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Conical Pendulum, find tension, radial force, speed, period, and angular speed

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A metal sphere is attached to the end of a string and then set in motion such that it rotates in a horizontal circle as shown in the sketch. The metal sphere has a mass of 0.5kg and the radius of the circle is .2m.

    1)Find the tension in the string
    2)Calculate the radial force
    3)Determine the speed of the rotating mass
    4)Determine how long it takes for the sphere to complete one rotation, i.e. determine its period
    5)What is the sphere's angular speed in rad/seconds?


    2. Relevant equations
    Assuming the metal sphere is on the left side of the circle;
    F_netx=Tcos(θ)=ma_x
    F_nety=Tsin(θ)-mg=ma_y=0
    θ(angle between x-axis and string)=arccos(.2)

    3. The attempt at a solution
    I believe I found everything except angular velocity, but the period I found also seems too short. Any input would be appreciated!
    1)T=mg/sin(θ)=5.001N
    2)Tsin(θ)=4.899N
    3)Tsin(θ)=mv^2/r, 4.899=.5v^2/.2, v=1.3999
    4)Δt=Δs/v=2∏(.2)/1.3999=.8977s
    5)I know ω=dθ/dt, but I'm not sure what to do with this...
     

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    Last edited: Oct 15, 2012
  2. jcsd
  3. Oct 15, 2012 #2

    cepheid

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    Gold Member

    Welcome to PF LibGirl17!

    This looks good. It doesn't really matter where in the circle the sphere is. You can always resolve the tension into two components, the "vertical" component, and the "radially-inward" component (the component of the tension that points towards the centre of the circle). The latter lies in the plane of the circle and is thus "horizontal" (although its direction changes, this doesn't matter, because it is always radially-inward).

    With a cone, you typically measure the angle from the vertical, in which case the horizontal component would have the sine factor, and the vertical component would have the cosine factor. I'm not sure where you're measuring the angle from here. EDIT: I see that you're using the angle from the horizontal, which means that the vertical component is Tsin(theta), and the horizontal component is Tcos(theta). That means you should use Tcos(theta) for the centripetal force below.

    This seems like the right general method. The radial component of the tension is equal to the centripetal force, so use that to solve for v.

    As for the angular velocity: in this case, the rotation rate is constant, so it's also true that ω = Δθ/Δt. You know how long it takes the object to go through an angular displacement of Δθ = 2π radians (one full rotation), because you computed the period.
     
    Last edited: Oct 15, 2012
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