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A simple Rigid Bodies Question

  1. Apr 15, 2004 #1
    This was a question on one of our past exams, and our professor said it was really simple. Of course, he also said that only one person got it of the 400-some people who took the course, possibly because the scenario was very unusual.

    Here's what I've done so far:

    [tex]\overline{I} = 1.686 kg m^2[/tex]

    When the body is at 30 degrees to the vertical:

    [tex] mgcos60 * (0.57) = [\overline{I} + (375/9.81)(0.57)^2]\alpha[/tex]

    Basically, I treated the body as though it were rotating about a fixed point A and found: α = 7.58 rad/s^2.

    I can then find the tangential component of acceleration at mass centre G relative to A: [tex]a_{(G/A)t} = (0.57)\alpha = 4.32 m/s^2[/tex]

    I figure that since the wheelchair is a rigidly connected, the acceleration at point a is equal to the acceleration of its mass center. So: [tex]a_A = 15 m/s^2[/tex] to the right.

    Then I use this equation to try and find the angular velocity:

    [tex]a_G = a_A + a_{(G/A)t} + a_{(G/A)n}[/tex]
    [tex]a_G = 15[/tex]
    [tex] + 4.32 [/tex][60 degrees to the vertical] + [tex](0.57)\omega [/tex][30 degrees to the vertical]

    This is where I get stuck, because I don't know the direction of [tex]a_G[/tex]. Is it simply directed towards the left?

    I'm not even sure if my previous steps were correct. Any help or corrections would be appreciated!​
     
  2. jcsd
  3. Apr 16, 2004 #2
    Aha! Solution found! (I hope!)

    I've figured out the solution (or so I hope), and thought I'd post it here in case anyone else might be interested. My previous approach was apparently completely off the tracks. The start was correct:

    [tex]a_{(G/A)t} = (0.57)\alpha[/tex]
    [tex]a_{(G/A)n} = (0.57)\omega^2[/tex]

    FBD Here - forces are in red. acceleration in blue.

    I basically broke down [tex]a_G = a_A + a_{(G/A)t} + a_{(G/A)n}[/tex]

    Then I took the moment about A:

    [tex] mgsin\theta * 0.57 = \overline{I}\alpha + m\overline{a}d[/tex]
    [tex] mgsin\theta * 0.57 = \overline{I}\alpha + m(a_{(G/A)t})(0.57) - ma_A cos\theta * 0.57[/tex]

    Subbing in the numerical values, I get:

    [tex] \alpha = 15.15 sin\theta + 23.17 cos\theta[/tex]

    Then using this relation:

    [tex] \alpha = \omega\frac{d\omega}{d\theta} [/tex]
    [tex] \int_{0}^{\theta} (15.15sin\theta + 23.17cos\theta)d\theta = \int_{0}^{\omega} \omega d\omega[/tex]
    [tex] 13.61 = \frac{1}{2}\omega^2[/tex]
    [tex] \omega = 5.22 rad/s[/tex]

    Does that look correct?
     
  4. Apr 16, 2004 #3
    Another Question

    So as to not spam the forum with too many of my posts, I'd like to ask another question I've been having trouble with.

    There is a Semi-Circle.

    Given:
    m = 10 kg
    ω = 4 rad/s [counterclockwise]
    R = 0.4 m
    OG = 4R/3π
    θ = 45 degrees, as indicated on the diagram.
    I = 0.51168 kg m^2
    coefficient of static friction = 0.5

    Question: Determine whether the disk slips at the instant.

    The procedure we've been taught is to assume that the disk is rolling without sliding, and then check our assumption that F <=F(max), where F is the friction.

    The thing I'm having trouble with is the geometry of the whole shape. I have no idea how far horizontally or vertically A is from G or O. So, how would I go about summing moments/torque about any point without that information? The angle given is probably crucial to solving it, but I can't seem to calculate any of the distances.

    If I knew the distances between points A-O and A-G, I'd set up an equation where the friction, normal forces and weight both contribute to create torque on the object around the point O, where the a(o) = r&alpha;.

    Can anyone help?
     
  5. Apr 17, 2004 #4

    arildno

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    Isn't O the origin of the circle while A is a point on the circumference?
    Furthermore, on the circle, the radius vector OA is normal to the tangent line, i.e the ground.
     
  6. Apr 17, 2004 #5
    Okay, thanks a lot! I really should work some more on my geometry skills. I just wasn't sure if I was allowed to assume OA is normal to the ground.
     
  7. Apr 17, 2004 #6

    arildno

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    I think there's a general theorem in geometry which states that if two differentiable surfaces meet in a single point, their normals at that point will be parallell to each other.

    However, just consider the following:
    1. A line segment meets a circle in a point P. If you draw the full line, you will see that most lines will cut a section off the circle.
    (Lines differ in their approach angle)
    In fact, there is only a single line that grazes the circle.
    (Showing that hones your geometry skills..)
    2. This must therefore be the tangent line to the circle at P.
    3. Since the ground grazes A, and only A, of the circle, it is the circle's tangent at A, which by definition of the circle is normal to OA.
     
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