5 - Circular Motion and Gravitation

A structured guide to the kinematics, forces, and gravitational applications that govern circular motion, including curves, loops, rotating systems, and circular orbits.

Motion Along a Circle

Circular motion is an application of Newton’s laws. The key distinction is between speed and velocity: speed is a scalar, while velocity includes both magnitude and direction.

At every point on a circular path, the instantaneous velocity is tangent to the circle and perpendicular to the radius. Even if the speed stays constant, the velocity changes because its direction changes. A change in velocity means that the object is accelerating.

The inward acceleration is

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

The subscript $c$ indicates the center-directed, or centripetal, direction. Using $v=r\omega$ , the same relationship becomes

a_c=r\omega^2.

A larger speed produces a much larger inward acceleration because speed is squared. For fixed speed, increasing the radius reduces the required acceleration.

When speed is constant, the acceleration is entirely centripetal. When speed changes, there is also a tangential component that changes the magnitude of the velocity:

\vec a=\vec a_c+\vec a_t.

Because the radial and tangential components are perpendicular,

|\vec a|=\sqrt{a_c^2+a_t^2}.

Takeaway: velocity is tangent to the path, while points inward and changes the velocity’s direction.

, , and Speed

The time and angular relationships of circular motion provide several equivalent ways to describe the same motion. One complete revolution covers the circumference of the circle, $2\pi r$ .

The $T$ is the time for one revolution, and the $f$ is the number of revolutions per unit time:

f=\frac{1}{T}.

The tangential speed is therefore

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

describes how quickly the object moves through angle. Since one revolution corresponds to $2\pi$ radians,

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

Combining these relationships with gives

a_c=\frac{v^2}{r}=\frac{4\pi^2r}{T^2}=4\pi^2rf^2.

For example, a car moving at $10\ \text{m/s}$ around a curve of radius $50\ \text{m}$ has

a_c=\frac{(10\ \text{m/s})^2}{50\ \text{m}}=2.0\ \text{m/s}^2.

The acceleration points horizontally toward the center of the curve.

Takeaway: , , , radius, and tangential speed are linked descriptions of .

Forces That Produce Circular Motion

The inward acceleration must be produced by the net inward force. Applying Newton’s second law in the radial direction gives

\sum F_r=ma_c=m\frac{v^2}{r}.

The expression

F_c=m\frac{v^2}{r}=mr\omega^2

is called the , but it does not represent a separate interaction. First identify the actual forces in a free-body diagram; then add their inward and outward radial components.

Common sources of the required inward force include:

Tension for an object attached to a string.
for a car on a level curve or an object on a rotating platform.
The normal force, or a component of it, on a curved track or banked road.
Gravity for a satellite in orbit.
Combinations of gravity and the normal force in a vertical loop.

A reliable problem-solving sequence is:

Identify the circular path and its center.
Determine the radius $r$ .
Draw a free-body diagram containing only real forces.
Choose the inward radial direction as positive or negative.
Write $\sum F_r=m\frac{v^2}{r}$ .
Add any independent equations, such as vertical equilibrium.
Check the direction and units of the result.

A force tangent to the path does not directly provide . It changes the object’s speed instead.

Takeaway: means the net inward radial force; it is never added as an extra force to a free-body diagram.

Vehicles on Level and

For a car on a level curve, the vertical forces balance:

N-mg=0, \qquad N=mg.

supplies the horizontal inward force:

f_s=m\frac{v^2}{r}.

Because cannot exceed $\mu_sN$ , the greatest speed without slipping satisfies

\mu_sN=m\frac{v^2}{r}.

Using $N=mg$ and solving for speed gives

v_{\max}=\sqrt{\mu_sgr}.

The car’s mass cancels, so the limiting speed depends on the coefficient of , gravitational field, and curve radius rather than on mass. For $\mu_s=0.50$ , $r=40\ \text{m}$ , and $g=9.8\ \text{m/s}^2$ ,

v_{\max}=\sqrt{(0.50)(9.8)(40)}=14\ \text{m/s}.

A banked road can redirect part of the normal force inward. For an ideal banked curve without friction,

N\cos\theta=mg, \qquad N\sin\theta=m\frac{v^2}{r}.

Dividing these equations gives

\tan\theta=\frac{v^2}{rg},

\theta=\tan^{-1}\left(\frac{v^2}{rg}\right).

Friction can also act on a banked road. Its direction depends on whether the vehicle tends to slide down the bank or up the bank.

Takeaway: level curves rely on available , while use a horizontal component of the normal force and may also involve friction.

Pendulums and Vertical Loops

In a conical pendulum, a mass moves in a horizontal circle while a string makes an angle $\theta$ with the vertical. The tension has a vertical component that balances weight and a horizontal component that supplies the inward force:

T\cos\theta=mg, \qquad T\sin\theta=m\frac{v^2}{r}.

Dividing the radial equation by the vertical equation gives

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

The entire tension is not the ; only its horizontal component points toward the center.

For an object moving inside a vertical loop, the radial direction changes from point to point. At the top, both the normal force and gravity may point toward the center:

N+mg=m\frac{v^2}{r}.

At the bottom, the normal force points toward the center while gravity points away from it:

N-mg=m\frac{v^2}{r}.

At the top of a loop, just maintaining contact means $N=0$ . Thus,

mg=m\frac{v^2}{r},

which gives the minimum speed

v_{\text{top,min}}=\sqrt{gr}.

The radial equation must always be written using the direction toward the center at the specific location being analyzed.

Takeaway: in rotating systems and loops, resolve forces into radial and tangential directions and account for how the radial direction changes.

Gravity and Circular Orbits

Gravity can provide exactly the inward force required for a . Newton’s law of universal gravitation gives

F_g=G\frac{m_1m_2}{r^2}.

For an object of mass $m$ orbiting a much larger central mass $M$ , set gravitational force equal to the required radial force:

G\frac{Mm}{r^2}=m\frac{v^2}{r}.

The orbiting mass cancels, leaving

v=\sqrt{\frac{GM}{r}}.

Thus, at a fixed orbital radius around the same central body, objects with different masses have the same circular orbital speed. A larger radius produces a lower circular orbital speed.

Using $v=\frac{2\pi r}{T}$ , the orbital is

T=2\pi\sqrt{\frac{r^3}{GM}}.

The increases as the orbital radius increases, consistent with the circular-orbit form of Kepler’s third law.

An orbiting satellite is not force-free. Gravity continually accelerates it toward the central body while its tangential velocity carries it along the path. If gravity suddenly disappeared, the satellite would move in a straight line tangent to the orbit. If its tangential speed suddenly became zero, it would fall inward.

Takeaway: circular orbits occur when gravitational attraction supplies the net inward force, linking orbital speed and to the central mass and orbital radius.