EN4: Dynamics and Vibrations
Division of Engineering
Brown University
2.6.4 Examples using Cylindrical Polar Coordinates
Example I: At a particular time, the polar coordinates of a point P moving in the (x,y) plane are r=4ft,
, and their time derivatives are
and
.
(a) What is the magnitude of the velocity of P?
The picture reminds you of the conventions associated with cylindrical polar coordinates. Recall our general expression for velocity in cylindrical polar coords
We are told that the particle moves in the (x,y) plane, so that dz/dt=0. Substituting the remaining values shows that
(b) What are the Cartesian components of the velocity of P?
We will solve this problem using two different approaches. First, note that we can convert our expression for velocity in to {i,j,k} components directly, be resolving each basis vector into components of the new basis. The picture below helps show how to do this
Note that all the basis vectors have unit length. Therefore
Substitute these results into our expression for v to see that
We can obtain the same result directly. Start by writing the position vector as
Differentiate to find the velocity
Substituting the numerical values given in the question shows that
as before.
Example 2: A boat seaching for underwater archaeology sites in the Aegean Sea moves at 4 knots and follows the path
m, where
is in radians. (a knot is one nautical mile, or 1852 meters per hour). When
radians, determine the boat’s velocity, in terms of polar coordinates.
Recall the general expression for the velocity of a particle in cylindrical-polar coordintes:
Here, dz/dt=0 because the boat remains in the (x,y) plane. We somehow need to determine the values of dr/dt and
from the information given in the question.
We are given that
Hence, the velocity may be expressed as
We are also given the magnitude of the velocity (4 knots); therefore, taking the magnitude of v and simplifying:
We can now substitute back in our expression for v to see that
Example 3: Compute the boat’s acceleration in terms of polar coordinates.
Recall that in cylindrical polar coodinates, the acceleration is
We know most of the values of the parameters in this expression, but we need to determine
.
In the preceding problem, we showed that
Hence
We now know all the data to determine the acceleration:
To check this result, we note that since the boat is moving at constant speed, its tangential component of acceleration (that is to say, the component of acceleration parallel to its path) must vanish. Recall that the boat’s velocity is tangent to its path, so this means that
You can evaluate this dot product to see that with our numbers this does indeed work out.
Example 4: The bar rotates in the x-y plane with constant rotation rate
. The radial component of acceleration of the collar C is
, where K is a constant. (It turns out that K=k/m, where k is the spring stiffness and m is the mass of the collar) When
the radial component of velocity of C is
. Determine the radial and transverse components of the velocity of C as a function of r.
Here is the general expression for acceleration in a rotating cylindrical-polar coordinate system, this time relating velocity components to acceleration components
We are given the radial component of acceleration and the angular velocity. Therefore
Now, we somehow need to integrate this expression to obtain a relationship between radial velocity and position. To do this, recall our `alternative expression for acceleration’ from straight line motion
Hence, substituting for dv/dt and rearranging
Integrate, using the given initial conditions
This is actually quite an interesting result. We observe that for
, the radial velocity decreases with r – as the collar moves away from r=0, it slows down, and will eventually reverse direction. For
, however, the radial velocity increases with r. This means that as the collar moves away from r=0, it picks up speed an eventually flies off the end of the rod.
To see this more clearly, we can compute r as a function of time. To keep things simple, let’s suppose that
(you can do the calculations for arbitrary initial velocity, but the algebra gets messy). From our expression for radial velocity
The following result is helpful (you should be able to do the integrals for yourself)
Hence, evaluating the integrals and simplifying shows that
For the first solution, r oscillates between
. For the second solution, r increases exponentially with time.
This is a very simple example of a dynamic instability. If the rod rotates slowly, the collar is stable at r=0. If the rod rotates too fast, the collar will fly off the end of the rod.
Dynamic instabilities like this are usually a bad thing, but sometimes they can be put to good use. For example, this phenomenon could be used as a safety device to regulate the speed of the rotating rod. If you monitored the electrical connection between the collar and the rotating rod, you would find that the connection was broken when the rod spun too fast, and could use this to shut down power. Many very clever speed regulating devices are based on this idea – we will discuss some of them shortly.
Example 5: A 2kg mass rests on a flat horizontal bar. The bar begins rotating in the vertical plane about O with a constant angular acceleration of 1 rad/s/s. The mass is observed to slip relative to the bar when the bar is 30 degrees above the horizontal. What is the static coefficient of friction between the mass and the bar? Does the mass slip toward or away from O?
We will solve this problem by applying Newton’s second law F=ma. We will assume that the mass is not slipping on the bar. If this is the case, we may determine its acceleration, and then deduce the force acting on the mass. The condition that the mass is about to slip when
will then tell us the friction coefficient.
Begin by working out the acceleration. The general result is
Here, we know r is constant (the mass is not slipping) and we are given
. We need to determine
, preferably as a function of
. To do this, recall the angular acceleration-angular velocity relation
where we have noted that the angular acceleration is constant. Substitute into the expression for a to see that
Next, turn to forces. The picture below shows a free body diagram for the mass
In polar coordinates, the force may be written:
Hence, Newton’s second law requires that
Hence, comparing components:
(Note that
must be in radians when you substitute numbers) If the mass is on the point of slipping, then
To decide which way the mass will slip, we merely need to look at the direction of the friction force. We have found that T is positive, and so acts away from O. The mass must be on the point of slipping in the opposite direction (friction force always opposes sliding). Therefore, the mass must slip towards O.
Finally, we conclude with a few applications where rotational motion is being used in a clever way.
Window blinds. Have you ever wondered how window shades work? You give the shade a little downward jerk, let it go, and it winds itself up. If you pull the shade down slowly, it stays down. The screen for the video projector in Barus-Holley 166 works the same way.
The picture below shows the neat little mechanism (which probably only costs a few dollars to manufacture) that achieves this remarkable feat of engineering.
The picture shows an enlarged end view of the window shade. The hub, shown in brown, is fixed to the bracket supporting the shade and cannot rotate. The drum, shown in peach, rotates as the shade is pulled up or down. The drum is attached to a torsional spring, which tends to cause the drum to rotate counterclockwise, so winding up the shade. The rotation is prevented by the small dogs, shown in red, which engage with the teeth on the hub. You can pull the shade downwards freely, since the dogs allow the drum to rotate counterclockwise.
To raise the shade, you need to give the end of the shade a jerk downwards, and then release it. When the drum rotates sufficiently quickly (we will calculate how quickly shortly) the dogs open up, as shown below. They remain open until the drum stops rotating, at which point the topmost dog drops and engages with the teeth on the hub, thereby locking up the shade once more.
We will estimate the critical rotation rate required to free the rotating drum. Idealize athe topmost dog as a particle on the end of a massless, inextensible rod, as shown below. Assume that the drum rotates at constant angular rate
. Find the rotation rate at which the dog will engage with the teeth on the hub.
From our general result, the acceleration of the particle is
The picture below shows a free body diagram for the particle. Three forces act on the particle: a reaction force N from the edge of the drum; a gravitational force and a reaction force P due to the rigid link. We have neglected friction.
The resultant force is
Apply Newton’s second law
Compare components to see that
The dog drops if
, i.e.
The Flyball Governor: This is another speed-regulating device that operates on the same principle. It was particularly common on steam engines, but is still used today where an inexpensive mechanical speed regulator is required.
The picture shows a very simple version of a flyball governor. The assembly is connected to the steam engine through a set of gears, which are not shown. It rotates about the steam pipe at an angular rate proportional to the engine speed. The lower collar slides freely up and down the steam feed pipe. We will see that as the rate of rotation of the governor increases, the collar is pulled up the pipe, and so opens up the hole. Less steam reaches the engine so it slows down. If the rate of rotation drops, the hole is closed up, so the engine speeds up. If properly designed, the governor will force the engine to run at a fixed speed (we need to know something about feedback control systems to design the rest of the assembly, but this is beyond the scope of this course).
For now, we will simply calculate the height of the lower collar as a function of the speed of rotation.
The picture below shows a simplified drawing of the geometry. The height h of the collar C is given by
We need to find the angle
as a function of the rotation rate
. We will do this by applying Newton’s second law. Begin by calculating the acceleration of the ball. We could do this by applying our formulae, but many 3D problems are best solved by going back to the basic definitions of velocity and acceleration, and we will follow this approach here. Note that O is a fixed point. We will describe position, velocity and acceleration using the basis shown in the picture. The position vector of the mass relative to O in this basis is
The basis rotates with the governor assembly at angular rate
. Since the basis rotates about a vertical axis, its angular velocity is
We will assume that the angle
We can compute the time derivatives of the rotating vectors using the standard formula
Substitute
Differentiate again to get acceleration
Next, we need the force acting on the mass. The free body diagram is shown below
T is the force exerted on the mass by the topmost rigid link. We know the lower rigid link is free of force, because we have neglected the mass of the collar C (if you don’t see this, draw a free body diagram for the collar). The resultant force on the mass is therefore
Apply Newton’s second law
As usual, this is two equations for two unknowns. Eliminate T to see that
This has two solutions
The second solution exists only if
Imagine steadily increasing the rate of rotation. The ball would hang vertically downwards (
) until the critical speed given in the preceding equation is reached, at which point the ball would start to fly out laterally, as shown in the picture. (You may wonder why the governor doesn’t stay at
at all speeds. This is a good question. It turns out that if the rotation rate exceeds the critical value, then
Finally, we are able to write down an expression for h as a function of the rotation rate
