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Components of Acceleration

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    One end of a cord is fixed and a small 0.250 kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.50 m as shown in the figure below. When θ = 20.0°, the speed of the object is 7.70 m/s. At this instant, find each of the following.

    (a) the tension in the string

    (b) the tangential and radial components of acceleration

    ar =

    at =

    (c) the total acceleration

    2. Relevant equations

    I have been trying to use ar = mgcos(20)
    at = mgsin(20)

    Also for the Tension, I was using T= mg(v^2/Rg + cos(20)

    3. The attempt at a solution

    My numbers are not coming out right, and I am afraid I don't no understad this problem in the slightest. Please help me. Thank You.
     
  2. jcsd
  3. Feb 13, 2008 #2
    Any help would be appriciated. I am new to this, please help.
     
  4. Feb 14, 2008 #3
    THIS SHOULD HELP YA AT ITS BEST...::

    Tension=mg*cos theta

    FOR Tangential accln...
    TAKE angular displacement as=(pi/180)*20 radian
    v=7.70 m/s
    Omega=v/R
    Use omega^2=0+2*alpha*(pi/180)*20 radian

    Calculate alpha...
    then calculate t using omega=0+alpha*t

    then Tangential accln=v/t

    Total accln={at^2+ac^2}^1/2

    Radial accln=v^2/r
     
  5. Feb 14, 2008 #4
    Newton's laws will always rescue us.:smile:

    That is incorrect. That is only so when the object/particle is at the extremes of oscillation and at such points v=0 but it is not so in this case.

    That is not right. Radial accelaration is T - mgcos(20) = m(v^2)/R for it is centripetally accelarating.( Fnet= ma! You've got to consider all the forces.)

    Thats right. How come you have used T-mgcos(20)=m(v^2)/R here? Tension should come out right.

    Feel free to ask anything else if you still feel unsure about something.
     
  6. Feb 14, 2008 #5
    I believe you meant centripetal acceleration.

    Radial acceleration comes from the other component of the weight
     
  7. Feb 14, 2008 #6
    Huh? It is radially accelerating inwards, yes? Then should radial accn. not equal Centripetal accn.?
     
  8. Feb 14, 2008 #7
    You said the tension part is incorrect, what is the correct way to find the tension?
     
  9. Feb 14, 2008 #8
    Radial acceleration means the acceleration that is linear to its movement,

    Centripetal acceleration means the acceleration that is perpendicular to its movement.

    EDIT:
    OK so sorry, radial acceleration is centripetal acceleration. The acceleration that is linear to movement is known as tangenial acceleration
     
    Last edited: Feb 14, 2008
  10. Feb 14, 2008 #9
    I figured out the tension and the tagnet acceleration, but I need to figure out the radial now. How might I do that, I don't understand all the stuff that guro did. I used at = gsin(20).
     
  11. Feb 14, 2008 #10
    Ok, I have solved all of the problem except for one piece:

    atotal = 23.9 m/s2 inward and below the cord at °

    What does below the cord mean? I know it's asking for an angle.
     
  12. Feb 14, 2008 #11
    Hi Jim,


    For an object moving in a citcle we know that the resultant acceleration is given as the centripetal force of acceleration. In the question, the tension and the weight component add up to this centripetal acceleration. Can you figure this out now?
     
  13. Feb 14, 2008 #12
    It's Joe. Yes I did! Thank You! I took the tan inverse of at/ar and that have me 8.04. Thank You!
     
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