Circular Motion

Karen Tennenhouse

Some Notes and Cautions about Circular Motion Problems

1. 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. In fact, it is accelerating with an acceleration ${\bar {a}}$ whose direction is towards the center of the circle, and whose magnitude is

$a=v^{2}/R$
2. What causes the object to accelerate towards the center?
   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. The net force causes the object to accelerate, in the direction of Net${\vec {F}}$ , with

${\vec {a}}$ = (Net${\vec {F}})/m$
Notice, the acceleration does not cause a force. Rather, NET FORCE causes acceleration.
3. Because it points radially towards the center, the acceleration is also called the “centripetal acceleration” and also sometimes called the “radial acceleration”. Likewise, because the net force is pointing towards the center, it is also called the “centripetal force.”

Centripetal force is NOT a “separate” or “extra” force; it is just another name for the NET force ∑${\vec {F}}$ in this situation.
4. Treat every circular motion problem as a Newton’s 2nd Law problem, just with a special formula for the ‘a’ and with special care about the direction of ${\vec {a}}$.
   • Set up the solution as for any dynamics problem, by drawing the isolation diagram.
   • But, very carefully figure out where is the plane of the circle, where is the centre, where is the object, and thus carefully determine the direction of acceleration. Draw it as a small double arrow.
   • It is strongly advised that you choose your positive axis in the same direction as the acceleration.

   [If you really prefer, you may choose the opposite axis; BUT beware: you must then remember to substitute ( – a) instead of (a) in the Net ${\vec {F}}$ equation. Most people forget.]

   • As usual, write: ∑${\vec {F}}_{axis}$ = ..... = m ${\vec {a}}$ but knowing that magnitude a = (v ² / R)
We never draw the net force in the isolation diagram, and we’re not going to start now! So, the isolation diagram should NOT have a ${\vec {F}}_{C}$ (Centripetal Force) in it.
(If you have drawn a force in your diagram and cannot say by what object this force is exerted, then probably this is not a real force.)


5. All the above notes deal with uniform circular motion, ie where the object is going in a circle but at constant speed. But, what happens if an object is travelling in a circle (thus changing direction) and is also changing its speed? We will look at that situation very briefly near the end of our NYA course (see Rotational Motion). For your interest, here is a mini-preview:
   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, ie it points towards the center and has magnitude a_R = (v² / R)
     
   • The object’s acceleration is the vector sum of its two components.

• Rotational Motion
  
  

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Related Videoclips

1. Introduction to uniform circular Motion




2. The rotor

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Exercises

Helena Dedic

Exercise 1

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

Solution:

False. The radial acceleration has a constant magnitude but its direction changes as the particle moves around the circle. The acceleration always points towards the centre.

Exercise 2

The electron in a hydrogen atom has a speed of $2.2x10^{6}$ m/s and orbits the proton at a distance of $5.3x10^{-11}m$. 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 $v=2*10^{6}m/s$ and orbits the proton with radius $r=5.3*10^{-11}m$. You are asked to find its centripetal or radial acceleration:

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

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 $10m/s^{2}$ 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 $4*10^{17}m/s^{2}$. 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 $r=2*10^{4}$ m. You are asked to find the centripetal acceleration of a point on its equator:

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

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

Exercise 3

Can a particle move at a constant speed and yet be accelerating? If so, give an example.

Solution:

Yes, a particle can move at constant speed around a circular or any other curved path and have an acceleration. Since the direction of the velocity changes the velocity vector is not constant and therefore the motion is an accelerated motion.

• Rotational Motion
  
  

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