KAHIBARO
Table of Contents
Applying Newton's Laws to Circular Motion
When an object moves in a circle, even at constant speed, its velocity is changing because its direction is changing. A change in velocity means there is acceleration. By Newton's second law, acceleration requires a net force. In circular motion, this force must point toward the center of the circle.
This inward force is called the centripetal force. The word centripetal means center-seeking. It is not a new kind of force like tension or gravity. Instead, it is the name for the net force required to keep an object moving in a circular path.
If the object has mass $m$, moves with speed $v$, and follows a circle of radius $r$, then the required centripetal acceleration is
$$
a_c = \frac{v^2}{r}
$$
Using Newton's second law, the net inward force must be
$$
F_c = m a_c = m\frac{v^2}{r}
$$
For circular motion, the net force toward the center must satisfy
$$
\sum F_{\text{radial}} = m\frac{v^2}{r}
$$
This is one of the most important equations for applying Newton's laws to circular motion.
Identifying the Real Force
A very common mistake is to think that centripetal force is an extra force added to the problem. It is not. You must identify which real forces provide the required inward net force.
For example, in different situations the centripetal force may be provided by tension, gravity, friction, the normal force, or a combination of forces.
| Situation | Force providing centripetal force |
|---|---|
| Ball tied to a string and swung in a circle | Tension |
| Car turning on a flat road | Friction |
| Satellite orbiting Earth | Gravity |
| Object moving in a vertical loop | Normal force, tension, gravity, or a combination |
So the correct method is not to write "centripetal force" as a separate force in a free-body analysis. Instead, draw all real forces and then set their inward component equal to $m v^2 / r$.
Do not add "centripetal force" as an extra physical force on a free-body diagram.
Centripetal force is the net inward force caused by real forces.
Radial Direction and Newton's Second Law
In circular motion problems, Newton's second law is usually applied along the radial direction, which points toward or away from the center of the circle. The acceleration points inward, so the net force in that direction must also point inward.
If we choose inward as the positive radial direction, then
$$
\sum F_r = m\frac{v^2}{r}
$$
The exact form depends on which forces point inward and which point outward.
This is especially useful because many circular motion problems involve forces that act along the radius.
Example, Car Turning on a Flat Road
Consider a car of mass $m$ turning on a flat circular road of radius $r$ with speed $v$. The car tends to keep moving in a straight line because of inertia, but static friction between the tires and the road pulls it inward and changes its direction.
If static friction is the only horizontal force, then
$$
f_s = m\frac{v^2}{r}
$$
This means friction provides the centripetal force.
If the maximum static friction is $f_{s,\max} = \mu_s N$, and on a flat road $N = mg$, then the maximum speed before slipping is found from
$$
\mu_s mg = m\frac{v^2}{r}
$$
So
$$
v_{\max} = \sqrt{\mu_s g r}
$$
This shows that a car can turn faster if the road is less slippery, if the curve radius is larger, or if gravity is stronger.
Example, Mass on a String
Suppose a small mass moves in a horizontal circle while attached to a string. If the string tension is the only inward force, then
$$
T = m\frac{v^2}{r}
$$
So the tension must increase if the speed increases or if the radius becomes smaller.
The velocity is tangent to the circle, while the tension points inward toward the center.
Example, Satellite in Circular Orbit
For a satellite orbiting a planet, gravity provides the centripetal force. If the planet has mass $M$ and the satellite has mass $m$, and the orbital radius is $r$, then
$$
\frac{G M m}{r^2} = m\frac{v^2}{r}
$$
After canceling $m$, we get
$$
v = \sqrt{\frac{GM}{r}}
$$
This result shows that the orbital speed does not depend on the mass of the satellite.
Vertical Circular Motion
Vertical circular motion is more subtle because gravity may help or oppose the required inward force depending on the position of the object.
At the top of the circle, the center is below the object, so any downward force contributes to the centripetal force. At the bottom of the circle, the center is above the object, so upward forces contribute inward.
Consider an object attached to a string moving in a vertical circle.
At the top:
$$
T + mg = m\frac{v_{\text{top}}^2}{r}
$$
At the bottom:
$$
T - mg = m\frac{v_{\text{bottom}}^2}{r}
$$
The tension is usually different at different points because the speed may change and gravity affects the radial direction differently.
This is why objects in vertical loops can lose contact if the speed becomes too small. For example, if a roller coaster reaches the top of a loop too slowly, the normal force may become zero.
Condition for Maintaining Contact
When an object moves along a circular track, the contact force cannot become negative. If the normal force becomes zero, the object is just about to lose contact.
At the top of a loop, for the minimum speed needed to stay on the track, the normal force is zero. Then gravity alone provides the centripetal force:
$$
mg = m\frac{v^2}{r}
$$
So the minimum speed at the top is
$$
v_{\min} = \sqrt{gr}
$$
This is an important result for loop-the-loop problems.
At the point where an object is just about to lose contact with a track, the normal force is
$$
N = 0
$$
Use this condition to find the minimum speed required to remain on the circular path.
Banked Curves
A road or track may be tilted so that the normal force has a horizontal component. This helps provide the required centripetal force.
For a frictionless banked curve with banking angle $\theta$, the normal force $N$ has components:
$$
N \cos\theta = mg
$$
$$
N \sin\theta = m\frac{v^2}{r}
$$
Dividing the second equation by the first gives
$$
\tan\theta = \frac{v^2}{rg}
$$
So the speed for which no friction is needed is
$$
v = \sqrt{rg\tan\theta}
$$
This explains why highways and racetracks are often banked on curves.
Solving Circular Motion Problems
In applications of Newton's laws, circular motion problems are solved by the same general strategy used elsewhere. First identify the object. Then draw the real forces acting on it. Next choose the radial direction carefully, usually toward the center. Finally apply Newton's second law in that direction.
In some problems you may also need Newton's second law in a vertical or tangential direction, especially if the speed is changing or if the path is inclined.
| Step | What to do |
|---|---|
| 1 | Identify the object |
| 2 | Draw all real forces |
| 3 | Choose inward radial direction |
| 4 | Write $\sum F_r = m v^2 / r$ |
| 5 | Solve for the unknown quantity |
The key idea in Newton's laws and circular motion is this:
An object moving in a circle must have a net inward force.
Without that inward force, it will not continue in a circular path.
Common Mistakes
One common mistake is confusing velocity and force directions. In circular motion, velocity is tangent to the path, but centripetal acceleration and net force point inward.
Another common mistake is treating centripetal force as a separate force. Instead, always ask which real force supplies it.
A third mistake is using the formula $m v^2 / r$ without choosing signs carefully in vertical circles. The radial direction changes from point to point, so the force equation must match the geometry.
Final Perspective
Circular motion is one of the clearest examples of Newton's second law in action. The motion continues along a curved path only because real forces keep pulling or pushing the object inward. Whether the force comes from friction, tension, gravity, or the normal force, the same rule applies, the net radial force must equal $m v^2 / r$. Once this idea is understood, many practical problems, from cars on curves to satellites in orbit, become much easier to analyze.