# Non-uniform Circular Motion

## Old Content of Uniform Circular Motion, subject to Change

**Uniform circular motion** occurs when an object travels along a circular path at a constant speed or angular velocity.

Velocity is defined by speed and direction, so although an object's *speed* is constant, its direction changes constantly as it moves around a circle. Any change in velocity necessitates a force according to Newton's second law. Thus an object undergoing uniform circular motion experiences a centripetal acceleration, \(a_c\), whose instantaneous direction is toward the center of curvature of the object's path, and whose magnitude can be expressed in terms of the radius, \(r,\) and either tangential velocity, \(v,\) or angular velocity, \(\omega:\)

\[a_c = \frac{v^2}{r} = \omega^2 r.\]

**Period**

Any object whose motion is periodic may be described in terms of the time it takes to complete one cycle or the **period**. In uniform circular motion, the constant speed produces a constant period.

\[ T=\frac{2\pi r}{v} = \frac{2\pi}{\omega}\]

To start, consider the definition of translation velocity:

\[v=\frac{\Delta x}{\Delta t}.\]

The time \(\Delta t\) will be equal to the period \(T\) only when the distance traveled \(\Delta x\) is equal to one circumference of the circular path \(2\pi r:\)

\[v=\frac{\Delta x}{\Delta t} = \frac{2\pi r}{T}.\]

Solving for the period gives

\[T=\frac{2\pi r}{v}.\]

Substitute \(v=r\omega\) to solve for the second definition.

A car travels \(40\frac{\text{m}}{\text{s}}\) along a turn of radius \(50 \text{ m}\) as the car turns \(90^\circ\). How long does this take?

The period is \[T= \frac{2\pi r}{v} \frac{2\pi(40)}{50} = \frac{8\pi}{5}\text{s }.\]

Since the car travels \(90^\circ,\) the turn will take \(\frac14\) of the period.

\[t = \frac{T}{4} = \frac{2\pi}{5}\text{ s}\]

**Swinging a ball on a string**

A basic example of uniform circular motion is a mass attached to a string is moving with a constant speed while the other end of the string remains fixed. Newton's 2nd Law of Motion, is proportional to acceleration) is also pointing toward the center of the circle.

The mass has a centripetal acceleration directed toward the fixed end of the string at the center of its circular trajectory. Because the acceleration is pointing toward the center of the circle, the net force on the mass (which, byA \(0.2 \text{kg}\) ball on the end of a \(2 \text{ m}\) string is swung in a horizontal circle at \(3 \frac{\text{m}}{\text{s}}\). Ignoring gravity, what is the tension in the string?

To solve this problem, first we calculate the acceleration of the mass: \[a_c = \frac{v^2}{r} = \frac{3^2}{2} = 4.5\frac{m}{s^2}\]. Then we can calculate the net force using Newton's 2nd Law: \[F_{\text{net}} = ma = (0.2 \text{kg})(4.5\frac{m}{s^2}) = 0.9 \text{N}\] Then, by looking at a free body diagram of the ball,

we conclude that the only force acting on the ball is the tension \(T\) in the string, since we are told to ignore gravity. Therefore,\[F_{\text{net}} = T = 0.9 \text{N}\].

The tension in the string is 0.9 N. Many problems in Newtonian mechanics provide information about the physical forces on an object and ask you to calculate acceleration. It is worth noting that in this example, we took an opposite approach. When an object is in uniform circular motion, its acceleration often is known or can be easily calculated, as in this example. We then used the acceleration to compute the net force, which we used to deduce information about a physical force, tension. This observation helps to explain why problems involving uniform circular motion can seem "backwards" compared to other problems involving forces and motion in Newtonian mechanics.

A \(5\text{ kg}\) object is tied to the end of a \(0.2\text{ m}\) string and swung in a horizontal circle while making an angle of \(\frac{\pi}{6}\) below the horizontal. If \(g=10\frac{\text{m}}{\text{s}^2},\) how fast is the object traveling in \(\frac{\text{m}}{\text{s}}?\)

**No work, just change**

The work done by a centripetal force is always \(0\text{ J}\).

In order to find the work done by a force, consider a differential piece of work, which is the dot product of a constant force with a differential displacement.

\[dW = \vec{F}\bullet d\vec{r} = F_c dr \cos(\theta)\]

The differential displacement is infinitesimally small, and will thus always point in the direction of the tangential velocity. Since the centripetal force is always directed radially inward, \(\theta = 90^\circ\) always for an object moving along a perfectly circular path.

\[dW = F_c dr \cos(90^\circ) = 0\text{ J}\]

The Earth takes 1 year to complete its orbit around the Sun. Treating the orbit as perfectly circular, find the net work done on the Earth by the Sun in \(\frac14\) of a year. Give your answer to three significant figures.

Assumptions:

Mass of the Earth = \(5.97\times 10^{24} \text{kg}\)

Distance from Sun to Earth = \(1.50\times 10^{11} \text{m}\)

1 Earth year = 365 Earth days

1 Earth day = 24 hours

**Cite as:**Non-uniform Circular Motion.

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