Section 3: Uniform Ciruclar Motion

Lesson 1: Introduction to Centripetal Motion

Angle of Rotation

Figure 1: The angle of rotation (Δθ), arc length (Δs), and radius of curvature (r) shown on a Frisbee as it is spinning in the air.

Consider a line from the center of the Frisbee to its edge. In a given time, each point on this line moves through the same angle. The angle of rotation is the amount of rotation and is the angular analog of distance. The angle of rotation Δθ is the arc length divided by the radius of curvature:

$$\Delta \theta=\frac{\Delta s}{r}$$

The angle of rotation is often measured by using a unit called the radian (radians are actually dimensionless, because a radian is defined as the ratio of two distances, radius, and arc length). A revolution is one complete rotation, where every point on the circle returns to its original position. One revolution covers 2π radians (or 360 degrees), and therefore has an angle of rotation of 2π radians, and an arc length that is the same as the circumference of the circle. We can easily convert between radians, revolutions, and degrees using the relationship

$$1 \text{ revolution}=2\pi=360\text{ degrees}$$

Angular Velocity

If we want to determine how fast a Frisbee can rotate as it is thrown, the concept of angular velocity can be used to answer this question. First let's discuss angular speed before we delve into the concept of angular velocity. Angular speed, ω, is the rate at which the angle of rotation changes. In equation form, the angular speed is

$$\omega=\frac{\Delta \theta}{\Delta t}$$

If an object rotates through a greater angle of rotation in a given time, it has a greater angular speed. The units for angular speed are radians per second (rad/s).

Now let's consider the direction of the angular speed, which means we now must call it the angular velocity. The direction of the angular velocity is along the axis of rotation. For an object rotating clockwise, the angular velocity points away from you along the axis of rotation. For an object rotating counter clockwise, the angular velocity points toward you along the axis of rotation.

$$v=\frac{\Delta s}{\Delta t}$$

From the definition of the angle of rotation, we see that Δs=rΔθ. Substituting this into the expression for v gives

$$v=\frac{r\Delta \theta}{\Delta t}=r\omega$$

The equation v=rω says that the tangential speed v is proportional to the distance r from the center of rotation. Consequently, tangential speed is greater for a point on the outer edge of the Frisbee (with larger r) than for a point closer to the center of the Frisbee (with smaller r). This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center. Note that both points will still have the same angular speed, regardless of their distance from the center of rotation.

Now, lets consider another example: the tire of a moving car. The faster the tire spins, the faster the car moves - large ω means large v because v=rω. Similarly, a larger radius tire rotating at the same angular velocity, ω, will produce a greater linear (tangential) velocity, v, for the car. This is because a larger radius means a longer arc length must contact the road, so the car must move farther in the same amount of time.

Figure 2: A car driving with the angular velocity and tangential velocity of the front tire shown.

However, there are cases where linear velocity and tangential velocity are not equivalent, such as a car spinning its tires on ice. In this case, the linear velocity will be less than the tangential velocity. Due to the lack of friction under the tires of a car on ice, the arc length through which the tire treads move is greater than the linear distance through which the car moves. It’s similar to running on a treadmill or pedaling a stationary bike; you are literally going nowhere fast.

Centripetal Acceleration

Figure 3: Directions of the velocity of a car at two different points.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let's discuss the magnitude of centripetal acceleration. For an object travelling at speed v in a circular path with radius r, the magnitude of centripetal acceleration is

$$\vec{a_c}=\frac{v^2}{r}$$

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that a_c is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than a 50 km/h.

We can also express a_c in terms of the magnitude of angular velocity. Substituting v=rω into the equation above, we get:

$$\vec{a_c}=r\omega^2$$

Lesson 2: Analysing Force in Circular Motion

Centripetal Force

Because an object in uniform circular motion undergoes constant acceleration (by changing direction), we know from Newton’s second law of motion that there must be a constant net external force acting on the object.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

Figure 12: Four examples of centripetal force in action: (a) As the string with a ball at its end rotates, it experiences tension causing a centripetal force; (b) The frictional force required for the turning of the wheels of a car is provided by the centripetal force; (c) In a roller coaster, the force provided to balance the normal force is the force provided by the centripetal force; (d) Planets resolving around the Sun also experience centripetal force because of the gravitational force (Geeks for Geeks).

By using the two different forms of the equation for the magnitude of centripetal acceleration, a_c=v²/r and a_c=rω², we get two expressions involving the magnitude of the centripetal force F_c. The first expression is in terms of tangential speed, the second is in terms of angular speed:

$$\vec{F_c}=\frac{mv^2}{r}=mr\omega^2$$

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton's second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get

$$r=\frac{mv^2}{F_c}$$

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature - that is, a tight curve, as seen in the graph in Figure 13.

Figure 13: Change in the radius of curvature with a change in the magntiude of the centripetal force.