##### 2.6.1 Draw a vector diagram to show that the acceleration of a particle moving with uniform speed in a circle is directed toward the centre of the circle.

**Review of basic kinematics:** If the acceleration and velocity of an object are parallel (or anti-parallel) then the object's speed will increase (decrease). If the acceleration and velocity of an object are perpendicular then only the direction of the velocity will change and the speed (i.e. the magnitude of the velocity) will remain constant.

If a ball is attached to the end of string and swung at a constant speed (i.e. only the direction of the velocity is changing not the magnitude) then there must still be an acceleration. The acceleration is directed towards the center of the motion. This acceleration is call centripetal acceleration!

##### 2.6.2 State the expression for centripetal acceleration.

The acceleration of any object moving in a circle at a constant speed is given by the equation:

(1)It is important to note that centripetal acceleration is very special. It is the acceleration required for an object to move in a circle at a *constant* speed.

The reverse is also true if an object's acceleration is equal to $v^2 \over r$ (and perpendicular to the velocity) then the object must be going in a circle.

If an object is moving in a circle at with a changing velocity, then the overall acceleration is not equation to the centripetal acceleration. However the acceleration perpendicular to the velocity (that is the part changing the direction) is still equal to $\frac{v^2}{r}$

##### 2.6.3 Identify the force producing circular motion in various situations

Sometimes people will make reference to the "centripetal force." This is not a real force, its a pseudo-force. In general the centripetal force is made up of many other forces and is the sum of those forces. This is not unlike the idea of a net force which is also generally the sum of multiple forces.

If you have a ball on the end of a string and you swing it in a vertical circle the "centripetal force" or the forces causing the acceleration will be a combination of the *tension* from the string and *gravity*. At the top of the circle the ball will be going slower than at the bottom (conservation of energy). Since the speed is lower at the highest point the centripetal acceleration will be reduced. The weight remains the same, so the tension must be reduced to maintain circular motion. While at the lowest point the ball will be moving faster and thus the centripetal acceleration will be increased, so the tension must have increased to maintain circular motion.

The Tension and Weight are the forces causing the acceleration. The ball is also moving in a circle so at the highest and lowest points $Tension + Weight = Centripetal \: Force$. Some care should be taken with the sign (+/-) of the tension force as its direction changes throughout the motion.

Take it a step farther:

(2)--

=( that's it?

ReplyOptionsThe page was a little skimpy… I added a good chunk. Hope that helps some.

ReplyOptionsGreetings

Some confusion evident in sections 2.6.1 and 2.6.3.

1. The centripetal force is nowhere constant - force is a vector, and the direction changes continually. Same for centripetal acceleration. For an object going around in a circle at constant speed the centripetal force and acceleration have constant magnitudes.

2. It is not possible for a ball on a string to go round a vertical circle at constant speed. It has more potential energy at the top and way less at the bottom, i.e. potential energy at the top changes to kinetic at the bottom hence the speed is minimum at the top and maximum at the bottom.

ReplyOptionsThanks for the feedback.

As to #1, I'm not seeing where I claimed the force is constant, but it is quite possible that I got sloppy and forgot to say "the magnitude of…"

#2, I was looking for a quick easy and concise example, but you are absolutely correct a ball on a string can not travel in a vertical circle at a constant speed. I did question this, I just didn't think of the very simple energy argument.

I'm glad to have another set of eyes of the site. I haven't taught this content in nearly 4 years. I know the site (badly) needs revision.

ReplyOptionsplz clear 2nd confusion of 2.6.3 i cant undrstand

ReplyOptionsYou're going to have to be a bit more specific. Certainly my explanation is not perfect, but you've got to give me more…

ReplyOptionsThis is a good page, thanks. If anyone is worried about a lack of content on this site until it is fully updated you might want to head towards

'The Open Door Website'with its fantastic section on IB Physics, both HL and SL. You can find it saburchill[dot]com/physics/chapters[dot]htmlReplyOptionshello. we did an experiment on circular motion using a rubber stopper whirled above a person's head. The aim of the lab is to find the relationship between the centripetal force and the frequency. I found out the force is directly proportional to the frequency squared.

Question: How will the graph of this relationship should look like? Will put the place frequency on the x or y axis?

Thanks a lot.

ReplyOptionsGenerally the data that goes on the horizontal axis is the variable that you have direct control of its value and were changing throughout the lab - did you change the force and then measure the frequency or did you change the frequency and then measure the force.

It's a bit hard to tell from your description which variable (force or frequency) you are changing.

ReplyOptionsWe whirled the rubber stopper for 20 revolutions, we changed the mass hanging on a string where the rubber stopper is connected on the other end of the string.

Sounds to me that mass ("converted" to a weight force if that's what is expected) is your independent variable and should be graphed on the horizontal axis.

Thanks. Now what is making me confuse is that some lab reports (same experiment) that I was able to find on the web are saying that the frequency squared is on the x-axis (independent) and the converted mass, the force, is on the y-axis (dependent).

Facts I know:

Tension force = mass x gravity — the force causing the rubber stopper to move in a circular motion

Centripetal force = mass x velocity^2 / radius -- v = 2 (pi) r / t

Tension force = centripetal force

substituting the variables …

T = 4 (pi)^2 m r / t^2 where frequency (f) = 1/t

therefore

T = 4 (pi)^2 m f^2

The independent variable is always on the horizontal. From the way you described the lab I would say that the mass (i.e. tension force) is the independent variable. It is very possible to run the lab in reverse -> spin the rubber stopper at different speeds and then measure via force probe the tension in the string. In this second case I would argue that the frequency is the independent variable.

Admittedly there are many labs, like yours, where it can be unclear which variable is independent. In those cases pick one and stick to it.

Ask your teacher too. In my opinion you are sweating a fairly small detail - that in some ways doesn't have a correct answer.. You just need to be consistent with the answer you choose.

ReplyOptionsIt was good

ReplyOptionscould you please explain how to derive the equation for acceleration. I don't understand how it goes from ∆V/T to v^2/r.

Cheers,

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