10-25-99

Sections 7.8 - 8.3

Center of gravity

The center of gravity of an object is the point you can suspend the object from without there being any rotation because of the force of gravity, no matter how the object is oriented. If you suspend an object from any point, let it go and allow it to come to rest, the center of gravity will lie along a vertical line that passes through the point of suspension. Unless you've been exceedingly careful in balancing the object, the center of gravity will generally lie below the suspension point.

The center of gravity is an important point to know, because when you're solving problems involving large objects, or unusually-shaped objects, the weight can be considered to act at the center of gravity. In other words, for many purposes you can assume that object is a point with all its weight concentrated at one point, the center of gravity.

For any object, the x-position of the center of gravity can be found by considering the weights and x-positions of all the pieces making up the object:

 

A similar equation would allow you to find the y position of the center of gravity.

The center of mass of an object is generally the same as its center of gravity. Very large objects, large enough that the acceleration due to gravity varies in different parts of the object, are the only ones where the center of mass and center of gravity are in different places.
Neat facts about the center of gravity

Fact 1 - An object thrown through the air may spin and rotate, but its center of gravity will follow a smooth parabolic path, just like a ball.

Fact 2 - If you tilt an object, it will fall over only when the center of gravity lies outside the supporting base of the object.

Fact 3 - If you suspend an object so that its center of gravity lies below the point of suspension, it will be stable. It may oscillate, but it won't fall over.
Rotational variables

We'll now switch the focus from straight-line motion to rotational motion. If you can do one-dimensional motion problems, which involve straight-line motion, then you should be able to do rotational motion problems, because a circle is just a straight line rolled up. To solve rotational kinematics problems, a set of four equations is used; these are essentially the one-dimensional motion equations in disguise.

If you spin a wheel, and look at how fast a point on the wheel is spinning, the answer depends on how far away the point is from the center. Velocity, then, isn't the most convenient thing to use when you're dealing with rotation, and for the same reason neither is displacement, or acceleration; it is often more convenient to use their rotational equivalents. The equivalent variables for rotation are angular displacement (angle, for short); angular velocity , and angular acceleration . All the angular variables are related to the straight-line variables
by a factor of r, the distance from the center of rotation to the point
you're interested in.

Although points at different distances from the center of a rotating
wheel have different velocities, they all have the same angular
velocity, so they all go around the same number of revolutions per
minute, and the same number of radians per second. Angles (angular
displacements, that is) are generally measured in radians, which is the
most convenient unit to work with. A radian is an odd unit in physics,
however, because it is treated as being unitless, and is often put in
or taken out whenever it's convenient to do so.

It is helpful to recognize the parallel between straight-line
motion and rotational motion. Writing down the four rotational
kinematics equations reinforces that. Any equation dealing with
rotation can be found from its straight-line motion equivalent by
substituting the corresponding rotational variables.

The straight-line motion kinematics equations apply for
constant acceleration, so it follows that the rotational kinematics
equations apply when the angular acceleration is constant. The
equations should look familiar to you:

The equations are the same as the constant-acceleration equations for 1-D motion, substituting the rotational equivalents of the straight-line motion variables.
A rotational example

Consider an example of a spinning object to see how the rotational kinematics equations are applied. Imagine a ferris wheel that is rotating at the rate of 1 revolution every 8 seconds. The operator of the wheel decides to bring it to a stop, and puts on the brake; the brake produces a constant deceleration of 0.11 radians/s2.

(a) If your seat on the ferris wheel is 4.2 m from the center of the wheel, what is your speed when the wheel is turning at a constant rate, before the brake is applied?
(b) How long does it take before the ferris wheel comes to a stop?
(c) How many revolutions does the wheel make while it is coming to a stop?
(d) How far do you travel while the wheel is slowing down?

(a) The wheel is rotating at a rate of 1 revolution every 8 seconds, or 0.125 rev/s. This is the initial angular velocity. It is often most convenient to work with angular velocity in units of radians/s; doing the conversion gives:

Your speed is simply this angular velocity multiplied by your distance from the center of the wheel:

(b) We've calculated the initial angular velocity, the final angular
velocity is zero, and the angular acceleration is -0.11 rad/s2. This allows the stopping time to be found:

(c) To find the number of revolutions the wheel undergoes in this 7.14 seconds, one way to do it is to use the equation:

This can be converted to revolutions:

(d) To figure out the distance you traveled while the wheel was slowing
down. the angular displacement (in radians) can be converted to a
displacement by multiplying by r:

Tangential velocity; tangential acceleration

In uniform circular motion (motion in a circle at a constant speed), there is always a net acceleration (the centripetal acceleration) towards the center of the circular path. In non-uniform circular motion the speed is not constant, and there are two accelerations, the centripetal acceleration towards the center of the circle, and the tangential acceleration. The tangential acceleration is tangent to the circle, pointing in the direction the object is traveling if the object is speeding up, and the opposite way if the object is slowing down. 

The two vector diagrams show an object undergoing uniform circular motion (constant angular velocity), and an object experiencing non-uniform circular motion (varying angular velocity). For uniform circular motion, the centripetal acceleration points towards the center of the circle, and the velocity points in the direction the object is traveling. This is tangent to the circular path, so we call it the tangential velocity. For non-uniform circular motion, the centripetal acceleration and tangential velocity are still there, and there is also a tangential acceleration in the direction the object is traveling. The net acceleration is the vector sum of the centripetal and tangential accelerations. Just as we separate everything out into x and y components when analyzing a projectile motion question, we can always separate things out into the tangential and radial (towards the center) directions in non-uniform circular motion.

Note that the centripetal acceleration is connected to the tangential velocity, through the usual v2 / r relationship, while the tangential acceleration is connected to any change in the tangential speed.
Rolling

When an object such as a wheel or a ball rolls, it does not slip where it makes contact with the ground. With a car or a bicycle tire, there is friction between the tire and the road, and if the tire is rolling then the frictional force is a static force of friction. This is because there is no slipping, so the point on the tire in contact with the road is instantaneously at rest.

This is somewhat counter-intuitive, but it comes about because the
velocity of each point on the tire is a sum of the linear velocity
associated with the car (or bike) moving, and the rotational velocity
associated with the tire rolling. For a point on the outside of the
tire, the rotational speed happens to be equal to the linear speed of
the car: this is because each time the tire makes a complete
revolution, the car will have traveled a distance equal to the
circumference of the tire, so the linear distance and the rotational
distance are the same for the same time interval. For a point on the
top of a tire, the two velocities are in the same direction, so the
total velocity at the top of a tire is twice the linear velocity of the
car; for a point at the bottom of a tire, the two velocities are in
opposite directions, so the total velocity is zero there.

Consider a bike on a flat road. You climb on, and start
pedaling, and the bike accelerates forward, with both tires rolling
along the road. As the bike accelerates, which way does friction act?
The answer depends on which tire you consider. Think about what would
happen if there was no friction between the tires and the road. When
you pedal, the chain causes the rear wheel to spin. With no friction,
the rear tire would spin on the road and the bike wouldn't move.
Friction opposes this tendency, so it points in the direction you're
accelerating on the bike; it's static friction, because the tire does
not slip, it rolls.

The front tire, on the other hand, is not being spun by the
chain, so with no friction it wouldn't spin at all. Friction is what
makes it spin, then, so it must point opposite to the way the bike is
accelerating, and, again, it's static friction because the tire does
not slip on the road.

Once you've accelerated the bike and you're going at a constant
speed, the frictional forces don't have to be as large. The friction on
the rear tire has to provide enough force to overcome the resistance
forces (rolling resistance, air resistance, friction in the wheel
bearings) tending to slow you and the bike down. The friction on the
front tire has to do even less, because all it has to do is keep the
front tire spinning at a constant rate.

 

 

11-1-99

Sections 8.7 - 8.9

Rotational work and energy

Let's carry on madly working out equations applying to rotational motion by substituting the appropriate rotational variables into the straight-line motion equations. Work is force times displacement, so for rotation work must be torque times angular displacement:

A torque applied through a particular angular displacement does
work. If the object is rotating clockwise and the torque is a clockwise
torque, the work is positive; a counter-clockwise torque applied to a
clockwise rotating object does negative work.

What about kinetic energy? A spinning object has rotational kinetic energy:

A rolling object has both translational and rotational kinetic energy.

Angular momentum

To finish off our comparison of translational (straight-line) and rotational motion, let's consider the rotational equivalent of momentum, which is angular momentum. For straight-line motion, momentum is given by p = mv. Momentum is a vector, pointing in the same direction as the velocity. Angular momentum has the symbol L, and is given by the equation:

 

Angular momentum is also a vector, pointing in the direction of the angular velocity.

In the same way that linear momentum is always conserved when there is no net force acting, angular momentum is conserved when there is no net torque. If there is a net force, the momentum changes according to the impulse equation, and if there is a net torque the angular momentum changes according to a corresponding rotational impulse equation.

Angular momentum is proportional to the moment of inertia, which depends on not just the mass of a spinning object, but also on how that mass is distributed relative to the axis of rotation. This leads to some interesting effects, in terms of the conservation of angular momentum.

A good example is a spinning figure skater. Consider a figure skater who starts to spin with their arms extended. When the arms are pulled in close to the body, the skater spins faster because of conservation of angular momentum. Pulling the arms in close to the body lowers the moment of inertia of the skater, so the angular velocity must increase to keep the angular momentum constant.

Parallels between straight-line motion and rotational motion

Let's take a minute to summarize what we've learned about the parallels between straight-line motion and rotational motion. Essentially, any straight-line motion equation has a rotational equivalent that can be found by making the appropriate substitutions (I for m, torque for force, etc.). 

Example - Falling down

You've climbed up to the top of a 7.5 m high telephone pole. Just as you reach the top, the pole breaks at the base. Are you better off letting go of the pole and falling straight down, or sitting on top of the pole and falling down to the ground on a circular path? Or does it make no difference?

The answer depends on the speed you have when you hit the ground. The speed in the first case, letting go of the pole and falling straight down, is easy to calculate using conservation of energy: 

In the second case, also apply conservation of energy. If you have
negligible mass compared to the telephone pole, just work out the
angular velocity of the telephone pole when it hits the ground. In this
case we use rotational kinetic energy, and the height involved in the
potential energy is half the length of the pole (which we can call h),
because that's how much the center of gravity of the pole drops. So,
for the second case:

For a uniform rod rotating about one end, the moment of inertia is 1/3 mL2. Solving for the angular velocity when the pole hits the ground gives:

For you, at the end of the pole, the velocity is h times the angular velocity, so:

So, if you hang on to the pole you end up falling faster than if you'd fallen under the influence of gravity alone. This also means that the acceleration of the end of the pole, just before the pole hits the ground, is larger than g (1.5 times as big, in this case), which is interesting.

Which way do these angular variables point, anyway?

Displacement is a vector. Velocity is a vector. Acceleration is a vector. As you might expect, angular displacement, angular velocity, and angular acceleration are all vectors, too. But which way do they point? Every point on a rolling tire has the same angular velocity, and the only way to ensure that the direction of the angular velocity is the same for every point is to make the direction of the angular velocity perpendicular to the plane of the tire. To figure out which way it points, use your right hand. Stick your thumb out as if you're hitch-hiking, and curl your fingers in the direction of rotation. Your thumb points in the direction of the angular velocity.

If you look directly at something and it's spinning clockwise, the angular velocity is in the direction you're looking; if it goes counter-clockwise, the angular velocity points towards you. Apply the same thinking to angular displacements and angular accelerations. 

 

 

11-3-99

Sections 9.1 - 9.3

Objects in equilibrium

We've talked about equilibrium before, stating that an object is in equilibrium when it has no net force acting on it. This definition is incomplete, and it should be extended to include torque. An object at equilibrium has no net force acting on it, and has no net torque acting on it.

To see how the conditions are applied, let's work through a couple of examples.

Example 1

The first example will make use of the hinged rod supported by a rope, as discussed above. The rod has a mass of 1.4 kg, and there is an angle of 34° between the rope and the rod.
(a) What is the tension in the rope?
(b) What are the two components of the support force exerted by the hinge?

The free-body diagram is shown below, with the support force provided by the hinge split up into x and y components. If you aren't sure which way such forces go, simply guess, and if you guess wrong you'll just get a negative sign for that force. 

Something we'll assume in this example is that the rod is uniform, so
the weight acts at the center of the rod (the center is the center of
mass, in other words). As usual, sum the forces in the x and y
directions:

There are too many unknowns here, and this is why summing the torques
can be so useful. To sum torques, choose a point to take torques
around; a sensible point to choose is one that one or two unknown
forces go through, because they will nt appear in the torque equation.
In this case, choosing the hinge as the point to take torques around
eliminated both components of the support force at the hinge. As with
forces, where you choose plus and minus directions, choose a positive
and negative direction for torques. In this case, let's make
counter-clockwise negative and clockwise positive.

This can be solved for T, the tension in the rope. Note that r, which represents the length of the rod, cancels out:

This can be substituted back into the force equations to find the components of the hinge force:

Example 2 - a step-ladder

A step-ladder stands on a frictionless horizontal surface, with just the crossbar keeping the ladder standing. The mass is 20 kg; what is the tension in the crossbar?

This is something of a tricky problem, because you have to draw the free-body diagram of the entire ladder to figure out the normal forces, and then draw the free-body diagram of one half of the ladder to complete the solution. This is also what makes it a good example to look at, however. 

 

Consider first the free-body diagram of the entire ladder. The floor is
frictionless, so there are no horizontal forces exerted by the floor.
The ladder is uniform, so the weight acts at the center of mass, which
is halfway up the ladder and halfway between the two legs. Summing
forces in the y-direction gives:

One way to approach this is to say that the ladder is symmetric, and
there is no reason for the two normal forces to be different; each one
should be equal to half the weight of the ladder. If you don't like
this argument, simply take torques about one of the points where the
ladder touches the floor. This will give you an equation saying that
one normal force is equal to half the ladder's weight, so the other
normal force must be equal to half the weight, too. Either way, you
should be able to show that:

Now consider the free-body diagram of the left-hand side of the ladder.
I'll attach a 1/2 as a subscript to the mass, to remind us that the
mass of half the ladder is half the mass of the entire ladder.

Taking torques around the top of the ladder eliminates the unknown
contact force (F) coming from the other half of the ladder, and gives
(this time taking clockwise to be positive):

This can be solved to find the tension in the crossbar:

 

10-27-99

Sections 8.4 - 8.6

Torque

We've looked at the rotational equivalents of displacement, velocity, and acceleration; now we'll extend the parallel between straight-line motion and rotational motion by investigating the rotational equivalent of force, which is torque.

To get something to move in a straight-line, or to deflect an object traveling in a straight line, it is necessary to apply a force. Similarly, to start something spinning, or to alter the rotation of a spinning object, a torque must be applied.

A torque is a force exerted at a distance from the axis of rotation; the easiest way to think of torque is to consider a door. When you open a door, where do you push? If you exert a force at the hinge, the door will not move; the easiest way to open a door is to exert a force on the side of the door opposite the hinge, and to push or pull with a force perpendicular to the door. This maximizes the torque you exert.

I will state the equation for torque in a slightly different way than
the book does. Note that the symbol for torque is the Greek letter tau.
Torque is the product of the distance from the point of rotation to
where the force is applied x the force x the sine of the angle between
the line you measure distance along and the line of the force:

http://buphy.bu.edu/~duffy/PY105/16c.GIF

In a given situation, there are usually three ways to determine the
torque arising from a particular force. Consider the example of the
torque exerted by a rope tied to the end of a hinged rod, as shown in
the diagram.

The first thing to notice is that the torque is a counter-clockwise
torque, as it tends to make the rod spin in a counter-clockwise
direction. The rod does not spin because the rope's torque is balanced
by a clockwise torque coming from the weight of the rod itself. We'll
look at that in more detail later; for now, consider just the torque
exerted by the rope.

There are three equivalent ways to determine this torque, as shown in the diagram below.

 

 

Method 1 - In method one, simply measure r from the hinge along the
rod to where the force is applied, multiply by the force, and then
multiply by the sine of the angle between the rod (the line you measure
r along) and the force.

Method 2 - For method two, set up a right-angled triangle, so
that there is a 90° angle between the line you measure the distance
along and the line of the force. This is the way the textbook does it;
done in this way, the line you measure distance along is called the
lever arm. If we give the lever arm the symbol l, from the right-angled
triangle it is clear that

Using this to calculate the torque gives:

Method 3 - In this method, split the force into components,
perpendicular to the rod and parallel to the rod. The component
parallel to the rod is along a line passing through the hinge, so it is
not trying to make the rod spin clockwise or counter-clockwise; it
produces zero torque. The perpendicular component (F sinq) gives plenty
of torque, the size of which is given by:

Any force that is along a line which passes through the axis of rotation produces no torque. Note that torque is a vector quantity, and, like angular displacement, angular velocity, and angular acceleration, is in a direction perpendicular to the plane of rotation. The same right-hand rule used for angular velocity, etc., can be applied to torque; for convenience, though, we'll probably just talk about the directions as clockwise and counter-clockwise.

oment of inertia

We've looked at the rotational equivalents of several straight-line motion variables, so let's extend the parallel a little more by discussing the rotational equivalent of mass, which is something called the moment of inertia.

Mass is a measure of how difficult it is to get something to move in a straight line, or to change an object's straight-line motion. The more mass something has, the harder it is to start it moving, or to stop it once it starts. Similarly, the moment of inertia of an object is a measure of how difficult it is to start it spinning, or to alter an object's spinning motion. The moment of inertia depends on the mass of an object, but it also depends on how that mass is distributed relative to the axis of rotation: an object where the mass is concentrated close to the axis of rotation is easier to spin than an object of identical mass with the mass concentrated far from the axis of rotation.

The moment of inertia of an object depends on where the axis of rotation is. The moment of inertia can be found by breaking up the object into little pieces, multiplying the mass of each little piece by the square of the distance it is from the axis of rotation, and adding all these products up: 

Fortunately, for common objects rotating about typical axes of rotation, these sums have been worked out, so we don't have to do it ourselves. A table of some of these moments of inertia can be found on page 223 in the textbook. Note that for an object where the mass is all concentrated at the same distance from the axis of rotation, such as a small ball being swung in a circle on a string, the moment of inertia is simply MR2 . For objects where the mass is distributed at different distances from the axis of rotation, there is some multiplying factor in front of the MR2.

Newton's second law for rotation

You can figure out the rotational equivalent of any straight-line motion equation by substituting the corresponding rotational variables for the straight-line motion variables (angular displacement for displacement, angular velocity for velocity, angular acceleration for acceleration, torque for force, and moment of inertia for mass). Try this for Newton's second law:

Replace force by torque, m by I, and acceleration by angular acceleration and you get:

 Example - two masses and a pulley

We've dealt with this kind of problem before, but we've never accounted for the pulley. Now we will. There are two masses, one sitting on a table, attached to the second mass which is hanging down over a pulley. When you let the system go, the hanging mass is pulled down by gravity, accelerating the mass sitting on the table. When you looked at this situation previously, you treated the pulley as being massless and frictionless. We'll still treat it as frictionless, but now let's work with a real pulley.

A 111 N block sits on a table; the coefficient of kinetic friction between the block and the table is 0.300. This block is attached to a 258 N block by a rope that passes over a pulley; the second block hangs down below the pulley. The pulley is a solid disk with a mass of 1.25 kg and an unknown radius. The rope passes over the pulley on the outer edge. What is the acceleration of the blocks?

 As usual, the first place to start is with a free-body diagram of each
block and the pulley. Note that because the pulley has an angular
acceleration, the tensions in the two parts of the rope have to be
different, so there are different tension forces acting on the two
blocks.

Then, as usual, the next step is to apply Newton's second law and write
down the force and/or torque equations. For block 1, the force
equations look like this:

For block 2, the force equation is:

The pulley is rotating, not moving in a straight line, so do the sum of the torques:

Just a quick note about positive directions...you know that the
system will accelerate so that block 2 accelerates down, so make down
the positive direction for block 2. Block 1 accelerates right, so make
right the positive direction for block 1, and for the pulley, which
will have a clockwise angular acceleration, make clockwise the positive
direction.

We have three equations, with a bunch of unknowns, the two
tensions, the moment of inertia, the acceleration, and the angular
acceleration. The moment of inertia is easy to calculate, because we
know what the pulley looks like (a solid disk) and we have the mass and
radius:

The next step is to make the connection between the angular
acceleration of the pulley and the acceleration of the two blocks.
Assume the rope does not slip on the pulley, so a point on the pulley
which is in contact with the rope has a tangential acceleration equal
to the acceleration of any point on the rope, which is equal to the
acceleration of the blocks. Recalling the relationship between the
angular acceleration and the tangential acceleration gives:

Plugging this, and the expression for the moment of inertia, into the torque equation gives:

All the factors of r, the radius of the pulley, cancel out, leaving:

Substituting the expressions derived above for the two tensions gives:

This can be solved for the acceleration:

Accounting for the mass of the pulley just gives an extra term in the
denominator. Plugging in the numbers and solving for the acceleration
gives:

Neglecting the mass of the pulley gives a = 5.97 m/s2.