 Vertical Circles and Non-Uniform Circular Motion Printer Friendly Version
Unlike horizontal circular motion, in vertical circular motion the speed, as well as the direction of the object, is constantly changing. Gravity is constantly either speeding up the object as it falls, or slowing the object down as it rises.

We will begin by looking at two special positions which are usually analyzed in problems: the very top of a vertical circle and the very bottom of the circle.

Suppose a block is being whirled on the end of a string in a vertical circle. Let's look at the freebody diagrams of the forces acting on the block at the top of the circle and at the bottom of the circle. Remember when drawing freebody diagrams for objects moving in circular motion that the net force on the object is ALWAYS directed towards the center of the circle, no matter where the object is located in its circular path.
 top bottom  Top of the circle:

Let's calculate the tension in the string at the top of the circle. Notice that both of the forces T and mg are directed downwards towards the center. Since our block is in circular motion, we know that the NET FORCE must act towards the center of the circle.

That is, the NET FORCE towards the center, or the centripetal force, is the resultant or SUM of these two REAL forces. You should never label Fc in a freebody diagram!

net force to the center = T + mg
Fc = T + mg
m(v2/r) = T + mg
T = m(v2/r) - mg

If we wanted to calculate the minimum or critical velocity needed for the block to just be able to pass through the top of the circle without the rope sagging then we would start by letting the tension in the rope approaches zero.

0 = m(v2/r) - mg
m(v2/r) = mg
v2/r = g
v2 = rg
v = √(rg)

Refer to the following information for the next three questions.

Consider a 1-kg brick being whirled in a vertical circle at the end of a 1-meter rope.
 What critical velocity must the brick achieve in order to pass safely through the top of its circular path?

 What is the tension in the rope as it is passing through the top of its circular path?

 How would be the critical velocity of the brick if it were to be whirled on the moon where the acceleration due to gravity is 1/6th that on earth?

Bottom of the circle:

Now let's calculate the tension in the string at the bottom of the circle. Since the block is maintaining a circular path, we take the direction towards the center as positive. The NET FORCE acting towards the center, Fc, is the resultant force or the difference between T and mg since they now point in opposite directions.

net force to the center = T - mg
Fc = T - mg
m(v2/r) = T - mg
T = m(v2/r) + mg

This formula will be used frequently to calculate the tension in the string in a simple pendulum as the pendulum bob swung through its lowest position - the equilibrium position, the point of greatest KE.

Refer to the following information for the next question.

Consider a 1-kg brick being whirled in a vertical circle at the end of a 1-meter rope.
 If the brick is being whirled so that it just passes through the top of its circular path without allowing the rope to sag, then what tension will be in the rope as the brick passes through the lowest point in its path?

Notice that the tension in the string is GREATEST as the block passes through the bottom of the circle and LEAST while it passes through the top of the circle.

If you were asked to calculate the tension at any intermediate point in a pendulum's swing, then the net force to the center would equal T - mg cos θ. net Fc = T - mg cos θ
mv2/r = T - mg cos θ
T = mv2/r + mg cos θ

Notice that components of the weight are taken, not components of the tension. It is a component of the weight vector that accelerates the bob towards equilibrium.

In order to solve for tension, you would have to use conservation of energy techniques to first solve for the velocity at the requested intermediate position.

Refer to the following information for the next two questions.

Suppose a 1-meter pendulum is released from a horizontal position as shown below. What would be the speed of the bob just as it reaches an arc equal to 37º?

 What would be the tension in the string just as the bob reaches an arc equal to 37º?

Now, consider an example of a person riding a roller coaster through a circular section of the track, a "loop-the-loop."   Let's look at the formulas needed to calculate the normal force, N, exerted on a object traveling on the inside surface of a vertical circle as it passes through the bottom and through the top of the ride.

At the top:

net force to the center = N + mg
N + mg = m(v2/r)
N  = m(v2/r) - mg

While at the bottom:

net force to the center = N - mg
N - mg = m(v2/r)
N  = m(v2/r) + mg

If we let the value of normal approach zero in the formula for the top of the roller coaster we would get the same value for the critical velocity that we got when solving for the tension in the string in our previous discussion, v = √(rg). This principle of critical velocity is used in many places. When you watch clothes drying in a dryer, they are being rotated in a vertical circle. But the rate of rotation does not allow the clothes to achieve this critical value as they pass through the top of the circle. Therefore the clothes fall away from the drum and are "fluffed" as they spin. In roller coasters, this critical velocity is a safety threshold. The coasters MUST exceed this minimum value in order to be certified. Obviously, no one would want the cars to fall away from the rails as the participants experience a thrill passing through a "loop-the-loop" section of the track!

The value of the normal at the bottom of the ride is equivalent to questions asking about the apparent weight of a pilot as he pulls out of a vertical drive. The expression a = v2/r + g is often called the "g's" a pilot is experiencing.

Note that the normal, N, appears to play the same role as the tension, T, in our equations for vertical circular motion.

Refer to the following information for the next question.

While driving to work you pass over a "crest" in the road that has a radius of 30 meters. How fast would you need to be traveling to experience apparent "weightlessness" while passing over the crest? Related Documents