1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stone revolving vertical

  1. Nov 2, 2006 #1
    Hello there!

    My study mate and I have been trying to figure this one out but have not had success. This is the problem we would appreciate to get some help with:

    A ball with mass of 20 grams is attached to a robe and is revolving vertikal in a circle. There is no energy beeing added to it. At the ball highest point, a force of 0.2 Newton is pulling on the robe. What is the force on the ball"s lowest point?

    The statet result says 1.4 Newton.

    The most logical way of solving this problem is that at the highest point, the sum of all forces is the the centripedal force minus the gravitational force:

    [tex] 0.2 N = - mg + m\omega^2r [/tex]

    At its lowest, the sum of the forces that act on the ball should be
    [tex] F = + mg + m\omega^2r [/tex]

    Is this assumption right? With those to equations we get 0.4 Newton, which sounds actually good to me. Maybe my professor is wrong ?!

    Any help will be appreciated!
  2. jcsd
  3. Nov 2, 2006 #2


    User Avatar

    Your method looks OK to me - but using that same method I don't get 0.4N

    Use the first eqn to find a value for mw^2 r

    Although neither do I get 1.4N - any suggestions anyone?
  4. Nov 2, 2006 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Two comments:
    (1) You assume the same angular speed at top and bottom. Not so: At the bottom, the ball will be going faster. (Remember conservation of energy.)

    (2) "centripetal force" is not a force! It's a term used to describe the net force acting towards the center in circular motion. So, the net force on the ball at the top is:
    [tex]F_{net} = -T -mg[/tex]

    This equals the "centripetal force", via Newton's 2nd law applied to circular motion:
    [tex]F_{net} = - m v^2/r[/tex]

    [tex]-T -mg = - m v^2/r[/tex]

    Where I take down to be negative. At the bottom, the tension and acceleration both point up. (And the speed and tension are different, of course.)
  5. Nov 2, 2006 #4


    User Avatar

    Hmm, When we do this at A-level we always have the stone travelling at constant speed.

    is it posible to solve this otherwise, with no further info?
  6. Nov 2, 2006 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Sure you can solve it: Use conservation of energy.
  7. Nov 2, 2006 #6


    User Avatar

    Yes, but when I tried that, I was left with r in the equation, which we don't have.

    What have I missed?

    .....I'll keep trying....
    Last edited: Nov 2, 2006
  8. Nov 2, 2006 #7


    User Avatar

    Cancel that, think I've got rid of the r and yes, it gives the answer required.

    fara, from the top of the circle, when you find a value for mv^2/r, you have to be able to convert this to kinetic energy. The at the bottom, after you've added the KE gained, you have to be able to convert it back to a new value of mv^2/r

    Unless there's a quicker way, Dr Al?
    Last edited: Nov 2, 2006
  9. Nov 2, 2006 #8
    thank you very much guys for your help but unfortunately, I am totally clueless about converting the [tex] \frac{mv^2}{r} [/tex] into kinetic energy. It just does not appear to me ! :(
  10. Nov 2, 2006 #9


    User Avatar

    (mv^2)/r and 0.5 mv^2

    So to convert mv^2/r to KE you multiply by r/2

    Use the data you've been given for the top pf the circle to find mv^2/r
    Convert it to ke.

    Add on the ke you gain (from GPE) from top to bottom of circle.

    Now you have the total ke at the bottom. This time, to turn it back to mv^2/r you multiply by 2/r - handy because otherwise you´d be left with an r in the expression.

    Now use your original T = mg + mv^2/r at the bottom to find the tension.

    I got 1.38 doing it this way - which is close enough to what you want.
  11. Nov 2, 2006 #10
    yeah! I have got 1.377 N but I do not understand why you can multiply it by [tex]\frac{r}{2}[/tex] to have it converted into KE.
    Here is what I figured out:

    Situation 1, top :
    Situation 2, bottom:


    So with that I get the velocities:

    Conservation of Energy says:


    After solving it to [tex]T_2[/tex] I got:
    and that is 1.37 N!

    We really appreciate your help guys!
    Last edited: Nov 2, 2006
  12. Nov 2, 2006 #11


    User Avatar


    as for the conversion, well they both contain mv^2 , one divided by 2 and one divided by r. So you use the r and the 2 to convert one to the other.
  13. Nov 2, 2006 #12
    well, sometimes it is so obvious that one cannot see it ;)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Stone revolving vertical