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Homework Help: Tangential and radial acceleration

  1. Jul 6, 2013 #1
    A ball tied to the end of a string 0.50 m in length swings in a vertical circle under the influence of gravity. When the string makes an angle x= 20 degrees with the vertical, the ball has a speed of 1.5 m/s. Find the magnitude of the radial component of acceleration at this instant.

    So i have a concept question. How do i know that tangential acceleration = gsin(20 degrees).

    Basically, prove the general formula tangential acceleration= gsin(x) for this problem.
    Last edited: Jul 6, 2013
  2. jcsd
  3. Jul 6, 2013 #2


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    Draw a free body diagram on the ball at the instant specified in the problem (when it it makes an angle of 20 degrees with the vertical). Then decompose the forces into the radial and tangential directions. That's a good rule of thumb for when you are solving problems using Newton's 2nd law.
  4. Jul 6, 2013 #3
    we havent read about newton's 2nd law yet -- thats the next chapter.

    there is already a drawning. i dont see why tangential acceleration=gsin(x)
  5. Jul 6, 2013 #4


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    Well you don't really need it here, I was just saying it's a good rule of thumb in general. But do what I said: draw a free body diagram on the ball; what are the forces acting on the ball at that instant? How can you decompose these forces into radial and tangential directions?

    If there's already a diagram, which part of it is troubling you?
  6. Jul 6, 2013 #5
    im not there yet. im on chapter 4, motions in two dimensions. The only force we think of is gravity.

    nvm, a teacher helped me with it already. he just moved the radial acceleration vector all the way to the top so that they tip of that vector touches the top of the string. i see the right triangle now.
  7. Jul 6, 2013 #6
    We assume that the string cannot change its length. That means that any force applied to the end of the string is effectively cancelled along the string; but its transverse component acts unopposed.
  8. Jul 6, 2013 #7
    I am not thinking about forces in this chapter yet, except for gravity.
    hmmm, in g2=at2+y2, what is y2? and what is the physical meaning of this? y doesnt seem to be ar.

    I mean pertaining to this triangle at=gsin(x).
    Last edited: Jul 6, 2013
  9. Jul 6, 2013 #8
    What I wrote fully applies to gravity.

    'y' could be the "radial" acceleration if the ball were allowed to fly freely. But it is not, so 'y', and the entire equation for that matter, does not make a lot of sense physically.

    What does make sense is ## (mg)^2 = (ma_t)^2 + P_r^2 ##, where ##P_r## is the radial component of the weight.
  10. Jul 6, 2013 #9
    If it does not make sense, then i am still confused about why at=gsin(x).

    I havent read about Mass/weight/newtons laws yet so can you explain it with just what I know.
  11. Jul 6, 2013 #10
    I did in #6. Just replace "force" with "gravity" there.
  12. Jul 6, 2013 #11
    I still don't get how this makes sense of at=gsin(x). The transverse component of gravity?
  13. Jul 6, 2013 #12


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    Gravity exerts a force of mg straight down. What is the component of gravity in the direction the ball is traveling when the angle of the string X is 20° from vertical?
  14. Jul 6, 2013 #13
    9.80 m/s2 South. I see a triangle but what is the 3rd side?
  15. Jul 6, 2013 #14
    Gravity acting on the ball on a string has the radial and tangential (transverse) components. Find them, that is simple trigonometry.
  16. Jul 6, 2013 #15
    Yes, I get hypotenose g and leg at. I want to know what this other leg is -- I dont understand that.
  17. Jul 6, 2013 #16
    You were given an explanation earlier. You rejected it because you did not know some concepts. I am afraid you will have to wait till you know those concepts.
  18. Jul 6, 2013 #17
    Why does Pr not include mass?
  19. Jul 6, 2013 #18
    The full equation should be ##(mg)^2 = P^2 = P_t^2 + P_r^2 ##. Tangentially, there is no other force except ##P_t##, so we can write ##P_t = ma_t## and then have ##(mg)^2 = (ma_t)^2 + P_r^2 ##. Radially, however, other forces may be present, ##ma_r = P_r + F##, so we cannot replace ##P_r## with ##ma_r##.
  20. Jul 6, 2013 #19
    Were you taught the formula for radial acceleration? V2/R. Note that, in this question, it does not depend on g, since, as stated in the problem description, the ball is traveling in a circle.
  21. Jul 6, 2013 #20
    Im only talking about the tangential acceleration.

    I think Pr=mar+F makes more sense because Pr is mar plus other possible forces. Why is it the other way?
    Last edited: Jul 6, 2013
  22. Jul 6, 2013 #21
    whoops xD
  23. Jul 6, 2013 #22
    No, ##P_r## is not ##ma_r## plus other possible forces. The basic equation is ## mg = P = P_r + P_t ##, where ##P_r## and ##P_t## are the radial and tangential components of weight; they are uniquely defined by the orientation of the radius and do not depend on any other force.

    In addition to that, Newton's second law requires that ##ma_r## be equal to the sum of all the forces, so ##P_r = ma_r + F ## would violate that law.
  24. Jul 6, 2013 #23


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    This is similar to a wedge or inclined plane where the hypotenuse is the direction the ball is moving, which would be perpendicular to the string which is 20° from vertical. The other two sides would be vertical and horizontal. For triangle based on the force from gravity, the hypotenuse is a vertical line (vector), the radial component of gravity would be a line (vector) 20° from vertical, and the tangental component would be a line (vector) perpendicular to the radial component.
    Last edited: Jul 6, 2013
  25. Jul 6, 2013 #24
    Oh, excuse me. My fault. You wanted to focus on the tangential acceleration. You would do well to follow rcgldr and wannabenewton's advice. When the mass is 20 degrees from the vertical, the gravitational force on the mass can be resolved into components radial and tangential to the circular path. The radial component of the gravitational force is mg cos 20, and the tangential component of the gravitational force is mg sin 20. This is the only force acting on the mass in the tangential direction (in the radial direction, there is also a tension force from the string). Since, in the tangential direction, the only force acting on the mass is mg sin 20, this must be equal to its mass times its acceleration in the tangential direction:

    mat = mg sin 20

    If you cancel the m's from the above equation, you get

    at = g sin 20
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