This entire packet should be read and understood by Monday. There are 3 problems to complete at the end of the reading. No solutions have been provided for you, although the answers are at the bottom of the page. However, after carefully reading the following pages, you should be able to answer the quesitons.
Circular Motion in “Vertical” Circles
In our previous lesson, we learned that a CENTRIPETAL FORCE makes an object move in a circular path (which, of course, it doesn’t want to do). The value of this centripetal force was found using
All of the problems that we have looked at so far have dealt with objects moving in “horizontal” circles. For example, when a car goes around a circular turn, it is traveling in a flat, horizontal circle. When a ball is swung in a circle above a man’s head, the circle is, again, flat. In these problems, the basic question that we had to ask ourselves was “What supplies the Centripetal force?”. In the case of the car, friction pushed the car into the circle. In the case of the ball on the string, the tension pulled the ball into the circle. In both cases, SOMETHING had to provide the centripetal force in order to make the object move in a circle.
Let’s look at a Free-Body-Diagram for each of the two situations that I just mentioned.
We have many different forces drawn on these FBDs. Why don’t all the forces SUPPLY the centripetal force”? For example, in the “car driving around the turn problem” why doesn’t the weight supply the centripetal force? Why doesn’t the Normal force help to supply the centripetal force? The answer is simple. Since the centripetal force must keep the object IN the circle, the only forces that supply the centripetal force must also point INTO the circle. Since weight and the normal force don’t point into the circle, they do not produce the inward acceleration (ie. the centripetal acceleration) and thus they don’t cause the centripetal force.
The next scenario is going to seem a little strange. But, it will help us to understand a VERY important point. I want you to imagine that a man is constantly trying to push the
car OUT of the circular turn as it travels around the turn. He isn’t strong enough to make the car leave the circle, but he is trying. What would the FBD look like in the situation now? Well, we would now have the weight, the normal force, friction, AND the push force acting on the car. This can be seen in the FBD below.
So, in this “slightly odd situation” that was described above, what supplies the centripetal force? Again, it’s the friction. BUT…….and here is the point of this example problem….the push force from the man works against friction. So, as usual, we play the “Winner minus Loser” game. The winner is friction (since the car stays in the circular path). The loser is the push. Therefore,
So, when multiple forces point into or out of the circle, we are going to follow a general rule. The question is still the same: “What supplies the centripetal force?”. Our answer will be:
The sum of the forces pointing into the circle MINUS the sum of the forces pointing out of the circle provides the centripetal force.
Or, in a simpler way, “INs – OUTs = Fc”. Understanding this concept is the key to understanding our next type of problem: Vertical circles. The easiest “Vertical Circle” problem to understand is the case of a man swinging a ball on a string. However, unlike the case shown in Figure A below, he will be swinging this ball in a circle in front of him, as shown in figure B.
We need to look at the FBD of the ball at 4 “key” points along it path. These points, and their
accompanying FBDs are shown at the right. At every point along the ball’s path, the tension acts inward. This should make sense, since the string is being held at the center of the circle. And, at every point along its path, the weight of the ball should point downward.
Now, let’s remember how to find the centripetal force. “INs – OUTs = Fc”. We can see that when the ball is at the top, we only have “INS”. Two of them, in fact. When the ball is at the bottom, we have an IN and an OUT. When the ball is on either the left or the right, we have only one IN for each.
So…..
at the top
at the bottom
at the side points
These equation can tell us something interesting. The tension in the string when the ball is at a side point is equal to . The tension in the rope when the ball is at the top of the circular path is less than this. The tension in the rope when the ball is at its lowest point is greatest of all.
So, let’s re-cap a few quick points. Since the tension is greatest at the bottom of the circular path, this is where the rope is most likely to break. It should make sense that the tension at the bottom is the greatest. Imagine watching the summer Olympics, specifically the gymnastics competitions. When a gymnast is doing the parallel bars, spinning themselves around in a vertical circle, when does she have to struggle the most to keep hold of the bar? Not at the top. In fact, at the top the gymnast feels relief. At the bottom, however, the gymnast’s arms will hurt. Gymnasts are most likely to lose hold of the parallel bar when reaching the bottom of their circular path.
However, what happens if a the ball on the string (or the gymnast, for that matter) isn’t going very fast. Without enough speed, the ball won’t be able to complete the circle. It will leave its circular
path at the top, because the string will lose tension. This can be seen in the diagram at the right. So, a MINIMUM VELOCITY is needed for objects to complete vertical circles. How do you find this minimum velocity? Well……if the ball “just barely” gets over the top of the circle, the tension in this “just barely” case will be “just barely” greater than zero at the top. After all, the key to swinging something in a vertical circle is to keep tension in the rope. Therefore, if there is tension in the rope, even an itsy, bitsy, teeny, weeny, yellow(?) ….bit of friction, then the ball will complete the loop. Therefore, using the equation we derived for the top of the loop, we know that
As long as the ball moves faster than “the square root of GRRRRRRRR”, then the ball will complete the circular loop. So, remember this: when talking about the rope breaking, use the bottom-of-the-loop equation. When talking about critical speed, use the top-of-the-loop equation, with tension set equal zero.
Practice Problems
1. A .5 kg ball is swung on a 1 m rope in a vertical circle with a constant velocity of 5 m/s.
a) Find the position of the ball (top, bottom, left, right, etc) where the rope’s tension is greatest.
b) Find the position of the ball (top, bottom, left, right, etc) where the rope’s tension is least.
c) Draw an FBD of the ball for the positions in parts “a” and “b” above.
d) Find the maximum tension in the string.
2. A .2 kg ball is being swung on a 0.5 m rope in a vertical circle. The rope can withstand up to 50 N of force.
a) Find the maximum speed of the ball before it breaks the rope.
b) Find the minimum speed (“critical velocity) of the ball so that it doesn’t “fall out of the circle” at the top)
3. A 35 kg boy is swinging on a rope 7 m long. He passes through the lowest position with a speed of 3 m/sec. What is the tension in the rope at that moment?