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Find the tension in the lower string and the rotation rate

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data

    A 700g ball rotates around a vertical shaft supported by two strings. If the tension in the upper string is 20.0 N, determine (I have attached the file)

    a. The tension in the lower string.

    b. The rotation rate in rpm of the system.

    2. Relevant equations

    F=ma

    F= m v^2/r

    3. The attempt at a solution

    Sum of forces in y direction = 0 (again, I am not sure if this is zero, but I assume zero acceleration in y direction)

    Tension in the y direction of 1 (T1y) - tension in the y direction of 2 (T2y) - weight (mg) = 0

    T1sin theta - T2sin theta - mg = 0

    20sin(theta) - T2sin(theta) = mg

    sin(theta) * (20 - T2) = mg

    T2 = 20 - mg/sin(theta)

    theta = tan^-1 (0.35/0.50) --> Theta = 35

    T2 = 20N - 11.96 N

    T2 = 8.04 N ---> Can anyone verify if this is right?

    b) w = v^2/r but I need v

    sum of forces in x direction = mv^2/r

    -T1cos(theta) - T2cos(theta) --> Theta is 35

    T1 = 20 N
    T2 = 8.04 N

    F = -20cos(35) - 8.04cos(35)

    F = -22.969 N

    -22.969 N = mv^2/r --> -22.969 = (0.7) v^2/(0.5)

    V^2 = -16.406 (not sure why this is negative/where to go from here...) I am stuck.

    Any help is greatly appreciated! Thank you!



     

    Attached Files:

  2. jcsd
  3. Aug 6, 2017 #2

    TSny

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    Your work looks good. The sign issue is related to choosing what you want to call the positive direction for the forces along the radius of the circular motion. If you want to take the positive r direction to be outward, then the acceleration is negative (centripetal). But then, your net force also acts toward the center, so it is also negative.

    If you want to choose the positive direction to be inward, towards the center, then both your force and your acceleration will be positive.
     
  4. Aug 6, 2017 #3
    Thank you so much, I am going to attempt this again in the morning with that though. Much appreciated!
     
  5. Aug 6, 2017 #4

    rude man

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    method is fine, nope, won't check your math :sorry:
    what is this???
    is this with the correct sign?
    EDIT: did not see that T scooped me ... but I agree totally.
     
  6. Aug 6, 2017 #5
    w = rotation rate, which I am asked to find... that is the equation for w.

    Thank you, I appreciate your second opinion.

    Cheers
     
  7. Aug 6, 2017 #6

    rude man

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    It's w = v/r, not w=v^2/r, but all in all you surely did a fine job IMO.
     
  8. Aug 7, 2017 #7
    Yes, you are right! Thank you for the correction.
     
  9. Aug 7, 2017 #8
    So I attempted the solution again. I decided to say that the inward direction (direction of acceleration and force) is positive. From this I got the same values, just without a negative. I got a velocity of +/- 4.05 m/s.

    I had a quick question about the velocity I was hoping you (or someone else) could answer. Since the velocity is +/-, I would choose the positive value because the velocity is related to the acceleration ( a = v^2/r) so it should have the same sign?

    After choosing the positive value I used the formula w = v / r where r = 0.5 m... I got 2 = (4.05 m/s)/ (0.5m) --> w is then 8.1 rad/s

    My issue here is: Do I assume I have radians here or would it be 8.1 revolutions per second?

    I assumed it was 8.1 radians/second and converted to revolutions per minute. Thus obtaining, 77.4 rpm.

    Is that the correct way or would I assume I have rps? Thank you.
     
  10. Aug 7, 2017 #9

    TSny

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    In the equation a = v2/r, v represents speed, not velocity. Speed is always positive, so you would chose the + sign and get v = 4.05 m/s.

    It's radians. ω = v/r is derived from θ = s/r where s is arc length, r is radius, and θ is subtended angle in radians.

    If you want access to some greek letters and various math symbols, click on the ##\Sigma## symbol on the shaded tool bar at the top of the posting window. This tool bar also has superscript and subscript tools.

    Good.
     
  11. Aug 7, 2017 #10
    Okay. That makes sense. Thank you so much for all your help.
     
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