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Centripedal motion

  1. Nov 26, 2006 #1
    This is not really a homework question...it's more like me begging for a link. I'm in physics 131...basic freshman college class. I can do most of the stuff ok. Think about what problem is about, draw a pciture, draw some arrows...label some forces, sum things up in perpendicular directions...sin/cos some forces if I must to make them break into components...etc

    But I can not do centripedal motion when the circles are vertical. at all.

    question 1:

    I need to know the following: if a circle has radius..."r" and no friction, how fast must it enter the circle in order to do a loop. (imagine a roller coaster...).

    -Problem: Gravity acts downwards. The centripedal acceleration dictates that the force that is supposed to fight gravity is gonig to the center of cirle. So if both forces go downwards...what makes the object stay on the circle? I can picture it in my head...I know how things work, but I can't find out why.

    question 2:

    I'm swinging from a rope (i had similar problem on midterm...result wasn't pretty). and some guy collides with me. By conservation of momentum I can find my velocity right after the collision. But how do I know how far I'll swing? If I try "conservation of energy" and set Kinetic initial = potential final...I always get an answer different than if i try basic F=ma equations. And once again...gravity...and centripedal...how do I sum them up?

    Plz help? link? direct answer?

    thank you anticipated.

    ~Robokapp
     
  2. jcsd
  3. Nov 26, 2006 #2
    Does conservation of energy tell you the velocity at all points along the trajectory?

    Do you know the centripetal force (hint: normal force plus radial component of gravitational force)?

    For the swinging rope, are you claiming you correctly integrated the vector forces in applying F=ma?
     
  4. Nov 26, 2006 #3
    Uh...I never had to use integral for it. That's exactly it...I'm thinking it's something that changes as rope changes position, but my teacher, TA and everyone else are using...simple algegra.

    One piece of info I got form my TA, and I memorised because I have no idea where it came out is that in order to have a cart loop on the inside of a frictionless circle, the entry speed must equal or exceed

    v=(gr)^(1/2)

    But my problem remains: if a cart at an instant is suspended at the top of an inner circle, what force will keep it from falling? In a circular motion acceleration goes towards the center (down) and gravity always goes down...so what keeps it up?


    Conservation of Energy...Well it should. there is no work due to friction...and the word "conservation" suggests yes. There is nothing but kinetic and potential. So Kinetic initial plus potential initial (zero because ground level) becomes kinetic final (zero because its maximum height, so speed equals zero) and potential final (max because the height is maximum).
     
    Last edited: Nov 26, 2006
  5. Nov 26, 2006 #4
    [tex]\Sigma F=ma[/tex]
    You know that. You also (hopefully) know that since this is uniform centripital motion, [tex]a=\frac{v^2}{r}[/tex], so Newton's Second law can be rewritten as [tex]\Sigma F=m\frac{v^2}{r}[/tex]
    We're looking for minimum velocity here, which means that we want the right hand side of the equation to be as small as possible. That means we want the left hand side, ie total force, to be minimum as well. So at the top of the loop, let's only allow what force we must--gravity. Now the eqation reads:
    [tex]F_G=m\frac{v^2}{r}[/tex] But F_G=mg, so

    [tex]mg=m\frac{v^2}{r}[/tex]
    The masses cancel, and solving for v gives [tex]v=\sqrt{gr}[/tex] which is your TA's equation.

    Incidentally, this may help explain the other enigma, as well:

    At the top of the loop, force of gravity actually acts as the centripital force.
     
  6. Nov 26, 2006 #5
    Yes I knew that formula for centripital (is it not spelled "centripedal"? not trying to be rude, just asking)acceleration.

    Okay...but why do you call it "uniform centripedal motion". what do you mean by "Uniform". It is descelerating and accelerating. Do you mean "continuous"?


    What I understand (and thank you for it, it's very useful) is that what decides the result of sum of forces is the type of motion, not the individual forces. My mistake was that I was trying to add both "ma" and "mv^2/r" in same equations.

    correct me if I'm wrong:
    But assuming we want some force to exist. The equation would look like this:

    [tex]F_G+F_c=ma[/tex]

    [tex]F_G+F_c=m\frac{v^2}{r}[/tex]

    [tex]F_c=m\frac{v^2}{r} - mg[/tex]

    Correct?

    I understand this now...and I've already scheduled a meeting with my teacher but I'd love to know the answer to my second question also. I know all the info, just not how to put it all together. I spent (not enough) time looking at examples of problems etc.

    So object A is at the end of a rope motionless, and object B collides with it and they stick together. (they dont have to stick, just an example) By conservation of momentum,

    [tex]m_1v_1+m_2v_2=m_(1,2)v_f[/tex]

    [tex]v_f = \frac{m_1v_1} {m_1+m_2}[/tex]

    But what do I do to decide how high they go? Do I use

    [tex]K_i=P_g_f[/tex]

    [tex]mv^{2}/2 = mgh[/tex]

    [tex]mv^{2}/2mg = h[/tex]

    [tex]h = \frac{mv^{2}} {2mg}[/tex]

    [tex]h = \frac{v^{2}} {2g}[/tex]

    But is this right? I mean...velocity is calculated in momentum part for horisontal not vertical direction. Somehow I have the feeling I should use

    [tex]v_f^{2}=v_0^{2}+2ad[/tex]

    But even here...does the "distance" represent distance as in Arclength or as in vertical distance?
     
    Last edited: Nov 26, 2006
  7. Nov 26, 2006 #6
    By the second law, Fnet = ma_c. At the top and bottom of the hoop, the only forces that would sum in Fnet are N (normal) and mg.

    At the top (assuming the roller coaster is underneath the hoop):

    N + mg = ma_c = mv^2/r

    At the bottom:

    N - mg = ma_c = mv^2/r


    It's spelled "centripetal" (I think, I am the wurst spellur myself.) This isn't an example of unform circular motion, since the speed changes around the loop. But the case of the very top and the very bottom are the same for both uniform and non-uniform circular motion.

    Sticking or not sticking makes all the difference. Energy is not conserved over the collision if they stick, but it is if they do not stick. (This is the difference between non-elastic and elastic collisions.)

    For your example, since the collision is not elastic, energy is not conserved before and after the collision.

    However, it will be conserved after the collision. So, you can use conservation of momentum to find the velocity of the combined body immediately after the collision. Then use conservation of energy to find the maximum height to which it will rise. The equation would be:

    (m1 + m2)v'^2/2 = (m1 + m2)gh

    where v' is the velocity after the collision, and m1, m2 are the masses of the two bodies.

    h is the vertical distance. Conservative forces are independent of path, so the only thing that affects potential gravitation energy is the distance it moves vertically.

    Hope this helped,
    Dorothy
     
    Last edited: Nov 26, 2006
  8. Nov 26, 2006 #7
    Yes I know sticking/bouncing matters. For clarification I picked one of the two situations so you guys don't have to guess. I expressed what I meant poorly...

    EDIT:

    so Energy conservation would indeed tell you the height.

    Thank you for the help. it did help a lot. A small question however

    Are you assuming the "positive" y-direction is Down? or are you considering g as being negative?

    signs of parts of equations have always been an issue. To avoid this, I always considered g in my equations as -|g| so I don't have to stick the - later and screw up.

    Off topic: funny story but my first projectile motion problem I was looking at a cannonball and question was when it lands. My "g" in the ...+gt^2/2 equation remained positive...and you can imagine me scrolling in my calculator, minutes after launch, at outrageous heights...out of orbit :D
     
    Last edited: Nov 26, 2006
  9. Nov 26, 2006 #8
    Hi Robokapp,

    Well, you picked the trickiest one (in my opinion), so good choice :-)

    Hope the rest of it was helpful,
    Dorothy
     
  10. Nov 27, 2006 #9
    heh heh. Been there, done that. For myself, I think it's easier to always have a positive g, and use the physics of the situation to adjust the sign of the terms, but whatever works for you is what you should use, of course.

    I was using down as negative. Probably you are concerned about the equation at the top of the loop. In that cases, all the forces are down, so all the terms would be negative. In a case like that, I find it encourages me to make fewer mistakes if I write it all as positive.
     
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