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Math Help - Physics Problem: stationary bead relative to a revolving wire

  1. #1
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    Physics Problem: stationary bead relative to a revolving wire

    I have been working on this problem for hours. I have been working on it on Google Docs with another user on this site for a while: https://docs.google.com/drawings/d/1...vre0Z6feA/edit

    The problem: A small bead with a mass of 100g slides without friction along a semicircular wire with a radius of 10cm that is rotating about a vertical axis at a rate of 2.0 revolutions per second. Find the value of angle for which the bead will remain stationary relative to the rotating wire.

    The answer: 52 degrees

    Method: 1.588cos(theta) - the weight of the bead = 0 solve for theta

    I'm not sure why that method works (gets the answer) and that's my biggest problem. I want to understand how this problem works.

    This problem is difficult to understand without looking at the picture in the textbook. I drew the picture on the Google Docs link you may view. Tell me if you need any more information.
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  2. #2
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    Re: Physics Problem: stationary bead relative to a revolving wire

    Quote Originally Posted by over9000 View Post
    The problem: A small bead with a mass of 100g slides without friction along a semicircular wire with a radius of 10cm that is rotating about a vertical axis at a rate of 2.0 revolutions per second. Find the value of angle for which the bead will remain stationary relative to the rotating wire.

    The answer: 52 degrees
    I will use \theta as the angle between the normal force of the wire and the horizontal as shown in the diagram ... it will be complementary to your angle.

    r = R\cos{\theta} , where R is the radius of the semicircle and r is the radius of the bead's rotation.

    N_y = N\sin{\theta} = mg

    N = \frac{mg}{\sin{\theta}}

    N_x = F_c

    N\cos{\theta} = mr \omega^2

    \frac{mg}{\sin{\theta}} \cdot \cos{\theta} = m R\cos{\theta} \cdot \omega^2

    \frac{g}{\sin{\theta}} =  R\omega^2

    \frac{g}{R \omega^2} = \sin{\theta}

    \theta = \arcsin\left(\frac{g}{R \omega^2}\right) \approx 38^\circ
    Attached Thumbnails Attached Thumbnails Physics Problem: stationary bead relative to a revolving wire-beadproblem.jpg  
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    Re: Physics Problem: stationary bead relative to a revolving wire

    what are does w stand for?
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    Re: Physics Problem: stationary bead relative to a revolving wire

    someone else explained this to me: It's the equation that omega is squared.
    angular acceleration (a_c) is velocity-squared divided by R, which is equal to omega-squared times R.
    He put the radius at the point of the bead in terms of R, the max radius of the circle with r = R*cos(theta)
    Ny = N*sin(theta) = mg, because the normal force on the wire is equal to the only counter force in that direction: weight.
    Thus, solving for N, we have N = mg/sin(theta)
    Nx = Fc = m*a_c = m*r*(omega-squared) = m*R*cos(theta)*(omega-squared)
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