Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net[latex]\boldsymbol{\textbf{F}=ma}.[/latex]For uniform circular motion, the acceleration is the centripetal acceleration— [latex]\boldsymbol{a=a_{\textbf{c}}}.[/latex]Thus, the magnitude of centripetal force[latex]\textbf{F}_{\textbf{c}}[/latex]is

By using the expressions for centripetal acceleration[latex]\boldsymbol{a_{\textbf{c}}}[/latex]from[latex]\boldsymbol{a_{\textbf{c}}=\frac{v^2}{r};\:a_{\textbf{c}}=r\omega^2},[/latex]we get two expressions for the centripetal force[latex]\textbf{F}_{\textbf{c}}[/latex]in terms of mass, velocity, angular velocity, and radius of curvature:

You may use whichever expression for centripetal force is more convenient. Centripetal force[latex]\textbf{F}_{\textbf{c}}[/latex]is always perpendicular to the path and pointing to the center of curvature, because[latex]\boldsymbol{a_{\textbf{c}}}[/latex]is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for[latex]\boldsymbol{r},[/latex]you get

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 3. The greater the angle[latex]\boldsymbol{\theta},[/latex]the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle[latex]\boldsymbol{\theta}[/latex]is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for[latex]\boldsymbol{\theta}[/latex]for an ideally banked curve and consider an example related to it.

For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

Figure 3 shows a free body diagram for a car on a frictionless banked curve. If the angle[latex]\boldsymbol{\theta}[/latex]is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight[latex]\textbf{w}[/latex]and the normal force of the road[latex]\textbf{N}.[/latex](A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude[latex]\boldsymbol{\textbf{mv}^2/\textbf{r}}.[/latex]Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,

Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is[latex]\boldsymbol{N\textbf{cos}\theta},[/latex]and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,

Now we can combine the last two equations to eliminate[latex]\boldsymbol{N}[/latex]and get an expression for[latex]\boldsymbol{\theta},[/latex]as desired. Solving the second equation for[latex]\boldsymbol{N=mg/(\textbf{cos}\theta)},[/latex]and substituting this into the first yields

Taking the inverse tangent gives

This expression can be understood by considering how[latex]\boldsymbol{\theta}[/latex]depends on[latex]\boldsymbol{v}[/latex]and[latex]\boldsymbol{r}.[/latex]A large[latex]\boldsymbol{\theta}[/latex]will be obtained for a large[latex]\boldsymbol{v}[/latex]and a small[latex]\boldsymbol{r}.[/latex]That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that[latex]\boldsymbol{\theta}[/latex]does not depend on the mass of the vehicle.

PHET EXPLORATIONS: GRAVITY AND ORBITS

Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!

Conceptual Questions

Conceptual Questions

1: If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.

2: Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?

3: If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.

4: Race car drivers routinely cut corners as shown in Figure 5. Explain how this allows the curve to be taken at the greatest speed.

5: A number of amusement parks have rides that make vertical loops like the one shown in Figure 6. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:

(a) The car goes over the top at faster than this speed?

(b)The car goes over the top at slower than this speed?

7: What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6 under the following circumstances:

(a) The car goes over the top at such a speed that the gravitational force is the only force acting?

(b) The car goes over the top faster than this speed?

(c) The car goes over the top slower than this speed?

8: As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.

9: Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 7 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

10: Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?

11: Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.