Module 6: Circular Motion

Karen Tennenhouse and Kevin Lenton

Learning Objectives

After reading this page, watching the videos and reviewing the exercises, you will be able to:

• Understand that a change in the velocity vector could indicate a change in the magnitude of velocity (speed), a change in the direction, or a change in both magnitude and direction.
• Understand why uniform circular motion requires centripetal acceleration.
• Know the direction of the centripetal acceleration.
• Know the magnitude of the centripetal acceleration.
• Apply the concepts of centripetal acceleration to some circular motion problems

Word Worksheet
pdfWorksheet

Openstax

Can a particle move at a constant speed and yet be accelerating?

The answer is YES.

The point is that the acceleration is the rate of change of the velocity vector with respect to time. A vector has both a value (magnitude) and a direction. You can change the velocity vector by changing the magnitude and/or by changing the direction.
An example of this is an object moving around a circle. The particle can move at constant speed around a circular or any other curved path and have an acceleration at the same time. Since the direction of the velocity changes, the velocity vector is not constant and therefore the motion is an accelerated motion.

A Special Case: Uniform Circular Motion

Consider an object travelling in a circle (radius R) at some instantaneous speed (v).

If the speed (magnitude of velocity) is constant, we say that this is Uniform Circular Motion.
Such an object does not have constant velocity; its direction is changing. Therefore, it is accelerating! This acceleration is called centripetal acceleration ${\vec {a}}_{c}$.

This centripetal acceleration ${\vec {a}}_{c}$ is also a vector and therefore has a magnitude and direction.

Direction of the centripetal acceleration ${\vec {a}}_{c}$

Because the definition of acceleration is
${\vec {a_{average}}}={\frac {\vec {\Delta v}}{\Delta t}}={\frac {{\vec {v_{f}}}-{\vec {v_{i}}}}{\Delta t}}\ \ \ or{\frac {{\vec {v_{2}}}-{\vec {v_{1}}}}{\Delta t}}$
The direction of the acceleration will be in the same direction as ${\vec {\Delta v}}$.

Watch this video to find out the direction of the centripetal acceleration.
Direction of the centripetal acceleration

Magnitude of the centripetal acceleration $|{\vec {a}}_{c}|$

The magnitude of the centripetal acceleration is equal to:

$|{\vec {a}}_{c}|={\frac {v^{2}}{R}}$

Where v is the magnitude of the velocity and R is the radius of the circle.

Here is a proof of this equation for the magnitude of the centripetal acceleration.
Magnitude of the centripetal acceleration

What causes the object to accelerate towards the center?
You will be seeing that what causes this acceleration is the same thing that ever causes any object to accelerate in any way: The real forces acting on it (gravity, friction, tension or whatever) are adding up to some nonzero net force.

What happens for non-uniform Circular Motion?

All the above notes deal with uniform circular motion, in other words the object is going in a circle but at constant speed.
If the object has any sort of change in the velocity vector, there will be an acceleration. For any motion in any circle, there must be a centripetal acceleration with direction towards the centre of the circle and magnitude $|{\vec {a}}_{c}|={\frac {v^{2}}{R}}$

However, as you know points on a rotating object can be speeding up or slowing down, that is to say, they can have an angular acceleration $\alpha$. That means that a point on the rotating object can also have a linear (or tangential) acceleration. We say tangential because this acceleration is a tangent to the circle.
Therefore the point is travelling in a circle (thus changing direction) and is also changing its speed.
In such a case, the object’s acceleration will have two (perpendicular) components:

• One component, along the direction of motion, is called the tangential acceleration, a_T . The tangential acceleration is related to the change in speed: it is in the same direction as velocity if the object is speeding up, or opposite to velocity if object is slowing down.)
  
• The second component, called the radial acceleration a_R , is exactly the centripetal acceleration: it points towards the center and has magnitude a_R = (v² / R)
  
• The point’s total acceleration is the vector sum of its two components.
  

$|{\vec {a}}_{total}|={\sqrt {a_{Radial}^{2}+a_{Tangential}^{2}}}$

The first 5 mins of this video explain this

Period of the Circular Motion

The period of the circular motion is the time for the object to complete one revolution, one circle.
The distance travelled in this time is the circumference of the circle $2\pi r$.
Therefore $v={\frac {2\pi r}{T_{period}}}$

Exercises

Helena Dedic

Exercise 1

True or False: When a particle moves in uniform circular motion its acceleration is constant.

Solution:

False. Trick Question! The radial acceleration is still a vector. The radial acceleration has a constant magnitude but its direction changes as the particle moves around the circle. The acceleration always pointing towards the centre, and is therefore constantly changing.

Exercise 2

(a)The electron in a hydrogen atom has a speed of $2.2\times 10^{6}m/s$ and orbits the proton at a distance of $5.3\times 10^{-11}m$. What is its centripetal acceleration?

(b) A neutron star of radius 20 km is found to rotate once per second. What is the centripetal acceleration of a point on its equator?

Solution:

a. It is given that the electron has speed $v=2\times 10^{6}m/s$ and orbits the proton with radius $r=5.3\times 10^{-11}m$. You are asked to find its centripetal or radial acceleration:

$a={\frac {(2\times 10^{6}m/s)^{2}}{(5.3\times 10^{-11}m)}}=9.1\times 10^{22}m/s^{2}$ 
 

It is always interesting to think about one's results. It is particularly intriguiging in this case because the magnitude of the acceleration is such an awesome number (compared to $9.8m/s^{2}$ on Earth). Now, suppose the Earth were to shrink and become a Black Hole; then at a distance of 3 cm from the centre, acceleration would still be only $4\times 10^{17}m/s^{2}$. Thus the large radial acceleration of the electron is testimony to the strength of the interaction between the electron and the proton in the atom.

b. You are told that a neutron star rotates once per second (i.e that it has period T = 1 s) and that it has a radius of $r=2\times 10^{4}$ m. You are asked to find the centripetal acceleration of a point on its equator:

$v={\frac {(2\pi )}{T}}={\frac {(2\pi \times 2\times 10^{4}m)}{1s}}=4\pi \times 10^{4}m/s$ 

$a=v^{2}/r={\frac {(4\pi \times 10^{4}m/s)^{2}}{(2\times 10^{4}m)}}=7.9\times 10^{5}m/s^{2}$