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Rock on stringsstuck on last part of problem

  1. Oct 20, 2003 #1
    Picture is attached as follows:

    A 480 gm rock is whirled on the end of a string 47 cm long which will break under a tension of 27 N.

    a) What is the highest speed the rock can reach before the string breaks? (Neglect gravity.)
    b) If two other strings identical to the first were attached to the rock, how fast could the rock be whirled before the three strings would break?

    Next the rock is held by two of the same 47 cm strings with ends 56 cm apart and whirled in a circle between them. Neglect gravity.
    c) What is the radius of the circle of motion?

    d) Now what is the maximum speed the rock can have before the string breaks?

    I'm stuck on part d, but I can go over a-c. Maybe that might help me solve c.
    a) Well, the first thing I did was convert grams to kilograms and cm to meters. Then by using this equation, F=(m*v^2)/R
    where F=force
    R=radius, i get V=5.14 m/s

    b) Same equation as above, except force now equals 81N (3*27)
    answer is 8.91 m/s

    c) This I drew a triangle. Since it is isoceles, I figured out that height (radius) = sqrt(47^2-.25*56^2) =>37.7 cm as my radius.

    d) Here I'm stuck. I know I can use the centripetal equations: a=v^2/R or F=(m*v^2)/R, but how to go about it I'm unsure about it.
    I did draw a triangle, where the hypothenuse was 47, the base is 37.7 cm..and since two angles are 70 and 90 degrees, the last angle is 20 degrees. So I did 47 sin 20 to get the x component, which turned out to be 16.07 cm. Whether this can be of any use to me, I don't know. The hint tells me to break the tensions into their components, but then what?

    Attached Files:

  2. jcsd
  3. Oct 20, 2003 #2


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    Yes, parts a- c you have done correctly.

    d is difficult. Once again, the centripetal force in toward the center of the circle is given by "F=(m*v^2)/R" as you had before.

    However, now you have two strings holding the rock, at an angle.

    If you drew a picture, then you saw that you have two right triangles. The hypotenuse of each right triangle is 47 cm long. The "base" is 56/2= 28 cm long and so the "opposite" side has length √(472-282) cm= 37.7 cm approximately, just as you got before.

    Now, each string will support half of the "m*v^2/R"= (0.48/0.377)v2= 1.27 v2.
    That is, the opposite side of the "force" triangle will be 1.2 v2. The hypotenuse of that triangle will be the tension in the string. Using "similar triangles", we can set up:
    T/(1.2v2= 47/37.7 or T= 1.2v2(47)/37.7= 1.52v2. Set that equal to the maximum value for T: 27 N and solve for v.
  4. Oct 20, 2003 #3
    Ok, so the three sides of the triangle are hyp = 47, x = 28, and y = 37.7. So what i did was do inverse sin of 28/47 or inverse cos of 37.7/56, and got the top angle (the angle between the string and the center-pointing tension force) to be 36 degrees. From there, I took the vertical component of the 27 N tension force, 27cos36, and got the vertical component of one string to be 21 N. I then multiplied this by two because there are two strings, and got 42, and used this as the tension. You set Fc = Ft, and I got the terminal velocity to be 5.7 m/s.
  5. Oct 21, 2003 #4


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    No, the vertical component is 27 sin(36.7) not cosine.

    Since you just used sin(θ)= 28/47 it is simpler to use similar triangles, as I suggested, to arrive at that (trigonometry is just a complicated way of working with similar triangles!).
  6. Oct 21, 2003 #5

    Chi Meson

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    Um. I think in this situation, the 36 degrees is made from the vertical axis, so the vertical component would be found using cosine.

    edit: in my picture I have the two strings being held from points separated horizontally, and the rock is traveling in a vertical circle (but without gravity)

    I agree that using similar triagle is eassier as it avoids this confusion.
    Last edited: Oct 21, 2003
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