Newton’s Laws II
4.7 Circular motion, loop-the-loop
Now we’ll turn to some interesting problems in circular motion of vari-ous kinds. The first, shown in part A of Figure 4.5, describes a ride in an amusement park. You sit in these baby rockets hanging from a rope along with other petrified victims. The whole thing begins to spin, and the rope, instead of remaining vertical, starts tilting at an angle θ, which we want to determine as a function of the tangential speed v and radius R of the circular orbit. We apply F= ma and start listing the forces on the baby rocket. Gravity provides the usual mg pointing down. The rope can only exert a force T along its length. Let us trade the tension along the rope for the sum of two equivalent forces, a vertical part, Ty= T cos θ, and a hor-izontal part, Tx= T sin θ directed toward the center of the circular orbit.
The equations are
T cosθ − mg = 0 vertical (4.33)
T sinθ = ma = mv2
R radial or horizontal. (4.34) In the first equation, we recall that by assumption the rocket’s orbit is a horizontal circle and it has no net acceleration in the vertical direction. In the second, we recall from Section 2.7 that a body moving in a circle of radius R at speed v has a centripetal accelerationv2R. So this is a case where we know a in F= ma. Eliminating T we find
tanθ = v2
Rg. (4.35)
Newton’s Laws II 65
Figure 4.5 (Left) An ”amusement” ride in which you go around in a horizontal circle of radius R in a baby rocket. The rope supporting you necessarily makes an angleθ = tan−1 v2Rg with the vertical. (Right) A car going around a circular racetrack of radius R at speed v. The road has a banking angle that obeys tanθ =v2Rg. The normal force N has a horizontal part that gives the necessary forcemv2R to bend the path into a circle. The symbol⊗ is used in both parts to indicate that the rocket or car is going into the page, away from you. The convention is based on how an arrow with feathers would appear going away from you. Likewise indicates an arrow coming out of the page toward you.
It is worth finding the tension on the rope since our lives may depend on it. It is
T= mg
1+ v2
Rg 2
(4.36)
using T=
Tx2+ Ty2.
Here is another interesting problem. You are driving on a circular racetrack of radius R at a speed v. Suppose the plane of the road is strictly horizontal, perpendicular to g. Some agency has to exert a force mv2R on the car to bend it into this circle. It is of course the road that does this, thanks to the frictional force f ≤ μsmg directed toward the center. (I use
66 Newton’s Laws II μs, the static coefficient, and notμk, the kinetic one, even though the car is moving, because we are discussing the force in the radial direction and the car has no velocity in that direction, unless it is skidding. Note also that the car does not really have to travel in a circle; all we need is that at this instant the trajectory is part of some circle of radius R.)
If you don’t have the requisite static friction, if μsmg< mv2R, your car will not be able to make the curve; it will fly off. But there is a clever way in which you can still make the turn without any friction, and that is to bank your road by an angleθ as shown in the right half of Figure 4.5. Imagine now you’re going into the paper. The frictionless road only exerts a normal force N. Let us resolve that force into a vertical part N cosθ and a horizontal one N sinθ directed toward the center of the circle. The equations are
N cosθ = mg (4.37)
N sinθ = mv2
R. (4.38)
Eliminating N we find the banking angle to be
tanθ = v2
Rg. (4.39)
Let me elaborate. You want the car to go around the bend at a certain speed. If you bank your road at that angle, you don’t need any friction to make the turn. Even though the frictionless road can only exert a normal force, thanks to banking, a part of the normal force points toward the cen-ter, providing the requisite centripetal force. Of course, when you drive on a real road, you do not have to travel at exactly this speed for a given R, for any small differences will be made up by the frictional force of the tire. It is just that you do not want to rely on friction for the entire radial force.
Finally, the famous loop-the-loop problem, which defies common sense, is shown in Figure 4.6. You come down on a roller-coaster track from some height H, you go on a vertical circle, and for a while you are upside down. The eternal question is, “Why don’t you fall down?”
We’ll find we can understand this phenomenon fully with Newton’s laws.
The forces on the coaster are mg acting down and the force of the track, which has to be normal to it since it is assumed to be frictionless. But
Newton’s Laws II 67
Figure 4.6 The roller coaster comes down from a height H and goes into a loop in the vertical plane. Why does it not not fall down? The forces on it are mg acting down and the track force N also acting down! It does fall, as explained in the text. The three arrows forming a triangle in the inset show the initial velocity v− v/2 just before it reaches the top, the change v in a small interval near the top, and the final velocity v+ v/2 just after it passes the top.
it too points down! We are doomed! Why don’t we fall? The answer is that we do fall, that is, we accelerate downward, but this does not mean we get any closer to the center. The two forces mentioned combine to bend the coaster into a circular path and produce the requisite downward centripetal acceleration:
N+ mg = mv2
R. (4.40)
Solving for N we find
N= m v2
R − g
. (4.41)
If N comes out positive, that is, points down in our convention, which happens ifv2R > g, we are safe. If it comes out negative, that is, if
v2
R < g, (4.42)
68 Newton’s Laws II
it means the track exerts an upward force, which is impossible, unless there is some other mechanism, like a T-bracket, that goes under the track and supports the coaster even if it is just hanging upside down. I believe such things exist in real roller coasters, in case they get stuck at the top or do not go fast enough. In our idealized coaster, without any of this backup, the speed v must obey
v2> Rg (4.43)
to safely make the loop.
We will figure out the minimum height from which it must be released to satisfy this condition when we derive the law of conservation of energy.
Let us be sure to understand again why accelerating down does not always mean gaining speed toward the earth. If you drop an apple, acceler-ating down means really picking up speed toward the ground, toward the center of the earth. It starts with zero vertical speed and picks up speed. In our example, the coaster is also accelerating, but the tiny change in veloc-ity in a tiny time, which points radially down in Figure 4.6 near the top of the loop, is now added to a huge horizontal velocity pointing to the left.
We will now see that this implies the velocity has a constant magni-tude and changing direction. Consider the impact of adding a tiny change
v to a velocity v on the magnitude of velocity. Assume v can be in any direction relative to v.The resultant velocity has a magnitude squared given by
(To find dv2dt , we need to keep just the term linear in v.) This equation generally implies a non-zero rate of change of the magnitude of v, unless
v and the acceleration a are perpendicular to v, as is the case at all times in circular motion.
Newton’s Laws II 69
Figure 4.7 You fire two bullets from a tower at increasing speeds, which land farther and farther away (points 1 and 2). Beyond a critical speed, the bullet would go into orbit. It has of course never ceased to accelerate toward the earth.
So a little later, the total velocity vector merely gets rotated with no change in length and becomes tangent to the circle at a slightly different point. Going around in a circle is an example of constantly accelerating toward the center but not getting any closer.
Here is another example of this phenomenon. Suppose you are on a tower and you fire a gun horizontally as shown in Figure 4.7. The bullet hits the ground at point 1, under the pull of gravity. If you fire another bullet at a greater velocity it will land a little further away, at point 2. While greater initial speed will extend the time of flight even on a flat earth, the flight is further enhanced by the earth curving under the bullet. There will be a certain speed at which the bullet will keep falling but will not get any closer to the center, because the earth is falling under it just as fast. It is in orbit, as shown by the circle. That is in fact how you launch a bullet into orbit. So, what’s the first thing you should do when you fire this gun?
Move away, because it’s going to come back in about 84 minutes and get you from behind at 17,650 miles per hour. This calculation assumes that there is no atmosphere, which of course leaves you with an even more pressing problem.