meters from the axis, then you would have more leverage on the system and the torque would be (.2 meter) * (30 Newtons) = 6.0 meter * Newtons.
The latter torque will give the system more angular acceleration than the former, as long as you can move your hand quickly enough to keep up with the system and maintain the torque (the system will build angular velocity more quickly than with the 3.0 meter*Newtons torque, and being further from the axis your would have to move faster just to achieve an equal angular velocity, so your ability to exert the torque is limited not by your hand's strength (you are easily able to exert a 30 N force) but eventually by hand speed).
Here's a definition of torque:
The moment arm of a force is the vector r from the axis of rotation to the point where the force is applied.
The magnitude of the torque is represented by the Greek letter tau; the torque is actually a vector so technically it needs to be represented as the vector tau. We aren't going to worry yet about its vector properties, but in order to provide a definition that will stand up to scrutiny, we will use the vector expression.
If the applied force is F then the torque is given by
tau = r X F.
The X actually represents a form of vector multiplication, something we aren't going to worry about just yet because we're going to simplify things a bit by assuming that r and F are perpendicular:
If r and F are perpendicular, then the torque is just
tau = r * F.
Examples: We have already seen examples of this calculation.
When in the preceding illustrations we exerted a force of 30 N at a distance of .1 m from the axis, the r vector ran from the axis to the point of application of the force. So the direction of r was parallel to the rod.
It was stated that the force was perpendicular to the rod.
So we were justified in using
tau = r * F = (.1 m) * (30 N) = 3.0 m * N.
r was also assumed to run along the rod when the force was applied at the .2 m position, and the force was still assumed perpendicular to the rod, so we had
tau = r * F = (.2 m) * (30 N) = 6.0 m * N.
Newton's Second Law
We are familiar with Newton's Second Law, which can be expressed for translational motion (i.e., point-to-point motion) in any of the following forms (the second and third forms being just algebraic rearrangements of the first):
For rotational motion, it isn't difficult to show that the following are completely equivalent to the translational form of the laws (the second and third forms again being just algebraic rearrangements of the first):
where I is the moment of inertia of a system about some axis, alpha is its angular acceleration about that axis, and tau is the net torque on the system.
Example: Suppose we exerted a 3.0 m * N net torque on the rod-and-domino system of our earlier example.
The net torque is 3.0 m * N, the moment of inertia is I = .0010 kg m^2, so the angular acceleration is
alpha = tau_net / I = 3.0 m * N / (.0010 kg m^2) = 3000 (m * N) / (kg * m^2).
It turns out that in this context, (m * N) / (kg * m^2) is equal to radians / sec^2, so our result is
alph = 3000 rad/sec^2.
The details of the unit calculation:
m * N / (kg * m^2) = m * kg m/s^2 / (kg * m^2). The kg and m all divide out, leaving the unit 1 / s^2.
So where does that 'radian' come from?
Remember that a radian is the angle for which the radius is equal to the arc distance on a circle. We could write our expression m * kg m/s^2 / (kg * m^2) as
(m / m) * (kg m/s^2) / (kg * m).
The kg m/s^2 / m becomes just 1/ s^2.
The (m / m) represents a meter of arc divided by a meter of radius (you'll probably just want to take my word for this), so it represents a radian.
Thus (m / m) * (kg m/s^2) / (kg * m) simplified to radian * (1 / s^2), which simplifies to rad / s^2.
Most students fail to understand this (in fact, it bothered me when I took the course but I never completely understood it), and end up simply understanding 'radian' as a phantom unit that appears when it's needed and disappears when it's not. You would do a little better by understanding that for some reason that's not all that easy to understand, (m / m) can give you radians when it's convenient, and m * radians will give you m.
Example: The strap-and-dominoes system in a preceding example coasts to rest while rotating through 8 radians in 4 seconds. What is the net torque acting on the system?
Solution: net torque = moment of inertia * angular acceleration; i.e.,
We have enough information to get a reasonable angular acceleration. We know angular displacement, we know the system comes to rest, and we know the time interval:
8 radians in 4 seconds implies an average angular velocity of 2 rad / sec.
Assuming constant angular acceleration, an average angular velocity of omega_Ave = 2 rad / s and a final angular velocity of omega_f = 0 rad/s imply an initial angular velocity of omega_0 = 4 rad / s.
The angular acceleration is therefore
alpha = (omega_f - omega_0) / `dt = (0 rad/s - 4 rad/s) / 4 s = - 1 rad/s^2.
We previously found that the moment of inertia of the system is .001 kg m^2, so we have
In this case kg m^2 / s^2 is taken to mean (m) * (kg m/s^2) = m * N, the unit of torque.
Direction of angular velocity
Angular velocity involves rotation about an axis. Every particle moves along a circular path, in a plane perpendicular to that axis. In that plane, the vector which represents the particle's direction of motion keeps changing. So there is no single vector in the plane which can represent the angular velocity. The only candidates for the direction of the angular velocity are therefore perpendicular to the plane, hence parallel to the axis of rotation.
The direction of the angular velocity vector is determined by the right-hand rule. Let the fingers of your right hand 'curl' in the direction of rotation. Your thumb points along the axis of rotation, in the direction of the angular velocity. (If you were to use your left hand, your thumb would again point along the axis, but in the opposite direction; the angular velocity is in the direction you get when you use your right hand).
Summary of rotational vs. translational analogies
The table below includes your familiar quantities for translational motion, and the analogous quantities for rotational motion. The reasoning by which the quantities for rotational motion are obtained from the basic definitions is identical at every step to the reasoning used for translational quantities. The equations which apply to rotational motion can be obtained from the equations for translational motion by simple substitution.
These equations are not given here in vector form. Most of these quantities (with the exception of mass, time, moment of inertia, work and energy) are in fact vector quantities. The vector nature of translational quantities is fairly easy to understand. We have seen above the vector nature of angular velocity, which is a little harder to visualize, and will consider the vector aspects of other rotational quantities in subsequent classes.
| translational (point-to-point in space) | rotational (rotating about a fixed axis) |
| basic quantities | basic quantities |
| position s or x in m | angular position theta in radians |
| velocity v in m/s | angular velocity omega in rad / sec |
| acceleration a in m/s^2 | angular acceleration alpha in rad/sec^2 |
| mass in kg | moment of inertia in kg * m^2 |
| force in Newtons | torque in meter*Newtons |
| basic definitions | basic definitions |
| vel is roc of position wrt clock time | ang vel is roc of ang position wrt clock time |
| accel is roc of vel wrt clock time | ang accel is roc of ang vel wrt clock time |
| immediate consequences of basic definitions | immediate consequences of basic definitions |
| v_Ave = `ds / `dt | omega_Ave = `dTheta / `dt |
| a_Ave = `dv / `dt | alpha_Ave = `dOmega / `dt |
| basic definitions if acceleration is uniform | basic definitions if acceleration is uniform |
| `ds = (v0 + vf) / 2 * `dt | `dTheta = (omega_0 + omega_f) / 2 * `dt |
| vf = v0 + a `dt | omega_f = omega_0 + alpha `dt |
| obtained by eliminating vf; then by eliminating `dt | obtained by eliminating omega_f; then by eliminating `dt |
| `ds = v0 `dt + 1/2 a `dt^2 | `dTheta = omega_0 `dt + 1/2 alpha `dt^2 |
| vf^2 = v0^2 + 2 a `ds | omega_f^2 = omega_0^2 + 2 alpha `dTheta |
| work, KE, impulse, momentum | work, KE, impulse, angular momentum |
| F = m * a | tau = I * alpha |
| KE = 1/2 m v^2 | KE = 1/2 I * omega^2 |
| `dW= F_ave * `ds | `dW = tau_ave * `dTheta |
| `dW_net = `dKE | `dW_net = `dKE |
| F_ave * `dt = `d( m v) | tau_ave * `dt = `d( I * omega) |
Brief experiment
Rotate a strap-and-domino system. Abruptly stop the strap, without interfering with the domino, and see in what direction the domino moves after it slides off the strap. This is the direction in which the domino was moving when the strap was stopped. The radial vector at the instant the system is stopped runs from the axis of rotation at the center of the strap to the center of the domino. The domino's velocity vector points in the direction of its motion as it slid off the strap. The radial and velocity vectors for the domino should be perpendicular.
Repeat and this time rotate the system at increasing velocity until the domino slides off the end of the strap. Observe the direction of its motion, and the direction of the radial vector at that instant. How do the two directions compare?
`q001. Describe what you observed in the 'brief experiment' above.
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When the strap is stopped abruptly the domino should come off with a velocity which is perpendicular to the strap at the instant the strap was stopped.
If the strap is accelerated until the domino comes off its end, the domino will not come off in a direction perpendicular to the strap. The velocity will have two components, one perpendicular to the strap and the other parallel to the strap.
In terms of vectors, as long as the domino is following its circular path its domino's velocity vector v is perpendicular to the radial vector r from the axis of rotation to the center of the domino. The radial vector is parallel to the strap, so the velocity vector is perpendicular to the strap. If the strap is suddenly stopped, the domino continues in the direction of the vector v.
The situation where the strap is accelerated until the domino slides off is more complicated:
At any instant the domino tends to move in a straight line (this according to Newton's First Law), parallel to the velocity vector. While in contact with the strap, the frictional force exerted on the domino toward the center of rotation continually diverts the domino from its straight-line path, resulting in its circular path.
As the original system is accelerated the centripetal force required to hold the domino in its circular path increases. This centripetal force is supplied by the static friction between domino and strap.
If the angular acceleration continues long enough, the required centripetal force will at some point exceed the maximum possible static friction, and the domino will begin sliding, now being subject to the force of kinetic friction. Kinetic friction is less than static friction, so the net force diverting the domino from its straight-line path will be less than before. Kinetic friction continues until the domino slides off the end. During the process it will have accelerated along the strap, towards the end of the strap. This added velocity is in the radial direction, so as it leaves the strap its velocity is a combination of its velocity perpendicular to the strap and its radial velocity parallel to the strap.
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`q002. Reasonably easy questions:
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The moment of inertia of each magnet is
- I_magnet = .030 kg * (.10 m)^2 = .0003 kg m^2.
There are two magnets so the total moment of inertia of the system is
- I_total = sum(individual moments) = .0002 kg m^2 + .0003 kg m^2 + .0003 kg m^2 = .0008 kg m^2.
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The required torque would be
tau = I * alpha = .0008 kg m^2 * 5 rad/s^2 = .004 kg m^2 * rad/s^2 = .004 m * N.
(units calculation: kg m^2 * rad / s^2 = (m * rad) * kg m / s^2 = m * kg m/s^2 = m * N.
Note, as explained earlier, that (m * rad) is the product of a meter of radius and a radian, which gives a meter of arc length.)
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Reasonable questions but challenging at this point:
Suppose that the magnets, positioned 10 cm from the axis of rotation, 'slip off' and hit you in the nose as soon as their centripetal acceleration reaches 10 000 cm/s^2.
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a_cent = v^2 / r, so
v = sqrt( a_cent * r ) = sqrt( 10 000 cm/s^2 * 10 cm) = 310 cm/s, approx.
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A radian of angle corresponds to 10 cm of arc. So 1 cm of arc corresponds to 1/10 rad.
The 310 cm/s velocity is along the arc.
v = 310 cm/s thus corresponds to
omega = 310 cm * (10 rad / cm) / s = 31 rad / s.
This illustrates the reason for the relationship
omega = v / r, which is equivalent to
v = r * omega.
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Commonsense reasoning:
5 rad/s^2 means the angular velocity increases by 5 rad/s every second.
- At this rate, it would take a little over 6 seconds to reach 31 rad/s.
The precise calculation:
- alpha = `dOmega / `dt so
- `dt = `dOmega / alpha = (31 rad/s - 0 rad/s) / (5 rad/s^2) = 6.2 s.
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310 cm/s is about 3 m/s, or about 7 mph. Not very fast, compared to a 98 mph fastball.
This corresponds to an easy underhand toss. Very unlikely to injure your nose, but it might hurt a little. Depends on how sensitive your nose is. You probably wouldn't want it to hit you in the eye or on a tooth, where it might cause injury.
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`q003. Imagine the last time you saw a Ferris wheel.
Standing in front of the wheel, was it rotating clockwise or counterclockwise (as you recall it; if you can't recall just pick one or the other and imagine it that way)?
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Standing in front of the wheel, how then would you describe the direction of its angular velocity?
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If you chose counterclockwise, then imagine reaching your right hand out so you fingers 'wrap around' the Ferris wheel in its direction of rotation. Your thumb will naturally point back toward you. The angular velocity vector is toward you. Fortunately this vector just represents the wheel's angular velocity; it has no translational velocity toward you.
If you chose clockwise, then to reach out and wrap your fingers around the wheel you would probably need to first walk around to the other side of the wheel, where you would find your thumb again pointing at you. From your original position, this vector would be away from you.
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Homework:
Your label for this assignment:
ic_class_091111
Copy and paste this label into the form.
Answer the questions posed above.
You have already seen most of the ideas in the qa's and Introductory Problem Set mentioned below. If you work through these documents as assigned, you will get plenty of practice and should develop good expertise with these concepts.
Do qa's #29 and 30 on radian measure and rotational motion.
It won't be assigned until next time, but if time permits consider doing qa #31 on torques and their effect on angular motion:
Introductory Problem Set 8 consists of 18 problems on angular motion, torques, moments of inertia, etc.. This won't be assigned until next week. If time permits you should take a look http://vhmthphy.vhcc.edu/ph1introsets/default.htm .