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View of the Gulf of Mexico from the Space Shuttle flying in a
nearly circular orbit around the earth.
Another type of motion that appears quite often in everyday life
is the motion of an object along a circular path. The path of
the object does not have to be a complete circle, even a car
going around a curve of constant radius constitutes circular
motion for the time the car is traveling the curve. As your
book explains, Newton was probably the first to describe a
special case of circular motion, that of
uniform circular motion, even though Christian
Huygens was the first to publish the correct mathematical
description of this kind of motion. As with most topics we
will cover, uniform circular motion makes certain simplifying
assumption to keep the mathematical description simple and
therefore focus us more on the physics phenomenon being studied.
In this case the description is of an object with constant mass
that travels in a perfectly circular path with uniform,
i.e. constant magnitude, velocity. To be circular, the path
must have a constant distance or radius from the center of
the circle, so the description of the direction of the velocity
is particularly simple: its magnitude is constant and its direction
is always tangential to the circular path at every position
of the object. With that description in words, we can readily
write down a mathematical description of where the particle is
as a function of time. First, the position of the particle must
be
It's obvious that theta is a function of time, hence, if we are
to calculate the instantaneous velocity, we will have to
keep in mind that d(theta)/dt is not zero. The displacement
can just be specified in terms of the position as a function of
time if we remember that we are looking at change relative to
the initial position (x0, y0).
This does give the expected behavior for counterclockwise motion in that the change in x is negative if 0 < theta < Pi. Only when we reach the most negative x position does counterclockwise motion turn around and head into the positive x direction. At theta = Pi, the y direction of motion is also down in the negative y direction as predicted by our velocity formula. We also note that the derivative of the displacement should be tangent to the path of displacement and that this also fits our notion that the velocity of the mass m particle should be tangent to the circular path.
However, what is d(theta)/dt? As it turns out, we can relate d(theta)/dt to the tangential velocity by going back to the definition for radians. First, we have to remember (Stewart textbook, pg. A24) that radian measure of an angle theta for a circular path is given by the relationship s = r * theta where s is the arc-length of an arc on a circle of radius r where the arc subtends an angle theta. We can ask Maple to draw out for us the length as a function of theta to show visually what this formula s = r*theta means. Check the Maple file.
So we can solve for theta and get theta = s/r and hence d(theta)/dt = 1/r * ds/dt since we have r as a constant with time. Now we spot that since s is the distance traveled by the particle around the perimeter of the circle, ds/dt must be v, the tangential velocity! Since we are going around the circular path with a fixed speed, the period is fixed and the angular velocity is usually defined as omega = d(theta)/dt. In this case, omega is constant with time. We can derive results that will be useful later by spotting the relationship, omega = v/r. We can also relate omega to the period (which we will call T) since, by definition, T = 2*Pi*r/v ==> v = 2*Pi*r/T ==> omega = v/r = 2*Pi/T. This is exactly what we expect. The particle takes time T to make an angular displacement of 2*Pi so the angular velocity should be 2*Pi/T.
The next thing we want to determine is the direction of the velocity. We know that the direction constantly changes for something that moves on a circular path and we know that the magnitude of the acceleration stays constant, but what is the direction of the velocity at any given instant? If we make a drawing of the velocity on a graph, we'd have to note that, assuming omega is positive, our previous derivatives show that the x component of the velocity depends on sin(theta) while the y component depends on cos(theta). If we draw this on the standard x-y coordinate system, we'd have to show the velocity vector as follows:
In relation to the position vector, r, you can see that the velocity must be at right angles to r, thus the instantaneous velocity's direction is at an angle of (90° - theta) + theta = 90° with respect to the direction of r and therefore is always perpendicular to r at any point in the particle's path around the circle.
The only thing left to fully describe the motion is to describe how
it changes. In this case only the direction changes, but we still
need to describe it mathematically. Again we resort to the definition
of instantaneous acceleration, namely
Remember that d(theta)/dt is constant in time so the derivative of this quantity with respect to time is zero. We can shorten the expressions a little by replacing d(theta)/dt with omega, but that still leaves us with a need to interpret what we got. For example, what is the direction of the acceleration as theta changes? Well, notice that the position vector for mass m is given by the vector r(t) which has components x(t) = r cos(theta) and y(t) = r sin(theta) at any given time t. We can find the value of theta as a function of time since theta should equal omega * t if omega is constant. If we look at the components of our acceleration, we see that ax = -x(t)*omega2 and ay = -y(t)*omega2, in other words, the acceleration has components that point opposite to the components of the position vector. The only conclusion we can reach is that the direction of the vector a(t) must always be opposite to the direction of the position vector r(t), so a(t) points inward along the radial line connecting the particle and the center of the circular path.
What about the magnitude of a(t)? That is easier to see
We also note that the magnitude of a(t) is constant with
time since r and omega are constant. Using our earlier relation,
we can state the centripetal acceleration
in its more familiar form
Therefore, for any object to travel in a uniform circular path, it must be subjected to a force which
Unless both these conditions are true, the particle is not travelling with uniform circular motion.
Let's do some examples where we make use of these ideas.
Solution:
Start with a picture:
The view shown is from above the mass. Since the tension is the only force acting in the horizontal plane, it must satisfy the centripetal condition of giving an acceleration which points always toward the center of the circle and has a magnitude of v2/r. Therefore,
| T | = mv2/r |
| = (2 kg)(10 m/s)2/ (0.5 m) | |
| T | = 400 Nt |
Solution:
We use the same picture, same setup. But now we solve for v
instead of T.
| T | = mv2/r ==> |
| v | = sqrt[(r*T)/m] |
| so | |
| vmax | = sqrt[(r*Tmax)/m] |
| = sqrt[(0.5 m)(500 Nt)/ (2 kg)] | |
| vmax | = 11.2 m/s |
Solution:
Draw a picture.
The forces acting on the penny are shown in the figure above. Note that the only force that can act in the horizontal direction is friction. We want static friction since we assume the penny does not move. To make this happen for the minimum coefficient of static friction, we want the static friction to satisfy the centripetal condition by assuming its maximum value. So
| fs, max | = mv2/r |
| muk, minFN | = mv2/r |
| muk, minmg | = mv2/r ==> |
| muk, min | = v2/(g*r) |
| = ((45 revs./min.)(1 min./60 s) (2*Pi*r/rev.))2/(g*r) | |
| = ((¾ )2(4 Pi2) (0.1 m)s-2)/(9.8 m/s2) | |
| muk, min | = 023 |
Solution:
Draw a picture.
The free-body diagram of the car at the top of the loop shows that the normal force and the gravitational force both point down. This may seem counter-intuitive. If both forces point downward, why doesn't the car immediately start accelerating downward and away from the track? The answer is: you mixed two ideas in the last sentence. The car does immediately start accelerating in the down direction, but it does not have to fall away from the track immediately because it already has some velocity to the left. You can't instantly change from going to the left to falling straight down (at least not without an infinite amount of force being applied), so, if things adjust themselves correctly, the forces lead to a change in the direction of the car's velocity but do not change the speed nor cause the car to lose contact with the track. Assuming that's true, we would say that we want the net force to satisfy the centripetal condition. Certainly, at the top of the track, the net force points straight down and that is toward the center of the circular path (if you are exactly above the center). So the directions are OK. The magnitude of the net force has to satisfy the condition:
| -FN - mg | = -mv2/R ==> |
| FN + mg | = mv2/R ==> |
where we note that the magnitude of the centripetal condition has a negative sign in front so as to be consistent with pointing down, which is the negative direction. Now let's ask what happens. The car tends to want to continue straight along rather than curving downward to follow the track. The track must respond by producing a normal force who's magnitude is strong enough to maintain the shape of the track and the position of the car relative to it. That means that the track adjusts FN so that it's magnitude satisfies the above equation. If the speed is high, FN just increases until the equation works. If the speed is low, then FN decreases. The smallest possible value for FN (which after all is a magnitude) is 0. So the speed must not fall below the value which has FN equal to zero else the equation can't be satisfied and, by definition of what we mean by normal force, if the normal force goes to zero, the track and car are no longer touching, so the car must leave the track at this point. Thus,
| FN, min + mg | = mvmin2/R |
| 0 + mg | = mvmin2/R ==> |
| vmin | = sqrt[gR] |
The final example can be seen through a digital video/audio file. Enjoy!
Next: Rotational Motion
Up: COMPLETING THE CIRCLE
Previous: Vectors
larryg@upenn5.hep.upenn.edu
A plane in circular motion