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Why is circular motion important? We use it every day! Many satellites move in a circular path around the earth and make it possible for us to have television, navigation like GPS, satellite radio, and much more. We rely on these things daily, so understanding the dynamics of circular motion and its applications is critical. This article will begin by going over uniform and nonuniform circular motion. We will then do a few examples to understand the applications of circular motion.
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Team Applications of Circular Motion Teachers
An object traveling in a circular path with a constant velocity and constant radius is experiencing uniform circular motion. As shown in the image below, the velocity vector for objects in uniform circular motion always points tangential to the circle. Since the velocity vector is changing direction as it moves around the circle, there is a centripetal acceleration and a centripetal force which are always perpendicular to the velocity vector. These point radially inward towards the center of the circle.
The velocity of an object in uniform circular motion can be described in terms of the period, T. The period is the time it takes an object to complete one revolution. Since the magnitude of the velocity is constant, we can find it by taking the circumference of the circle divided by the period:
\[\text{Velocity} = \frac{\text{distance}}{\text{time}}\]
\[v = \frac{2 \pi r}{T}\]
We can then write the centripetal acceleration as:
\[\begin{align} a_c &= \frac{v^2}{r} \\ &=\frac{(2 \pi r/T)^2}{r} \\ &= \frac{4 \pi^2 r}{T^2}\end{align}\]
If the velocity of the object is not constant, the object experiences nonuniform circular motion. An example of nonuniform circular motion is shown in the article, "Circular Motion and Free-Body Diagrams". This example discusses a ball being swung in a vertical circular path. In this example, the force of gravity causes the magnitude of the velocity to change as the ball is being swung. At every point along the circle except for the top and bottom, there are components of the net acceleration vector and the net force vector that are parallel to the velocity vector. This means that the net acceleration vector no longer points toward the center of the circle, as shown below.
For objects experiencing nonuniform circular motion, we can only solve for the velocity and centripetal acceleration at certain positions on the circle using Newton's second law; however, it is possible to use energy to relate the speeds at different positions.
Let's do an example of an object in uniform circular motion.
Consider a satellite orbiting the earth in a uniform circular path 4500 km from the surface of the earth. What is the centripetal acceleration of the satellite? Use 6371 km for the earth's radius and 5.98 · 1024 kg for its mass.
The centripetal force acting on the satellite is the force of gravity. So, we can start with Newton's second law and solve for the centripetal acceleration.
\(F_{net} = ma\)
\(F_{g} = ma_c\)
\(G \cdot \frac{m_{sat}m_e}{r^2} = m_{sat}a_c\)
\(\begin{align} a_c &= G \cdot \frac{m_e}{r^2} \\ &= \frac{(6.67 \cdot 10^{-11} Nm^2/kg^2)(5.98 \cdot 10^{24} kg)}{(4500 km + 6371 km)^2} \\ &= 3.375 m/s^2 \end{align}\)
Now let's consider an example of an object in nonuniform circular motion.
A 2.5 kg ball moves along a circular loop of radius 2 m, as shown in the image below. What is the magnitude of the normal force acting on the ball at the top of the loop? Take the velocity at the top of the loop to be 5 m/s and ignore friction.
First, we must determine all the radial forces acting on the ball to get the centripetal force. We have the force of gravity pointing downward and radially inward at this location. We also have the normal force to consider. Which direction does the normal force point? The normal force is the force from the surface of the loop acting on the ball, so it points from the surface of the loop to the ball, which is radially inwards in any position on the loop.
We can now use Newton's second law to find the magnitude of the normal force at the top of the loop.
\(F_{net} = ma\)
\(F_g + F_n = ma_c\)
\(\begin{align} F_n &= ma_c -mg \\ &=m \frac{v^2}{r} - mg \\ &=m \Big(\frac{v^2}{r} -g \Big) \\ &=(2.5 kg) \Big( \frac{(5 m/s)^2}{2} - 9.8 m/s^2 \Big) \\ &=6.75 N \end{align}\)
Remember that the velocity of the ball changes as it's going around the loop. If the ball were on the left side of the loop, the ball's velocity would be greater due to gravity. The normal force on the ball in that location would also be greater because the force of gravity would no longer have a centripetal force component. Thus, the normal force would just be the mass multiplied by the centripetal acceleration at that point.
Another example of vertical circular motion is a person on a Ferris wheel. Unlike the example of the ball going around the loop above, this vertical circular motion is uniform because the velocity is constant.
A Ferris wheel of radius 16 m is traveling in uniform circular motion with a velocity of 5 m/s. What is the normal force from the seat acting on a 60 kg person at the top and bottom of the Ferris wheel?
We will consider upwards to be the positive direction for both locations. In both locations, the normal force from the seat will point in the positive direction, while the force of gravity points in the negative direction. Since the centripetal acceleration vector points radially at both points, it will be in the negative direction at the top, and in the positive direction at the bottom.
Using our free-body diagrams for the person at the top, we can write our centripetal force equation as:
\(F_{net} = ma_c\)
\(F_{nt}-F_g = -ma_c\)
Notice that we've put a negative sign in front of the centripetal acceleration to account for its direction at the top. Now we can solve for the normal force.
\(\begin{align} F_{nt} &= mg-a_c \\ &=m \Big(g-\frac{v^2}{r} \Big) \\ &= (60 kg)\Big(9.8 m/s^2 - \frac{(5 m/s)^2}{16 m} \Big) \\ &= 494N \end{align}\)
Notice that it is possible for our normal force at the top of the Ferris wheel to be negative if the velocity is big enough. If this were the case, a downward force like from a seatbelt would be necessary in order to keep the person in the seat.
Our centripetal force equation at the bottom of the Ferris wheel is:
\[F_{nb} - F_g = ma_c\]
Our acceleration vector points upwards, so the direction is positive. Solving for the normal force at the bottom, we get:
\(\begin{align} F_{nb} &= ma_c + mg \\ &=m\Big(g + \frac{v^2}{r} \Big) \\ &= (60 kg) \Big(9.8 m/s^2 + \frac{(5m/s)^2}{16 m} \Big) \\ &= 682 N \end{align}\)
Thus, the magnitude of the normal force is greater at the bottom of the Ferris wheel compared to the top.
A car rounding a curve is an example of horizontal, uniform circular motion that we see every day.
A 1300 kg car is rounding a flat curve with radius 60 m. Find the velocity of the car if the friction coefficient is 0.8.
Since the curve is flat, the normal force and the force from gravity balance each other out so that \(F_n = F_g = mg\). This means that the only force that contributes to the centripetal force is friction. Thus, we have:
\(F_{net} = ma_c\)
\(F_f = m \frac{v^2}{r}\)
\( \mu F_n = m \frac{v^2}{r}\)
\(\mu \cancel{m} = \cancel{m} \frac{v^2}{r}\)
\(\mu g = \frac{v^2}{r}\)
\(\begin{align} v &= \sqrt{\mu gr} \\ &= \sqrt{(0.8(9.8 m/s^2)(60m)} \\ &= 21.7 m/s \end{align}\)
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What are some applications of uniform circular motion?
Some applications of uniform circular motion are satellites in circular orbits around the earth, a car going around a circular curve, and a ball being swung horizontally on a string.
Is acceleration constant in uniform circular motion?
The magnitude of the centripetal acceleration is constant in uniform circular motion because the magnitude of the velocity is constant.
What are some examples of circular motion applications?
Some examples of circular motion include an ice skater traveling in circular motion, a person riding a Ferris wheel and a ball being swung vertically on a string.
Is velocity constant in applications of uniform circular motion?
In applications of uniform circular motion, the magnitude of the velocity is constant but the direction of the velocity vector changes.
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