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Angular speed; horizontal circle

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A student ties a 400g rock to a 1.0m long string and swings it around her head in a horizontal circle. What angular speed does the string tilt down at a 10° angle?

    m = 400g = .4kg
    r = 1m

    2. Relevant equations
    [tex]F = \frac {mv^2}{r}[/tex]

    [tex]\omega = \frac {|v|sin \theta}{|r|}[/tex]

    [PLAIN]http://img31.imageshack.us/img31/7350/circ1.png [Broken]



    [PLAIN]http://img513.imageshack.us/img513/513/circ2.png [Broken]

    3. The attempt at a solution
    First I solved for the tensions:

    [tex]T_y = 9.8 * .4 = 3.92N[/tex]
    [tex]tan 10 = \frac {3.92}{T_x}[/tex]
    [tex]T_x = 22.23N[/tex]

    Then for velocity:
    [tex]22.23 = .4 * \frac {v^2}{1}[/tex]
    [tex]v = 7.45 m/s[/tex]

    Then for angular speed:
    [tex]\omega = \frac {7.45 sin 10} {1}[/tex]
    [tex]\omega = 1.29°/sec[/tex]

    I wasn't sure if I set this up correctly or used the right formulas. I also wasn't sure if my answer was in the correct units. I also wanted to know if the picture for this motion is like the 1st picture or the 2nd picture? I used the first picture to solve this problem.
    [tex][/tex]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 2, 2011 #2

    cepheid

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    In the first picture, the string is 10 degrees below horizontal, which is correct.

    In the second picture, the string is 10 degrees above horizontal.
     
  4. Oct 2, 2011 #3

    cepheid

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    I didn't look too closely at it, but it seems like your overall method is right. The y-component of the tension has to support the weight, and the x-component of the tension has to provide the centripetal force that keeps the thing moving in a circle.

    Your answer of degrees per second is in the wrong units. If you look carefully at your actual equation, you'll see that the result is in (m/s)/m = 1/s. However, we usually express this as radians/second in order to make it clear that it is an angular velocity. In the radian system, angles are dimensionless quantities (being expressed as the ratio of two lengths). Hence radians are a dimensionless unit, which is why they don't appear in the units unless if you explicitly put them in.

    If you want the answer in degrees per second, you'll have to convert from radians to degrees.

    In general this is something you have to be careful with. The formula [itex] v = \omega r [/itex] implicitly assumes angles measured in radians.
     
  5. Oct 2, 2011 #4

    cepheid

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    Another thing that I just noticed is that in the equation F = mv^2 / r, r is the radius of the circular path being swept out. This is NOT the same thing as the length of the string here. So that might be another source of error. Check it out.
     
  6. Oct 2, 2011 #5

    gneill

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    Be careful with the radius of the circle!

    attachment.php?attachmentid=39546&stc=1&d=1317600349.gif

    EDIT: Hmph. cepheid got in ahead of me!
     

    Attached Files:

    Last edited: Oct 2, 2011
  7. Oct 2, 2011 #6
    Ah yes I forgot the radius was different from the length of the rope.

    [tex]r = 1 * cos 10 = .98m[/tex]
    [tex]22.23 = .4 * \frac {v^2}{.98}[/tex]
    [tex]v = 7.38 m/s[/tex]

    [tex]\omega = \frac {7.38 sin 10} {.98}[/tex]
    [tex]\omega = 1.32/sec[/tex]

    So at this point, is the answer in radians/sec? If not, at which point do I convert it?

    [tex][/tex]
     
    Last edited: Oct 2, 2011
  8. Oct 2, 2011 #7

    gneill

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    Okay so far
    Can you explain the above? Where did 7.45 come from, and why the sine function?

    If you have the speed of the rock and the radius of the circle it travels in you can work out the circumference and time it takes to make one circuit of the path (the period). One circuit of the path also covers [itex]2\pi[/itex] radians...
     
  9. Oct 2, 2011 #8
    Oops, v is supposed to be 7.38 m/s. I used the sine function because that was part of the angular velocity formula. But keep in mind that I wasn't sure if I was using the right formulas.
     
    Last edited: Oct 2, 2011
  10. Oct 2, 2011 #9

    gneill

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    I can't place 7.38 m/s from your calculations either :frown:

    The quickest road from velocity to angular velocity is via the relationship [itex] v = \omega r[/itex].

    If I may make the suggestion, you might want to keep a few more decimal places in your intermediate results so that rounding error doesn't slowly erode accuracy over multiple calculation steps. By all means round final results to the correct number of significant figures, but hold on to accuracy till the end!
     
  11. Oct 2, 2011 #10
    So I use
    [tex]F = \frac {mv^2}{r}[/tex] to solve for velocity, and use [itex] v = \omega r[/itex] to solve for the angular velocity?
     
  12. Oct 2, 2011 #11

    gneill

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    Yup. Quick and easy!
     
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