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I Uniform Circular Motion and Centripetal Acceleration (non-intuitive)

  1. Aug 9, 2018 #1
    I'm in the chapter of Uniform Circular Motion and I have a hard time understating centipetal acceleration. Until now I knew that acceleration describes "how fast velocity changes in magnitude" except projectile motion because when it reaches maximum height then g is going to change the direction on the Vy and make it move downwards but now in this kind of motion, acceleration describes only "how fast the direction of the velocity changes" and not the magnitude. Ok I understand that but what I really don't understand is when I'm in a car and take a turn I feel pushed in the opposite direction of the circles center. But acceleration is pointing at the center of the circle
     
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  3. Aug 9, 2018 #2

    vanhees71

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    Any change of the velocity (a vector!) means that there's non-zero acceleration, because by definition
    $$\vec{a}(t)=\dot{\vec{v}}(t)=\ddot{\vec{x}}(t).$$
    For the uniform motion on a circle (for convenience I put it in the ##xy##=plane of the coordinate system) you have
    $$\vec{x}(t)=r [\vec{e}_x \cos (\omega t) + \vec{e}_y \sin(\omega t)].$$
    Now you can easily evaluate velocity and acceleration.

    Then, using Newton's 2nd law you immediately get which force you have to apply to the particle to make it move in the described way along a circle.
     
    Last edited: Aug 11, 2018
  4. Aug 9, 2018 #3

    Ibix

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    No - acceleration is how fast velocity vector is changing (or, more precisely, the rate of change of the velocity vector). The magnitude only changes sometimes; the direction only changes sometimes. The vector always changes. I remember feeling weird about that too, but it must be that way. There's a force acting on a mass (e.g. a string pulling on a stone being swung around in a circle), and ##F=ma## tells you there must be an acceleration.
    There are two ways of looking at this, and you are mixing them up.

    The first way to look at a car going round a corner is from outside it. I'm standing on a bridge above the road, looking down on your car as you start to corner. What I see is that the car is cornering, but your body is trying to carry on in a straight line (Newton's First Law). But if you carry on in a straight line, the car door curves into you and presses you into a curve. That's the centripetal force, and it's my explanation for why you turn.

    The second way to look at it is from inside the car. From this perspective, the world is turning and the car isn't moving. But you still need an explanation of why the door is pressing into you - mine doesn't make sense because the door doesn't turn in this perspective. Your explanation is the centrifugal force (sometimes called a fictitious force, or an inertial force) that just comes out of nowhere and presses you to the door.

    The book is using the first perspective. It's mathematically a lot simpler to handle, but it's a bit counter-intuitive since we're seldom doing detailed analysis of the kinematics of a car passing underneath a bridge. The second explanation is the natural one for us since we're used to travelling in cars, but it involves more complex maths to explain what inertial forces are so physics textbooks tend to avoid it until further on in your education. I think I was taught it in my second year at university (although I don't think I grasped the significance until much later...)
     
  5. Aug 9, 2018 #4

    vanhees71

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    The problem is that many books do not use the really simple mass, as I suggested to use in #2. Some textbook make very simple things very complicated by a misguided attempt to "avoid math". It should be very clear that math is helping to make things much simpler and cannot be avoided anyway. So one should introduce the appropriate math right away when it's useful for a given problem, rather than avoiding it. For mechanics you need vector algebra, derivatives and integrals as well as ordinary differential equations.
     
  6. Aug 9, 2018 #5

    Delta2

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    That's true , you feel LIKE there is a force that is pushing you to the opposite direction of the centripetal acceleration.

    But what you actually feel, is a force from the car's seat mainly, that is pushing you towards centripetal acceleration (so the force on your body from the car's seat is on the same direction as centripetal acceleration), and due to Newton's 3rd Law, your body is pushing the car's seat with an opposite force, that is your body is pushing the car's seat in a direction opposite to centripetal acceleration. That push from your body to the car's seat is what your wrongly understand as "being pushed in direction opposite of the centripetal acceleration".
     
  7. Aug 9, 2018 #6

    A.T.

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    When the car accelerates forward, you end up in the back of the car, if you are not fixed an thus fail to accelerate with the car. The same applies to accelerating left or right in a turn.
     
  8. Aug 9, 2018 #7
    I'm a little confused... Well I still haven't reached the chapter about forces and Newton's laws. Now I'm in the relative motion chapter and next is the forces and motion chapter and Newton's first law. Do you thing I will have a better understating after that chapter???
     
  9. Aug 9, 2018 #8
    Best reply!!!! Now I see!! Well I can't wait to reach the chapter of forces and Newton's laws to get a deeper understanding!!!
     
  10. Aug 9, 2018 #9

    Delta2

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    Yes, in my opinion you absolutely need a good understanding of newton's 2nd and 3rd law to fully understand what's going on.
     
  11. Aug 9, 2018 #10
    Thanks for the advice!
     
  12. Aug 11, 2018 #11

    Your answer make's the most sense now (i finished reading about uniform circular motion), but still i can't understand a few things. It's weird, i reached so far learned about motions, forces, Newton's laws and now I've stuck... When a particle move's in a uniform circular motion there is a Centripetal force towards the center.

    Question 1:
    So if we assume there is no gravity and no other forces applied on the particle, why the particle is not moving towards the center of the circle??? So if I name r the axis where the Centripetal force is lying , according to Newton's second law:

    ∑ƒr = m⋅a ⇔ -ƒc = m ⋅ ( -v^2 / R ) ⇔ fc = m ⋅ ( v^2 / R ) but still i can't understand why the parcicle is moving in a circle and not towords the center. If you can use maths to explain use them.

    Question 2:
    Let's say i have a rope tied in a particle and I'm spinning it (around a y axis) with my hand. Then the particle is entering a uniform circular motion but also it does not fall to the ground and according to Newton there must be another force with the same magnitude as Fg (weight) in the opposite direction so the particle won't move at all on the y axis. Also if i speed faster and increase the constant speed of the uniform circular motion, the rope will break. Consider the following sketch:
    IMG_20180811_174958.jpg

    I have a really hard time understanding the cause of these forces. It must F = -Fg so the particle won't fall to the ground or actually won't move to the y axis at all! There must be also a T force that when it gets high enough it will break the rope. Can someone explain them to me? In this scenario does the T force of the rope being related to the centripetal force?
     
  13. Aug 11, 2018 #12

    vanhees71

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    What's wrong with my suggestion in #2? It's really not more to the kinematics of circular motion than this very simple mathematical exercise!
     
  14. Aug 11, 2018 #13
    x(t)=r[excos(ωt)+eysin(ωt)]

    I can't understand how to relate the formula above with the forces. Also its a little hard for me to understand what the formula actually describes. Probably an orbit around the circle (position)? Maybe because I don't think "multidimensionaly"? I see a force (pulling the particle towards the center of the circle) and i'm wondering why it does not moving towards the center? Is it because i'm thinking that the force is being applied in on dimension? Please also consider the sketch i uploaded in #11
     
    Last edited: Aug 11, 2018
  15. Aug 11, 2018 #14

    vanhees71

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    Ok, so maybe I did not understand at which level you are studying. You marked the thread as to be I level. So I assumed you know vectors and calculus.

    ##\vec{x}(t)## describes the trajectory of the particle in 3D space, i.e., it gives the position in terms of a vector pointing from some fixed origin of your reference frame as a function of time. If you take a Cartesian basis system ##\vec{e}_j## (##j \in \{1,2,3 \}##), i.e., three unit vectors that are mutually perpendicular to each other, you can describe the vector by three real numbers, its components with respect to this Cartesian basis, which are functions of time:
    $$\vec{x}(t)=\sum_{j=1}^3 x_j(t) \vec{e}_j.$$
    Now the motion along a circle with constant angular velocity ##\omega## is described by the above given formula
    $$\vec{x}(t)=r [\vec{e}_1 \cos(\omega t) + \vec{e}_2 \sin(\omega t)], \quad r>0 \quad \text{constant}.$$
    Now the 1st derivative with respect to time gives the velocity,
    $$\dot{\vec{x}}(t)=\vec{v}(t)=r \omega [-\vec{e}_1 \sin(\omega t) + \vec{e}_2 \cos(\omega t)].$$
    The second derivative is the acceleration
    $$\ddot{\vec{x}}(t)=\dot{\vec{v}}(t)=\vec{a}(t)=-r \omega^2 [\vec{e}_1 \cos(\omega t) + \vec{e}_2 \sin(\omega t)]=-\omega^2 \vec{x}(t).$$
    Now according to the 2nd law, this implies that you need a force
    $$\vec{F}=m \vec{a}=-m \omega^2 \vec{x}.$$
    This tells you that the force has magnitude $$F=|\vec{F}|=m \omega^2 r=\text{const}$$, and at any instant of time points radially inwards, i.e., the force pulls the mass instantly towards the origin of the coordinate system, around which it rotates, to keep it on its circular orbit.
     
  16. Aug 11, 2018 #15
    LOL. I'm a university student and I know calculus and vectors. The problem was in the symbols (I'm such an indiot :p ). Because in greece we use i,j and k as the normalize vectors of the Cartesian system I did not realize that e1 and e2 was actually the normalize vectors (did't even noticed the arrows above them) :-p and thus i did not understand that the formula was actually producing a vector... And i was saying to myself "how can a scalar size give the position of an orbit?" Now i understand!!! I really appreciate your help! I can't understand why the book is hiding this formula, everything seems much easier now.
     
  17. Aug 11, 2018 #16
    So, can you know explain why when you are spinning a particle that is tied in a rope (so the particle is doing a uniform circular motion) why it does not fall to the ground due to it's weight? Also if the speed of the uniform circular motion is big enough it can break the rope, why is that? Also i want to ask something else. In the book: a = v^2 / R (this is the magnitude only) .The ω which u use in the formula above is something else than V? Why the book does not mention something about ω when it should?
     
  18. Aug 11, 2018 #17

    vela

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    Consider a projectile. The only force on it, the force of gravity, points down, but it doesn't fall straight down as you seem to suggest the spinning particle should. Why is that?
     
  19. Aug 11, 2018 #18
    I don't know :sorry: The projectile moves horizontally with a constant velocity but vertically accelerates with a = -g, so it will hit the ground sometime. The spinning particle accelerates all the time and moves in an orbit where it's velocity changes due to the acceleration. Maybe this is the reason? But there is no way to see it with mathematics? I can't just believe words... There's must be a force on the particle except it's weight (Fg) so ∑ƒy = 0 ⇔ -m⋅g + F = 0 ⇔ F = m⋅g and maybe this F has to due something with the Centripetal force. Probably I'm wrong but this the most logical thing to think of.

    Or better, take this example:
    %CE%91%CE%9D%CE%91%CE%9A%CE%A5%CE%9A%CE%9B%CE%A9%CE%A3%CE%97-2.png
    When the ball is upside down then the forces which are being applied to it are it's weigh, the normal force and the centripetal force so : -ƒc - m⋅g - ƒn = m⋅ay, in other words if the ball does not fall it means ay ≠ 0 and actually ay > 0. According to Newton if there is an ay > 0 (towards the sky) then it must be also a force in the same direction in such a way -ƒc - m⋅g - ƒn + F = 0 <=> F = ƒc + m⋅g + ƒn

    Honestly I can't think of something else...
     
    Last edited: Aug 11, 2018
  20. Aug 11, 2018 #19

    CWatters

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    Here is how I like to explain it...

    Consider a particle moving in a straight line. To deflect it off course slightly you need to give it a brief sideways force/nudge. Afterwards its still moving in a straight line but in a different direction. Give it another nudge and it changes direction again. With care you could make it move in a square pattern, pentagon, hexagon or octagon. The more sides the shape has the more nudges you need to give it and the more frequently you need to give them. In the extreme (eg a circle) you need to give it a constant sideways force. This is called the centripetal force.

    You can calculate the centripetal force _required_ to move in a circle of radius r at velocity v....

    F = mv2/r

    If you apply a force greater than mv2/r to the particle it WILL move toward the centre. You can rearrange the equation to..

    r = mv2/F

    If F increases the radius r reduces. If F reduces the radius r increases.

    Note that F is the net force on the particle. In some cases the net force has two components, gravity might be one and string tension or normal force from a track the other.
     
  21. Aug 11, 2018 #20

    CWatters

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    You mean spin it in a vertical circle?

    At all times the net force must be mv2/r otherwise it won't move in a circle.

    At the top "Falling down" requires the radius to reduce. In order for that to happen the centripetal force must be greater than mv2/r.

    That will happen if gravity provides too much centripetal force. Another way to say the same thing is.... Gravity will provide too much centripetal force (and it will fall) if the velocity is too slow.
     
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