Circles
Drive Faster Lyrics

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Circles running all through my life but I'll be just fine I just need some time to grow into something unknown before there's an unlocked door to somewhere I want to go people say not to waste my life its not worth the fight and this I say to you there is no wrong there is no doubt take me now before I finally fall apart she broke my heart and now I know what it's like to be far apart Circles running all through my life but I'll be just fine I just need some time to grow into something unknown before there's an unlocked door to somewhere I want to go people say not to waste my life its not worth the fight and this I say to you there is no wrong there is no doubt take me now before I finally fall apart she broke my heart and now I know what it's like to be far apart I'm moving on and on and on and on and on without you.

Overall Meaning

In Drive Faster's song "Circles," the lyrics describe the feeling of being stuck in a cycle or pattern that runs throughout the singer's life. They acknowledge that they need time to grow and change before they can move forward and find their way to somewhere they want to go. Throughout the song, the singer seems to be grappling with the idea of whether or not life is worth fighting for, and ultimately decides that there is no wrong or doubt in taking risks and making the most of their time.

In the chorus, the singer repeats the phrase "Circles running all through my life but I'll be just fine" as a sort of mantra, reminding themselves that despite feeling stuck in a loop, they have the power to break free and make a change. The song then shifts to focus on a relationship that has ended and the pain that comes with that loss. The singer expresses their desire to move on without the other person, but acknowledges the difficulty of that task.

Overall, "Circles" is a reflective and introspective song that explores themes of self-discovery, growth, and loss. The lyrics encourage listeners to embrace uncertainty and take risks in order to break free of the patterns that keep them stuck.

Line by Line Meaning

Circles running all through my life but I'll be just fine
My life is full of patterns that I can't seem to escape, but I know I'll be okay.

I just need some time to grow into something unknown before there's an unlocked door to somewhere I want to go
I need some space to figure out who I want to become before I can find the right path to take.

People say not to waste my life its not worth the fight
Others may tell me to take the safe route, but I believe that it's worth it to fight for what I want.

And this I say to you there is no wrong there is no doubt
I am confident in my choices and have no regrets or doubts about them.

Take me now before I finally fall apart
I'm afraid of what happens if I don't make a change soon.

She broke my heart and now I know what it's like to be far apart
I've experienced heartbreak and the pain of being distanced from someone I care about.

I'm moving on and on and on and on and on without you.
I'm letting go of someone who is no longer a positive presence in my life and moving forward without them.

Contributed by Sadie R. Suggest a correction in the comments below.

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Most interesting comment from YouTube:

@carultch

Here's how to prove it.

Consider a body moving CCW around a circular path of a fixed radius R, centered on the origin, and starting on the x-axis. The arc length it travels is given by a generic function L(t). We will call its speed V(t), which is also a general function. We expect that the centripetal acceleration will equal V(t)^2/R, and that the tangential acceleration will equal dV(t)/dt.

We can relate arc length L(t) to V(t) by the definition of speed:
V(t) = L'(t) = dL(t)/dt

The angular position in radians directly relates to the arc length it travels:
L(t) = R*theta(t)

Its position vector is determined by:
r(t) = R*<cos(theta(t)), sin(theta(t))>

The unit radial vector is:
ur(t) = <cos(theta(t)), sin(theta(t))>

And the unit tangential vector is perpendicular to this one:
ut(t) = <-sin(theta(t)), cos(theta(t))>

First derivative, which is velocity:
v(t) = r'(t) = R*<-theta'(t) * sin(theta(t)), theta'(t) * cos(theta(t))>

Second derivative, which is acceleration:
a(t) = r"(t) =

x-term:
d/dt rx'(t) = -R*theta"(t) * sin(theta(t)) - R*theta'(t)^2*cos(theta(t))

y-term:
d/dt ry'(t) = R*theta"(t) * cos(theta(t)) - R*theta'(t)^2*sin(theta(t))

Notice that the first term of the acceleration vector components has the same trig functions as the tangential vector, and likewise, the radial unit vector has the same trig functions as the second term, but negated.

This means we can express this with the unit vectors ut(t) and ur(t):
a(t) = R*theta"(t)*ut(t) - R*theta'(t)^2*ur(t)

Recall theta(t) = 1/R*L(t), which means:
theta'(t) = 1/R*L'(t) = 1/R*V(t)
theta"(t) = 1/R*V'(t)

Which substitutes as the following:
a(t) = V'(t) * ut(t) - V(t)^2/R^2 * u(t)

The component of a(t) that is in the tangential direction, as you can see, is equal to the rate of change in speed. This is the tangential acceleration.

The radial component of a(t) is equal to -V(t)^2/R. The negative of the centripetal acceleration.

All comments from YouTube:

@alondragonz2009

4y↑2

Thank you so much!! you explain everything slowly and clearly and take your time to derive and make it clear to the students! which some professors forget and just derive everything in their head and skip steps constantly! Even my own text book only gives you a brief description of the math but no reason WHY they're using those numbers or derivations instead. Which made it completely confusing even while following the book! Thank you so much! you are saving my semester!!! Please keep making videos!!!

@Noname2000100

your video helped me so much understand the concept better, thank you so much for providing us with this awesome and quality videos !!!!!!

@IGJTHOMAS

Thank you for all of your videos. These videos make more since than what my professor teaches. The examples with real world application retains much better than my professors way of going through a variable riddled proof and ending class with everyone stumped.

@yoprofmatt

Justin Thomas,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A

@josemarie2007

6y↑6

Sending you my regards from Singapore. Physics Mid term this saturday. You are a life-saver Prof Anderson.

@nephilimdeath9940

5y↑1

Hello sir. I'm Rakesh from India . I really appreciate your work Sir. I really liked your lectures .Thank you so much Sir . Ur lectures were there when I had hard times clearing my entrance exam for med . I would like you to know that I solved few questions in exam (All thanks to you sir) and I'm in med school now . THANK YOU SIR . ILL REMEMBER U TILL I DIE.

@representative1103

6y↑3

it is a helpful example, now i know what is the uniform circular motion means. thankyou sir!

@yoprofmatt

Great to hear. Keep up with the physics.
Cheers,
Dr. A

@GOATaro_

29w

My professor blows straight through this stuff but this guy makes it like baby food. Thank you so much.

@manuboker1

GREAT PHYSICS LECTURES !!! :))

@anirudhyasarkhel8247

It should be also applicable to an electron revolving around a
neutron in an atom

@carultch

At one point that was the model of the atom (the Bohr model), but it is no longer an accepted model of the atom due to not being consistent with experimental results. Other quantum mechanical effects come in to play.

@gardenmenuuu

sir I have a question silly question but yet please reply...If a body is moving in a circle and I apply torque to it,won't it contribute to the centripetal force too?Because the velocity's magnitude is increasing ?

@carultch

3y↑2

It turns out that only the existing velocity magnitude contributes to the centripetal acceleration. The torque, and correspondingly linear force that causes the torque, will contribute to its tangential acceleration instead. That is, the acceleration component that is parallel to its velocity, which results in a change in speed. After some time, when the tangential acceleration causes its speed to increase, the new speed will contribute to its centripetal acceleration. But at the first instant, all that mattered was its existing speed and the radius of curvature.

@carultch

@AbdArahman

@Carl Hansen
thank you carl for all your comments, you helped me out.

@aysarabdulbaki2769

6y↑2

Like around a 25m radius, we can ever get a speed of 100 kmph... and I usually do it... may you please explain the reason

@IppokratisPoimenidis

5y↑1

YOU GOT A GREATER FRICTION (TIRES ON DRY ASPHALT)

@hozwayy2594

they were just taking assumptions in the video, since we have no exact clue whether 45mph would be the max speed you could go in a 25m radius road. and it would make sense that if you have a curved road with a higher radius, you could go a lot faster and vice versa

@rosspetersen699

26w

So good.

@12HedmanLane

3y↑2

Why is it I understand all this math but not how he is writing it so I can read it?

@giselledepakakibo8643

5y↑3

Why is it static friction instead of kinetic friction? Isn't the car in motion?

@yoprofmatt

5y↑5

Yes, but rolling tires = static friction, skidding tires = kinetic friction. You have to consider where the rubber meets the road.
Cheers,
Dr. A

@carultch

5y↑1

The bottom surface of the tires are "locally stationary" relative to the road, when the wheels are rolling. It is known in dynamics as the instantaneous center of zero velocity or ICV. The bottom point of contact is stationary, while the top point on the tires has a velocity twice the car's velocity. That is why the static friction (traction) coefficient matters for vehicles in motion, rather than the kinetic friction coefficient. The kinetic friction coefficient matters for your brake pads.

Keeping your tires in static friction with the road is important, because otherwise skidding sideways would be no different than skidding forward. A steered tire "tells" the road to exert static friction forces that act sideways. A powered tire "tells" the road to exert a forward friction force. A braked tire "tells" the road to exert a backwards friction force. The friction force would only act directly opposite your velocity, if it were only kinetic friction. That's what happens when brakes lock up, and you loose your ability to steer. Thanks to ABS brake systems that pump the brakes for you, modern cars mitigate this risk.

@NinjatoBlade

So the mass falls out? Does this actually make sense? Under the assumption that whatever vehicle I decide to spin around in a circle is using a rubber wheel on concrete, I could get the same speed on a bicycle as I would a hummer or even an airplane, and as long as I didn't go over 32 mph I wouldn't spin out, but I would spin out on all three if I went over?

Thats a very strange concept, I imagine its different in real life as you have an issue with air perhaps lifting any of these objects as a result of going over them, but I still feel like mass should play more of a role, no?

@yoprofmatt

Correct.

@gabee8535

6y↑2

Well weight and center of gravity aren't included so no all three objects would be different.

@carultch

It is only for a first order understanding of friction, that this is true. I.e. that friction coefficients only depend on the identity of the surfaces interacting, and not on geometry, force or kinematics. Useful for introductory physics, and consistent with experimental results for many situations. Part of the difference among different vehicles, is not all tire rubbers are an identical recipe of rubber. The durometer (a measure of stiffness) of the rubber will play a role in different friction coefficients, although that would be explained as being a different "identity of the surfaces".

In reality, other factors will come in to play, when they happen to an extreme enough degree. Tread pattern, contact pressure, temperature from history of frictional heating will all play a role. Contact pressure will govern what category of contact is between the two surfaces, and contact pressure is not necessarily uniform across the whole contact patch.

@psychomind3105

6y↑1

why we take friction perpendicular to the car towards left????

@yoprofmatt

6y↑6

Something has to hold the car in its circle. The only thing between the tires and the ground is friction. Since the car "wants" to go straight, you turn the wheels to make it go left. This means the rubber molecules are pushing to the right on the asphalt. The asphalt then pushes back to the left on the rubber (Newton's 3rd). This is the static friction force that keeps the car in its circle. (Obviously, if we drive the car in a circle around to the right, then the static friction force will point to the right.)
Cheers,
Dr. A

@the_eternal_student

Stop using metric units, use english units.

@hemendraande9265

Wassup

@yoprofmatt

Hemendra Ande,
Wassup wid you?
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A

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