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Tension, speed, in a circle

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A 68.8 g steel ball is attached to the end of a string. The ball is swung in a vertical circle 86 cm in diameter. (a) What is the minimum speed the ball must have at the top in order to complete the circle without falling? (b) If this speed is constant, what will be the tension in the string at the bottom? Draw a separate FBD for both parts. Show all work. Hint: minimum speed to complete the circle is the velocity needed to keep the ball from falling when the tension at the top is just zero

    2. Relevant equations

    3. The attempt at a solution

    We know mass...(68.6g or .0686kg) and we know the radius (86/2 is 43)

    I just dont know where to go from here???

    for Part a)

    at the top of that circle....we know that mg (force of gravity is acting downwards) and centripetal force is also acting downwards

    but is this really a centripetal acceleration problem?

    A_c = V² / R

    we know what R (radius) is....but we need V (velocity)....and how do we figure out what A_c (centropetal acceleration) would be??

    any hints here?
  2. jcsd
  3. Mar 19, 2013 #2
    Actually wait...isnt

    F_tension = F_centripetal + mg?
    since the F_tension at the top is 0...

    0 = F_centripetal + .67424kg

    so the Centripetal force would be


    now with that

    F_c = V²/R

    V² = F_c * R
    V = √(F_c * R)

    wait no that cant be right....WOULD the centripetal force be "-"? it has to be right because it's acting in the direction opposite to what the ball wants to travel in....but we cant have a √ of a negative number.... :/

    inspiration gone lol

    any hints?
  4. Mar 19, 2013 #3
    The acceleration at the top, just like elsewhere depends on the forces acting on the ball. That is why you are required to do the FBD. Do it, use the hint provided, and you will find out what the acceleration at the top is.
  5. Mar 19, 2013 #4

    so there's that....I realized that since there is NO tension force acting at the top of the circle....the only forcing acting on the ball are

    F_g = mg


    F_c = m V² / R

    so in order for the ball not to fall....the F_c would have to equal mg


    *******radius = 84cm or .84m

    mg = mV² / r

    so the m's cancel out

    g = v² / r

    v² = r * g

    v = √(r * g)

    v = √(.42 * 9.8)

    v = √4.116

    v = 2.0288 m/s

    is this correct?
    Last edited: Mar 19, 2013
  6. Mar 19, 2013 #5
    Your result is correct. However, the reasoning in the beginning is a little wrong.

    ma = mg + T

    Now you have been told that T = 0 (no tension).

    So ma = mg, or a = g.

    On the other hand, a = v^2 / R, so g = v^2 / R.
  7. Mar 19, 2013 #6

    Ohh okay that does make sense...thank you....

    Now for part b) the tension at the bottom of the circle....

    we would have F_g , F_c , and Tension acting on the ball

    we want Tension

    based on the FBD i drew ....tension would equal the sum of F_g and F_c (makes sense) so

    T = F_g + F_c

    F_g = mg = -.67424N ***"-" becuase it is acting down right?

    F_c = mV² / R
    = (.0688 (2.0288²)) / .42 (since it says assume speed is constant..i used the speed I got from the previous part)

    = -.6742448226 ***also "-" because the string is being pulled down by the ball

    so Tension would equal the sum of both of these

    T = (-.67424 + -.6742448226) = -1.348484823N

    is THIS correct? I'm very not sure
  8. Mar 19, 2013 #7
    You cannot just postulate T = F_g + F_c. What you can say is that there are two REAL forces: F_g and T. The RESULTANT force is then their sum. We know that the motion is circular, so we can call this force "centripetal", so we should have F_c = F_g + T. Note this is different from what you postulated.

    Now, F_c is always directed toward the center - that's why it is called centripetal to begin with. At the bottom that direction is UP, so it has to be positive.
  9. Mar 19, 2013 #8

    That does make sense.... what I was thinking through that whole thing was

    at the bottom of the circle

    gravity is pushing down on the string..therefor making it feel a force pushing down....that's why It was "-"

    and the centripetal force acting on the ball...is pulling the ball toward the center...so I was thinking that would make it also "-" since the ball wants to stay in the straight line and the centripetal force wants to pull it in a circular motion....wrong I know...the positive force would be the centripetal...and negative being the gravity

    so the equation you posted of

    F_c = F_g + T

    does make sense because the tension would depend on how big the force of gravity and how big the centripetal force...if the 2 forces are nearly equal...there would be little tension right?

    T = F_c - F_g

    makes sense

    So now plugging in all the numbers from before

    T = .6742448226N - .67424N

    T = .0000048N ***if this is correct then what I said above about the forces being nearly equal would result in barely any tension in the string

    now look good?
  10. Mar 19, 2013 #9
    Wait a minute....

    The net force to the center...wouldn't it be mg - T???

    the mg is acting down...while T is acting up isn't it???

    wouldn't the formula be

    F_c = T - mg???


    m(V²/r) = T - mg

    so T would indeed equal the sum of the centripetal force and the force of gravity

    so wouldn't my answer be

    .6742448226 + mg = T

    the only thing I'm confused on here....is....i know mg is the negative force here...but would I use


    if it is the prior

    .6742448226 + -.67424 = T
    so T would equal the .0000048

    or the latter

    .6742448226 + .67424 = T
    making T = 1.348484823N

    the latter confuses me because...if the tension of the string is greater than the weight of the ball...wouldn't the string snap?? I'm sorry for all the "beginner" questions I'm just trying to understand the reasoning behind all this
  11. Mar 19, 2013 #10
    You are getting confused by signs. That happens a lot, so don't worry.

    What works best is just pretending at the FBD that all the vertical components point up, and all the horizontal components point right. Then the equations are simple:

    ma = F1 + F2 + F3 + ...

    All the plus signs.

    THEN, as you get to plugging in the givens, pay attention if any given component should be negative or positive. It should be negative if YOU KNOW FOR SURE it points down or left. For example, F_g = -mg, always.

    For unknown components, do not do anything - leave the signs they already have to them. Then, after you solve for them, they will automatically come out negative if they should point left or down, or positive otherwise.

    Now try to apply this logic to the problem at hand.
    Last edited: Mar 19, 2013
  12. Mar 19, 2013 #11
    ohh yes of course

    ƩF_y = ma_y

    Tension + F_c + F_g = 0 (no acceleration right?)


    T + m(V²/R) - mg = 0

    T + m(V² / R) = mg

    T = mg - m(V² / R) ***considering I use 9.81m/s² for g

    T = .674928 - .6742448226

    T = .00068317740

    is what I get?
  13. Mar 19, 2013 #12
    Wrong. The motion is circular, so there is acceleration. The centripetal acceleration.

    F_c is not an independent force, as remarked above. It is the sum of all the other forces, so you should either have

    F_c = ƩF


    ma = ƩF

    and F_c = ma
  14. Mar 19, 2013 #13
    ohh, the centripetal ACCLERATION... the things that's throwing me off is centripetal ACCELERATION and centripetal FORCE

    so then like you said...it would be

    F_c = T + F_g

    m(V²/r) = T - mg

    T = m(V² / r) + mg

    so then yeah......if this is true.....it should be that first answer I got...because since we had to change -mg to mg for the equation...it would make it positive...and adding that to the centripetal force would make

    T = 1.348484823N

    It's just not clicking in my head what so ever...was this correct?
  15. Mar 19, 2013 #14
    Yes, now it is exactly correct.

    Observe that the tension is greater than the weight in magnitude. That's known as a "g-force". You have probably experienced in a roller coaster drive that when the car is going in a "valley" you are pressed down, while when it is going over a "hill", you feel much lighter, sometimes almost weightless. The same thing is happening here.
  16. Mar 19, 2013 #15
    and let me just clarify this.....if the weight of an object on the string...is greater than the tension in the string.....the string will "or at least has the potential to" snap correct?

    just like if a picture was hanging via 2 strings....and the tension in both those strings were greater than the weight of the picture....the strings would have no chance of snapping...but if the weight were increased....the strings would be more "able" to snap?

    are those correct statements?
  17. Mar 20, 2013 #16
    If the weight of an object on the string is greater than the tension in the string, that means there is LESS load on the string than in a static configuration. Statically, the tension would be equal to the weight.

    At the top of the circle, the tension is less; at the bottom, it is greater.
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