Centripetal Force (problems at bottom)

Lesson 1: Background

Centripetal Force is the force required to change the direction of a moving object.

To understand this we first recall…

Newton’s
1^{st} Law: an object at rest tends to stay at rest and
an object in motion tends to continue in straight line motion
(unless acted upon by a force)

So how do we get something to "turn"?

We apply a force like this:

This is
called the Centripetal Force. Let’s see (mathematically
speaking) where this comes from…we first note that *whenever*
we have a net force, we have an acceleration (from Newton’s
2^{nd} Law: F = ma).

So, put a ball on the end of a string and spin it around your head with a constant speed, say 2.0 meters/sec. Now the speed doesn’t change, but it’s direction does…

And because of that change in direction we end up with an acceleration! Why? Recall our definition of acceleration:

It says
that a change…ANY change in velocity results in
acceleration. Remember velocity is a vector: it has speed and
direction. So if we change __speed__, we accelerate (or
decelerate). This is something we’re all familiar with.
However, if we change __direction__, we must, by definition,
accelerate as well (because a part of the velocity has changed)!

Exactly what does it mean to accelerate when changing direction?

Let’s take a second look at that ball…

At the bottom of the picture is the expanded equation for acceleration…the acceleration is the difference between the velocity at position 2 and the velocity at position 2.

Numerically,
there is no difference! It’s 2.0 meters/sec always. However,
there is a *directional* difference…let’s use some
vector graphical addition. We create a –v1 vector (reverse
it’s direction) and add it to v2:

The result is the vector (delta)v. If we knew the lengths of these vectors we could measure the length of (delta)v and get a numerical result. For now, note that (delta)v is directly proportional to the acceleration!!!!. We only need to divide (delta)v by the time to find it.

More importantly, look at the direction of the accelerated motion…I’ll paste the (delta)v vector directly onto the picture of the spinning ball, halfway between velocity1 and velocity2:

It’s pointed directly at the center! That tells us the acceleration is acting inward (radially inward, to be exact). In fact, it’s pointing in the same direction as the Force we showed at the beginning of this discussion.

We divide (delta)v by the time, t, to get the acceleration and we draw…

Where a_{c}
is the "centripetal" acceleration (designated by the
small subscript ‘c’).

Now, the **Centripetal
Force, **F_{c}**,** is found using Newton’s 2^{nd}
Law by simply multiplying a_{c} by mass:

F_{c}
= ma_{c}

Lesson 2: Derivation of centripetal acceleration equation:

We use our circling ball again, as well as our vector addition. However, this time we introduce features from the circle itself, like radius and the "chord":

__Examples:__

- A 4.0 kg ball is attached to 0.7 meter string and spun at 2.0 meters/sec. What is

a)the centripetal acceleration

b)the centripetal force

a)Since

a_{c}
= v^{2}/R

a_{c}
= (2.0 m/s)^{2}/(0.7 m)

a_{c}
= 5.71 m/s^{2}

b)For centripetal force,

F_{c}
= ma_{c}

F_{c}
= (4.0 kg)(5.71 m/s^{2})

F_{c}
= 23 Newtons

2. The same setup is now spun at 4.0 meters/sec,.

a)find the centripetal acceleration

b)find the centripetal force

c)divide your answer to (b) by 4.45 to find out how many pounds of force

a)As in the first problem,

- a
_{c}= v^{2}/Ra

_{c}= (4.0 m/s)^{2}/(0.7 m)a

_{c}= 23 m/s^{2}

- a

b)Now,

- F
_{c}= ma_{c}F

_{c}= (4.0 kg)(23 m/s^{2})F

_{c}= 92 N

- F

c)92/4.45 = 21 lbs of force

note that in problem 1, it was only 5.2 lbs of force (23/4.45). By doubling the speed we increase the centripetal force by 4 times!

Why? How did that increase sneak in?