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Toppling cylinder angular velocity

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    I am approximating an object as a cylinder (h=6, r=1, mass=200lb). The cylinder is in the process of toppling over from a vertical position. I will assume that the cylinder can pivot freely on the bottom surface. At some angle theta, the cylinder will strike an object, we'll assume with the top edge of the cylinder. I would like the estimate the impact load exerted on that object.

    2. Relevant equations
    F=m*a
    [itex]\tau[/itex]=r*F*sin([itex]\theta[/itex])
    [itex]\tau[/itex]=I*[itex]\alpha[/itex]
    Erot=[itex]\frac{1}{2}[/itex]*I*[itex]\omega[/itex]2

    3. The attempt at a solution

    I would like to use the impact loading case (basic summary here http://www.pdhcenter.com/courses/s164/s164content.pdf) to help find an equivalent load on impact.
    Basically, in this process I will approximate the object as a spring do an energy balance between the rotation energy of the cylinder and the energy stored in the spring.

    However I needed to find the angular velocity to calculate the rotational potential energy.
    I figured the gravitational force exerts a torque on the cylinder from the center of mass
    F=m*g
    [itex]\tau[/itex]=r*F*sin([itex]\theta[/itex])
    r in this case is the distance from the pivot point to the center of mass of the cylinder
    [itex]\tau[/itex]=[itex]\frac{h}{2}[/itex]*m*g*sin([itex]\theta[/itex])
    Since, [itex]\tau[/itex]=I*[itex]\alpha[/itex]
    [itex]\alpha[/itex]=[itex]\frac{h*m*g*sin(\theta)}{2*I}[/itex]
    This is the angular acceleration. I thought perhaps I could integrate it with respect to t to find the angular velocity. However, it would appear that with [itex]\theta[/itex] in the equation I cannot integrate since theta is not a constant with respect to time.

    Am I missing something important here, or just going about solving this problem totally the wrong way? Any help would be great, and I apologize in advance in case I put this in the wrong forum.
     
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2

    Simon Bridge

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    You have to put the terms with theta in them on the LHS before integrating.[tex]\frac{1}{\sin\theta}\frac{d^2\theta}{dt^2} = \frac{hmg}{2I}[/tex]... if you want. Second-order non-homogenious DE,
     
    Last edited: Jan 17, 2012
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