# Physics - Classical mechanics - Dynamics of Circular motion

Hello it's a me again drifter1! First of all, I want to point out that we reached 500 followers and I want to thank you all for your support! Today we continue with **Physics **by getting into more specific things of** circular motion dynamics** and **applications**! We covered a lot of things in the post about smooth circular motion here. If you haven't checked it don't worry at all, cause I will make a small recap/introduction first to give you the basic knowledge needed! So, without further do, let's get started!

## Recap

In my post about smooth circular motion we said that it is a 2d or **plane motion**, where we don't separate the movement for each axis like in projectile motion, but follow another approach.

We suppose that the **velocity is constant in meter **(value) but changes direction all the time in a perfect circle! That way we can use the simple equations of linear motion.

Using **polar coordinates** (angle θ, radius r) we end up with an **angular acceleration**:

**a rad = u^2/R**

This can be easily proven by triangle "similarity", using a avg = |Δu|/Δt and finding the limit to Δt -> 0. I covered it in my post!

The **velocity** is calculated using:

**u = 2πr / T**

where 2πr is the **circumference **of a circle and is the **period **or time taken to do a full circle.

You can clearly see that we use the basic x = ut => u = x/t.

Because of t = x/u the **period **can be calculated using:

**T = 2πR / u**

The **angular velocity **is noted as ω and looks like this:

ω = Δθ / Δt => **ω = 2πf = 2π / T**

So, angular velocity is how many radians (other term for degrees where 2π rad = 360 degrees) per second we are moving. It's calculated in rad/s.

That way another equation for** angular/radional acceleration** is:

**a rad = 4π^2R / T^2**

So, angular acceleration is calculated in rad/s^2.

## Circular motion Dynamics

Here we are now to apply Newton's laws to see what we can get from it.

It is very clear that the **object is not at rest**, but also** doesn't move in a constant linear motion**.

So, because the velocity changes direction there is a force applied to the object and** Newton's second law applies**.

This means that **ΣF != 0** and by using the angular acceleration we end up with:

ΣF = m * a rad =>

**ΣF = m * u^2 / R**

The force that makes the object follow a circular path is** always applied to the center of the circle** and **always has a constant meter** (value). Because it's applied to the center we call this force the **centripetal force** (Fc) and because of that** a rad is also called centripetal acceleration **(a c).

Don't fall into the trap thinking that the centripetal force Fc is a part of ΣF! It's ma!

This will help you when drawing the free-body diagram!

## Applications

Circular dynamics are used in the following:

- Conical Pendulum
- Horizontal (flat) curve/turn
- Inclined or banked (with degree) curve/turn
- Vertical circle

A **conical pendulum** looks like this:

Which is the force of the thread (tension) when the sphere is moving with a "constant" velocity u?

By analyzing the horizontal and vertical component vectors of the tension we end up with:

T * sinθ = m u^/R (2nd law)

and

F cos θ = mg (1st law)

By diving the two equations we end up with:

tanθ = u^2/gR (1)

The radius R = L * sinθ and so the circumference is 2πL sinθ.

That way:

u = 2πL sinθ / T (2)

Using those two equations (1 and 2) we end up with:

cosθ = gT^2 / 4π^2L

which gives us the thread tension:

**T = 2π root(Lcosθ / g)**

The **horizontal/flat curve motion** of a car into a corner comes next!

Here the **friction **of the tires and the ground (with a friction factor μs) is used!

Let's find the maximum velocity so that the car doesn't drift into the corner!

I waited for ages to say that

Supposing that the car doesn't drift into the corner (no slipping) we have:

F = m u^2/R

and

N - mg = 0

where F is the friction and the only force that turns the car and N is the vertical ground force (action/reaction).

Because the maximum friction is F = μs * N = μs * mg we have:

μs * mg = mu^2 / R =>

**u = root(ms * g*R)**

which is the maximum velocity we can travel at for a specific radius R and a "slip" factor μs.

Now let's get into **banked curves** that look like that:

Suppose that we want to find the degree θ of the slope so that there is no friction needed!

The vertical ground force N now has a degree θ to weight force and so we find the components and use the 1st and 2nd law again like before.

N * sinθ = mu^2 / R (2nd law)

and

N * cosθ - mg = 0 (1st law)

By diving the two equations we end up with:

tanθ = u^2 / gR =>

**θ = arctan(u^2 / gR)**

This is the degree needed for a constant velocity u with a radius R slope turn.

For the **vertical circular motion** I did some great examples at the end of this post!

The centripetal force Fc = mu^2/R must be "constant".

This means that there must be a upward force Ft at the top so that:

Ft - mg = - mu^2 / gR =>

Ft = m(g - u^2/R)

In the same way there must be a upward force Fb at the bottom so that:

Fb - mg = + mu^2 / gR =>

Fb = m(g + u^2/R)

The force Ft is smaller then the weight and the force Fb is greater then the weight!

### Image Sources:

https://qph.fs.quoracdn.net/main-qimg-93da58735e7ec7e183d99f52cc4018a6

https://cnx.org/resources/3b9073156ddf0f5f8e6fe674990b44228050b7cc/unnamed.jpg

https://i.stack.imgur.com/a7CAm.jpg

https://www.physicsforums.com/attachments/yf_figure_05_76-jpg.130139/

And this is actually it for today!

From next time on we will start getting into examples and exercises!

Bye!

ritikabindal (40)7 years agoWell, I haven't seen such informative post in a while yet. Lol I'm quite weak in physics but this is clearly with some great detail.

Following you for more :') Thanks!