Circular motion

This subchapter takes a close look at circular motion. You’ll learn the core ideas and formulas for:

• Newton’s law of Universal Gravitation, the fundamental concepts and key formulas.
• Concepts of uniform circular motion, centripetal acceleration and force, the relationship between angular speed and linear speed, and the period of motion.

Newton’s law of universal gravitation

Newton’s Law of Universal Gravitation says that any two point masses attract each other. The force is:

• proportional to the product of their masses
• inversely proportional to the square of the distance between their centers

Mathematically:

$$F=G\frac{m_1{m}_2}{r^2},$$

where

• $F$ is the magnitude of the gravitational force between two masses,
• $m_1$ and $m_2$ are the masses of the two objects,
• $r$ is the distance between the centers of the two masses, and
• $G$ is the gravitational constant, approximately $6.67\times 10^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$.

Explanation

Newton connected everyday falling motion to orbital motion: the same interaction that pulls an apple downward can also keep the Moon in orbit. Kepler’s work had already described planetary motion from observations, and Newton proposed gravity as the underlying cause.

Later, when we study circular orbital motion, you’ll see a step-by-step derivation that links gravitational force to the requirements for circular motion.

Gravitational field and acceleration

The gravitational field strength, $g$, at a distance $r$ from the center of a mass $M$ is defined as force per unit mass. Starting from $g=\frac{F}{m}$ and using Newton’s law of gravitation gives:

$$g=\frac{F}{m}=G\frac{M}{r^2}.$$

For Earth, if we take $r\approx R_E$ (Earth’s radius, approximately $6.37\times 10^6\text{m}$), then:

$$g\approx G\frac{M_E}{R_E^2}\approx 9.8{\text{m/s}}^2.$$

Fundamentals of circular motion

Uniform circular motion

In uniform circular motion, an object travels around a circle at constant speed $v$. Even though the speed stays the same, the velocity changes because its direction changes continuously. A changing velocity means the object is accelerating.

That acceleration points toward the center of the circle and is called centripetal acceleration.

Centripetal acceleration

For an object moving at constant speed$v$ in a circle of radius $r$, the centripetal acceleration $a_c$ is:

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

Key idea: the acceleration is radially inward at every point on the path. It comes from the continuous change in the direction of the velocity vector.

Derivation of centripetal acceleration

Suppose an object moves from point $P$ to $Q$ on the circle in a time interval $\Delta t$. During that time, its velocity changes direction by an angle $\Delta \theta$ (in radians). If the speed stays constant, the magnitude of the change in velocity is approximately:

$$\Delta v\approx v\Delta \theta .$$

Now connect the angle change to the distance traveled. The arc length is $\Delta s=v\Delta t$, and for a circle $\Delta \theta =\Delta s/r$, so:

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

Acceleration is change in velocity per unit time:

$$a_c\approx \frac{\Delta v}{\Delta t}\approx \frac{v\Delta \theta }{\Delta t}=\frac{v\cdot (v\Delta t/r)}{\Delta t}=\frac{v^2}{r}.$$

Centripetal force

Newton’s Second Law says the net force on an object equals mass times acceleration. For uniform circular motion, the net inward force required to produce the centripetal acceleration is:

$$F_c=m{a}_c=m\frac{v^2}{r}.$$

This net inward force is called the centripetal force. It’s not a new kind of force; it’s the name we give to whatever combination of real forces points toward the center (tension, gravity, friction, the normal force, or a combination).

Example scenario: Mass on a string in circular motion

Consider a mass$m$ attached to a string of length $r$ being swung in a horizontal circle. The tension $T$ in the string points toward the center, so it provides the centripetal force:

$$T=m\frac{v^2}{r}.$$

Angular velocity and time period

Angular velocity

Angular velocity $\omega$ measures how quickly the object sweeps out angle (in radians) per second. It relates to linear speed $v$ by:

$$v=\omega r.$$

So, for the same $\omega$, a larger radius means a larger linear speed.

Time period

The time period $T$ is the time for one complete revolution. One revolution covers a distance equal to the circumference, $2\pi r$, so:

$$T=\frac{2\pi r}{v}.$$

Substitute $v=\omega r$ to write the period in terms of angular velocity:

$$T=\frac{2\pi }{\omega }.$$

Applications

Banked curves (frictionless)

On a frictionless banked curve, the normal force is tilted. Its horizontal component points toward the center of the circular path and supplies the centripetal force. For a curve banked at an angle ${\theta }_b$ with radius $r$, the ideal speed $v$ is:

$$v=\sqrt{rg\tan {\theta }_b}.$$

This relationship is used in road and track design when you want the banking to reduce reliance on friction.

Conical pendulum

In a conical pendulum, a mass $m$ hangs from a string of length $L$ and moves in a horizontal circle at constant speed. The string makes an angle $\theta$ with the vertical.

The tension $T$ has:

• a vertical component that balances the weight
• a horizontal component that provides the centripetal force

That gives:

$$T\cos \theta =mg\text{and}T\sin \theta =m\frac{v^2}{r},$$

where the radius of the horizontal circle is $r=L\sin \theta$. Combining these relationships leads to:

$$v=\sqrt{gL\tan \theta }\sin \theta .$$

This model is used to analyze conical pendulums and other rotating systems where tension provides the inward force.

Example Problem 1

A $2\text{kg}$ object is attached to a $1.5\text{m}$ string and swung in a horizontal circle at a constant speed of $6\text{m/s}$. Calculate the centripetal force acting on the object.

Solution:

(spoiler)

Using the formula:

$$F_c=m\frac{v^2}{r},$$

substitute:

$$F_c=2\text{kg}\times \frac{6^2}{1.5\text{m}}=2\times \frac{36}{1.5}=2\times 24=48\text{N}.$$

Thus, the centripetal force is $48\text{N}$.

Example Problem 2

A conical pendulum consists of a $1\text{kg}$ mass attached to a $2\text{m}$ long string. The string makes an angle of $30^{\circ }$ with the vertical as the mass moves in a horizontal circle. Calculate the period $T$ of the motion.

Solution:

(spoiler)

First, determine the horizontal radius of the circular motion:

$$r=L\sin \theta =2\text{m}\times \sin (30^{\circ })=2\text{m}\times 0.5=1\text{m}.$$

The vertical component of the tension balances the weight:

$$T\cos \theta =mg.$$

We use the relation for centripetal acceleration:

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

and the centripetal force provided by the horizontal component of the tension:

$$T\sin \theta =m\frac{v^2}{r}.$$

Dividing the second equation by the first gives:

$$\tan \theta =\frac{v^2}{rg}.$$

Solve for $v^2$:

$$v^2=rg\tan \theta .$$

Substitute $r=1\text{m}$, $g=10{\text{m/s}}^2$ (for simplicity), and $\tan (30^{\circ })\approx 0.577$:

$$v^2=1\times 10\times 0.577=5.77,v\approx 2.4\text{m/s}.$$

The period $T$ is the circumference divided by speed:

$$T=\frac{2\pi r}{v}=\frac{2\pi \times 1}{2.4}\approx \frac{6.28}{2.4}\approx 2.62\text{s}.$$

Example Problem 3

A frictionless banked curve has a radius $r=50\text{m}$ and is banked at an angle ${\theta }_b=25^{\circ }$. Determine the ideal speed for a car to navigate the curve without relying on friction.

Solution:

(spoiler)

For a frictionless banked curve, the ideal speed is given by:

$$v=\sqrt{rg\tan {\theta }_b}.$$

Using $g=10{\text{m/s}}^2$, $r=50\text{m}$, and $\tan (25^{\circ })\approx 0.4663$:

$$v=\sqrt{50\times 10\times 0.4663}\approx \sqrt{233.15}\approx 15.27\text{m/s}.$$