6.2 Uniform Circular Motion

Section Key Terms

Centripetal Acceleration

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine $\text{Δ}s$ becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration a_c because centripetal means center seeking.

Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity $\text{Δ}v$ is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of $\text{Δ}s$, $\text{Δ}v$ points exactly toward the center of the circle (but this is hard to draw). Because $a_{\text{c}}=\text{Δ}v/\text{Δ}t$, the acceleration is also toward the center, so a_c is called centripetal acceleration.

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 traveling at speed v in a circular path with radius r, the magnitude of centripetal acceleration is

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 at 50 km/h.

We can also express a_c in terms of the magnitude of angular velocity. Substituting $v=r\omega$ into the equation above, we get $a_c=\frac{{(r\omega )}^2}{r}=r{\omega }^2$ . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

Centripetal Force

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.

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.

The component of any net force that causes circular motion is called a centripetal force. When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to F_net = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = a_c. Therefore, the magnitude of centripetal force, F_c, is $F_{\text{c}}=m{a}_{\text{c}}$ .

By using the two different forms of the equation for the magnitude of centripetal acceleration, $a_{\text{c}}=v^2/r$ and $a_c=r{\omega }^2$, we get two expressions involving the magnitude of the centripetal force $F_{\text{c}}$. The first expression is in terms of tangential speed, the second is in terms of angular speed: $F_{\text{c}}=m\frac{v^2}{r}$ and $F_{\text{c}}=mr{\omega }^2$ .

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

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.

Figure 6.8 In this figure, the frictional force f serves as the centripetal force F_c. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force F_c, the smaller is the radius of curvature r and the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force F_c produces a smaller radius $r^{\prime }$ .

Solving Centripetal Acceleration and Centripetal Force Problems

To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

Worked Example

Comparing Centripetal Acceleration of a Car Rounding a Curve with Acceleration Due to Gravity

A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity (g).

Strategy

Because linear rather than angular speed is given, it is most convenient to use the expression $a_{\text{c}}=\frac{v^2}{r}$ to find the magnitude of the centripetal acceleration.

Solution

Entering the given values of v = 25.0 m/s and r = 500 m into the expression for a_c gives

$$\begin{array}{ccc}\hfill a_{\text{c}}& =& \frac{v^2}{r}\hfill \\ & =& \frac{{\left(25.0\text{m/s}\right)}^2}{500\text{m}}\hfill \\ & =& 1.25{\text{m/s}}^2\text{.}\hfill \end{array}$$

Discussion

To compare this with the acceleration due to gravity (g = 9.80 m/s²), we take the ratio $a_{\text{c}}/g=(1.25{\text{ m/s}}^{\text{2}})/(9.80{\text{m/s}}^2)=0.128$ . Therefore, $a_{\text{c}}=0.128g$, which means that the centripetal acceleration is about one tenth the acceleration due to gravity.

Worked Example

Frictional Force on Car Tires Rounding a Curve

1. Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s.
2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
3. If the car would slip if it were to be traveling any faster, what is the coefficient of static friction between the tires and the road? Could we conclude anything about the coefficient of static friction if we did not know whether the car could round the curve any faster without slipping?

Strategy and Solution for (a)

We know that $F_{\text{c}}=m\frac{v^2}{r}$ . Therefore,

$$\begin{array}{ccc}\hfill F_{\text{c}}& =& m\frac{v^2}{r}\hfill \\ & =& \frac{\left(900\text{kg}\right){\left(25.0\text{m/s}\right)}^2}{600\text{m}}\hfill \\ & =& 938\text{N.}\hfill \end{array}$$

Strategy and Solution for (b)

The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve.

$$f=F_{\text{c}}=938\text{N}$$

Strategy and Solution for (c)

If the car is about to slip, the static friction is at its maximum value and $f={\mu }_s\text{N}={\mu }_{s}mg$. Solving for ${\mu }_s$, we get ${\mu }_s=\frac{938}{900\times 9.8}=0.11$. Regardless of whether we know the maximum allowable speed for rounding the curve, we can conclude this is a minimum value for the coefficient.

Discussion

Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since $f={\mu }_{\text{s}}\text{N}={\mu }_{\text{s}}mg$. The static friction is only equal to ${\mu }_s\text{N}$ when it is at the maximum possible value. If the car could go faster, the friction at the given speed would still be the same as we calculated, but the coefficient of static friction would be larger.