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A ball with some energy and yes a string

Am I correct?

Poll closed May 11, 2015.
  1. Yup.... you are..

    25.0%
  2. Is this a joke? no you are wrong like you are always

    50.0%
  3. None of these(democratic one)

    25.0%
Multiple votes are allowed.
  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    There is a question that says:
    A small bal of mass m is attached to the end of the string of length ##l=1m## whose other end is fixed. From its lower position, the ball is given a kinetic energy ##\frac {mgl}{5}##. Find the net acceleration (in ##m/s^2## ) of the ball at the instant when the string makes an angle ##\theta## of##37^{¤}##.

    2. Relevant equations
    ##\frac {mgl}{5}=mgh+1/2mv^2##

    3. The attempt at a solution
    since at an angle 37 degrees the ball will cover a distance ##\frac {l}{5}## because of which the kinetic energy at this point is zero. This it's velocity at that point is also zero. So the net acceleration should be equal to ##g## I. E. ##10m/s^2## since the centripetal force is equal to zero. But the answer is ##6m/s^2## by taking ##gsin\theta##. But then ##gcos\theta## is gravity's component in the radial direction, why is it ignored?
     
  2. jcsd
  3. May 4, 2015 #2
    As I understand it, the radial component of the force of gravity on the ball is countered by the tension in the string. Since the ball has zero velocity when it makes an angle of 37° with the vertical, it has no radial acceleration. That only leaves the tangential acceleration.

    This is all assuming the ball is moving in circular motion, which it does not specify in the problem statement. Saying that something has a certain kinetic energy gives no clue as to what direction that something's velocity is in.
     
  4. May 4, 2015 #3
    So since the question doesn't say that the ball leaves the circular path can we assume that tension is still present as a general rule?
     
  5. May 4, 2015 #4
    Not really. For instance, the ball could have that kinetic energy, but the velocity could be directly upwards. There would be no tension involved, and the acceleration due to gravity would just be 9.8 m/s2

    The question is too vague. Did you get it out of a textbook? Was there a picture with the question?
     
  6. May 4, 2015 #5

    SammyS

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    With no tension in the string, it's impossible to have any meaningful relationship between the angle the string makes with the vertical and the position of the ball.

    It would have been clearer had the question specified that the string remains under tension. However, it can reasonably assumed that the string is under tension when θ < 90° .
     
  7. May 4, 2015 #6
    I understand that. What I am saying is that since we are not told that the ball moves in circular motion nor are we told the direction of the velocity of the ball, no meaningful conclusion can be reached whatsoever.
    I agree that it can be assumed that while the ball is in its lowest position that there is tension in the string. It cannot, however, be assumed that the velocity of the ball is directly horizontal. If it was, then the resultant motion of the ball would be circular motion. There is nothing in the problem statement to indicate this.

    I don't mean to be argumentative, by the way. This is just a poorly worded question. (no fault of mooncrater's)
     
    Last edited: May 4, 2015
  8. May 4, 2015 #7

    SammyS

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    So, I suppose that we disagree.

    Now what I wonder is, what is OP trying to accomplish by making this a poll ?
     
    Last edited: May 4, 2015
  9. May 4, 2015 #8
    Wait, don't just say that! You are probably right. Can you tell me what is wrong with my previous statement?
     
  10. May 4, 2015 #9

    SammyS

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    The problem does state that the string makes an angle of 37° with the vertical (well, someone mentioned vertical).

    A limp string can't very well do that.
     
  11. May 4, 2015 #10
    Hmmm. You have a point.
     
  12. May 4, 2015 #11
    No.... there wasn't any picture with it.
    I was just trying out that option... no other reason...
     
  13. May 4, 2015 #12
    Is that sufficient to say that the the string will have a tension?
    I have done some questions and they asked the angle at which the ball leaves the circle... so if this angle is definable then why this limp or no limp matters?
     
  14. May 4, 2015 #13

    SammyS

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    If the kinetic energy at the bottom is sufficient for θ to exceed 90°, then the ball might leave the circle. The string doesn't push the ball outward, does it?

    What's the complete statement of this problem?

    As we learn more about this, some of the early mystery may disappear.
     
  15. May 5, 2015 #14
    The given question is the whole question .... I have omitted nothing.
     
  16. May 5, 2015 #15
    They should have given what is the tension in rope or some other information.
    If we consider tension 0 N and centrifugal force at that point is zero then your answer of 10ms-2 is correct.
    Is your book having hints section? (Obviously you would not have asked if it would have been given in your textbook but is there something in hints that you are not understanding? ). Otherwise you would have experienced earlier that sometimes there are misprintings.
     
  17. May 5, 2015 #16
    Why would you do that?

    There is absolutely no reason to involve fictitious forces in this problem.

    They didn't need to. The string has to be tense in order for it to make an angle, as Sammy already said.

    If you follow this line of thinking you should get an answer of about 6 m/s.
     
  18. May 6, 2015 #17
    But SammyS is saying that the problem statement is not complete.
    I'm a bit confused here.
     
  19. May 6, 2015 #18

    SammyS

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    It's complete enough to work this out.

    It could have been more complete, to remove some speculation. But it can be worked out with reasonable assumptions.

    Aleph is doing a fine job helping you.
     
  20. May 6, 2015 #19
    Okay, but how can we take radial component of gravity?
    Gravity is always working vertically down.
    We can take components for Tension only.
     
  21. May 6, 2015 #20
    Sorry for my above reply.
    Did some drawing, getting radial component countered by tension as the ball has no radial acceleration .
    So only we have tangential acceleration which is gsinθ.
    It should be 6ms-2. :-p
     
    Last edited: May 6, 2015
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