A turntable is set rotating at a constant frequency of 45 rpm (revolutions per minute). A nickel is placed on the turntable near the axis of rotation at a distance r from that axis. The nickel is observed to rotate around in a circle of radius r with a frequency of 45 rpm. The nickel is pushed slightly away from the axis of rotation, thereby increasing the value of r. This is repeated until r is so large that the nickel just slides right off of the turntable. The largest value of r such that the nickel remains in circular motion on the turntable without slipping of is found to be rmax = 7.8 cm. From these data, determine the coefficient of friction between the nickel and the turntable.
First of all, before we even start worrying about the solution to this problem, let’s first get our data in order. We are told the maximum value of the radius of the nickel’s circular motion. We are also told the frequency of that circular motion: f = 45 rpm. The units of rotations per minute are not proper units that will be useful in this problem. To make sure that we at least start off consistently, let’s put all of our data in MKS units. Doing so yields:
rmax = 0.078 m and f = 45 rev/min = 0.75 rev/s = 0.75 Hz .
We also know that, if the frequency of the motion is f, then the period of the circular motion must be
T = 1/f = 1.3 s .
Also, since the speed (not velocity!!) is a constant, it must follow that speed = distance / time, or
Now, since the question above involves a force (the friction force), we know that we will have to draw a good FBD for this problem. Such a FBD for the nickel undergoing circular motion is shown below. Keep in mind that we will call the direction in towards the center of the circular motion the c-direction:
As always, now that we have a FBD, we apply Newton’s
2nd law in component form to this diagram. First of all, we
look at the y-direction (this will give us the normal force,
which we will then use in the c-direction equation to give us
the friction force):
Note that we have used the weight equation W = mg. Now, applying Newton’s 2nd law to the c-direction, and keeping in mind that we know something special about the c-component of acceleration (do you remember what that is?), we get that
Note that we have used the maximum value of the radius throughout this solution, since we know that the maximum value of static friction occurs when the radius is a maximum (which is why the nickel slides when r gets any bigger than rmax — static friction can no longer supply the force required to hold the coin in circular motion!)
In the two examples we’ve done so far, we’ve seen that the force required to hold the object in question in circular motion (that is, the force supplying the so-called centripetal force) has been supplied by the tension force in a string, and by the friction force. In the next and last example, we will see the required force supplied by yet another force — the normal force.