1. Uniform Circular Motion
Centripetal force and acceleration
What force must be applied to Helen to keep her moving in a circle?
How does it depend on the Helen’s radius r ?
 How does it depend on Helen’s velocity v
 How does it depend on Helen’s mass m? r m v
On the next pass, however, Helen failed to clear the mountains.

2. Uniform Circular Motion Zitzewitz Section 6.2
Circular motion is the name for an object moving in a circle at constant speed, such as a stone being swung in circles at the end of a string or a fixed horse on a merry-go-round.
Are these objects accelerating if they are moving in a circle at constant speed?
What is the definition of acceleration? 𝑎 𝑎𝑣𝑔 = ∆𝑣 ∆𝑡
But v is a vector quantity and that means speed and direction are both parts of it.
We said the speed of a horse on a merry go round isn’t changing, but what about its direction?
Since its direction is changing, therefore it is accelerating!!

3. Uniform Circular Motion
Uniform circular motion is defined as an object moving at constant speed (note I didn’t say constant velocity!) around a circle with fixed radius.
As the object moves around the circle, the length of its position vector (the radius) doesn’t change, but the direction does. Same thing with its velocity vector (constant speed).

4. Uniform Circular Motion
Δr is the resultant displacement between r1 and r2
The velocity vector is perpendicular to the displacement vectors (r) and tangent to the circle.

5. Uniform Circular MotionΔv
Now let’s look what direction the object’s acceleration is in. The diagram above shows two velocity vectors separated by time Δt.
Using vector addition, v1 + Δv = v2
Recall that aaverage = Δv/Δt so that Δv vector when divided by Δt gives the acceleration.
Note that acceleration a, points the same direction as Δv, towards the center of the circle = “centripetal acceleration”

6. Centripetal Acceleration
The velocity tangent line is always a 90 degree angle to the radius, so this angle is preserved.
So ∆𝑟 𝑟 = ∆𝑣 𝑣
Now if you divide both sides by Δt: ∆𝑟 𝑟∆𝑡 = ∆𝑣 𝑣∆𝑡
Since by definition: 𝑣= ∆𝑟 ∆𝑡 and 𝑎= ∆𝑣 ∆𝑡
𝑣 𝑟 = 𝑎 𝑣 𝑜𝑟 𝒂= 𝒗 𝟐 𝒓

7. Centripetal Acceleration
How can we measure the speed of an object moving in a circle?
Measure T (period) which is the time for the object to make one complete revolution around the center.
One complete revolution = circumference = 2𝜋𝑟
So 𝑣= ∆𝑥 ∆𝑡 = 2𝜋𝑟 𝑇 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑
And then 𝒂= 𝒗 𝟐 𝒓 = 𝟒 𝝅 𝟐 𝒓 𝑻 𝟐

8. Centripetal Force
Because there is always an acceleration, a, towards the center, there is also a force directed inwards towards the center.
Newton’s 2nd law for circular motion:
ΣFnet = mac where ac = centripetal acceleration. This force is called centripetal force.
Note that the centripetal acceleration always goes on the right side of the equation! (not as a force on the left)

9. UCM Example Zitzewitz p.155 #2
A rubber stopper is being swung in a horizontal circle on a string.
m=13g, length of string = l = 0.93m, T=1.18 sec, find centripetal force
𝒂 𝒄 = 𝒗 𝟐 𝒓 = 𝟒 𝝅 𝟐 𝒓 𝑻 𝟐 = 𝟒 𝝅 𝟐 𝟏.𝟏𝟖 𝟐 .𝟗𝟑 =𝟐𝟔 𝒎/ 𝒔 𝟐
𝑭=𝒎 𝒂 𝒄 = .𝟎𝟏𝟑 𝒌𝒈 𝟐𝟔 𝒎 𝒔 𝟐 =𝟎.𝟑𝟒 𝑵

10. Centrifugal Force (not real!)
Let’s say you are in a car and you go around this corner to the left too fast. The passenger on the right side might get thrown against the right (or outer) door.
Is there an outward force on the passenger? Some people say it is centrifugal force, pushing you outward, but they are wrong – there is no such thing, it is fictitious.

11. Centrifugal force
The person being “forced” against the outside door is simply following Newton’s 1st law, saying an object will continue moving with the same velocity in the same direction until a force acts to change it. The person has no force acting to change his motion until the seat belt and/or door he’s leaning against do that job.
Common misconception!

12. Vertical UCM

13. Vertical UCM
Newton’s 1st : An object will travel in a straight line unless a force acts on it to change that.
In other words, in the absence of a modifying force, it is not natural for an object to travel in a circular path.
So for a ball swung in vertical circles on a string, the modifying force is the tension in the string that forces it to move in a circle. It pulls the ball off its straight path onto a circular path.
Centripetal force is therefore a pulling force which acts towards the center of the circle.

14. Centrifugal force – not real
Centrifugal force, an outward pushing force is a “phantom” force. Here’s proof.
If you cut the string when the ball is at A, the balls flies off in a direction of B, not C. If there had been an outward force, it would have gone towards C (or between B and C) but it doesn’t. B A C

15. Vertical UCM
But, you say, when you whirl a ball around vertically on a string, you can feel the string/ball “pulling” on your hand outwards.
What you ARE feeling is the resistance of the ball to moving in a circular path. Remember at every point, it wants to just go straight. To keep the ball going in a circle, the person must constantly PULL the ball towards the center.
This creates the tension in the string which is the centripetal force.

16. Demonstration – water in cup on platform
Note that the water in the cup always stays parallel to the platform, not to the ground like most water does – why?
This is because I’m creating a pseudo-gravity greater than the Earth’s gravity.
This pseudo gravity is always perpendicular to the surface of the tray. What causes this?
It’s the centripetal force being exerted by the tension in the string.

17. Continued…
The centripetal force is directed to the center of the circle at all times. Therefore the tray constantly pushes on the cup towards the center of the circle at all times.
Fg Fc
ocean water
The water level in the ocean and the water level in the cup are both perpendicular to the force.
But wait, those arrows look like the go in opposite directions…..

18. Continued… ocean water Fg Fc Earth Vertical circle
So really, they are both going in the same direction, towards the center of rotation!!

19. Now let’s consider the top and bottom of a vertical circle
top top T mg T mg

20. Top At the top, we know net force
y T mg
At the top, we know net force
must act towards the center of the circle.
The net force is the centripetal force.
You never label Fc in a free body diagram, it comes in as the mac part.
𝐹=𝑇+𝑚𝑔=𝑚 𝑎 𝑐
𝑇+𝑚𝑔= 𝑚𝑣 2 𝑟
So 𝑻= 𝒎𝒗 𝟐 𝒓 −𝒎𝒈 that – sign is why tension is less at the top of the swing
To calculate the minimum velocity so the string won’t sag, set T=0 and solve for v.

21. Bottom Now 𝑭=𝑻 −𝒎𝒈=𝒎 𝒂 𝒄 𝑻 −𝒎𝒈= 𝒎𝒗 𝟐 𝒓 So 𝑻= 𝒎𝒗 𝟐 𝒓 +𝒎𝒈mg y
Now 𝑭=𝑻 −𝒎𝒈=𝒎 𝒂 𝒄
𝑻 −𝒎𝒈= 𝒎𝒗 𝟐 𝒓
So 𝑻= 𝒎𝒗 𝟐 𝒓 +𝒎𝒈
Unlike the tension up at the top, the tension at the bottom is now the weight PLUS the centripetal force, so it feels like “it’s pulling harder” or “it weighs more”.

22. Rotational Inertia Video from Space
Ok, this isn’t exactly UCM, but I couldn’t resist this really cool 1 minute video anyway!
Here’s some Professor Lewin from MIT with 5’ video on what centripetal acceleration looks like in space.