2 thoughts on “5.28”

  1. When integrating theta-double-dot in this problem, why are we able to assume the constant of integration is zero? This is in contrast to Problem 5.29, where we do not assume it is zero and solve for the constant C. Both problems start from rest.

    Also, why we do we assume the constant is equal to (theta_o-dot^2)/2 in this problem, and assume it is just C in Problem 5.29?

    1. This is a very good question! I'm glad you asked it.

      For starters I was wrong to assume the constant of integration is zero (nor should I have called it \frac{\dot \theta_0 ^2}{2}). Instead I should have used C and solved in the same way as problem 5.29, or perhaps I could have used a definite integral like so:

      \frac{\dot \theta^2}{2} = \int_0^\theta -\frac{3g}{2l}\sin(\vartheta) + \frac{3a}{2l}\cos(\beta+\vartheta) \, d\vartheta


      = \left.\frac{3g}{2l}\cos(\vartheta) + \frac{3a}{2l}\sin(\beta + \vartheta) \right|_0^\theta


      = \frac{3}{2l}\left(g(\cos(\theta)-\cos(0)) + a (\sin(\beta + \theta) - \sin(\beta+0)) \right)


      = \frac{3}{2l}\left(g\cos(\theta) + a \sin(\beta + \theta) \right) - \frac{3}{2l}\left(g + a\sin(\beta)\right)

      Using the definite integral gives us the same thing we would find by plugging the initial conditions into the indefinite integral (with constant C). Now we can think about the maximum angle for \theta. Noting that \theta_m will occur when \dot \theta = 0, so:

      0 = \frac{3}{2l}\left(g\cos(\theta_m) + a \sin(\beta + \theta_m) \right) - \frac{3}{2l}\left(g + a\sin(\beta)\right)


      0 = g\cos(\theta_m) + a \sin(\beta)\cos(\theta_m) + a\cos(\beta)\sin(\theta_m) - \left(g + a\sin(\beta)\right)


      0 = (g + a \sin(\beta))\cos(\theta_m) + a\cos(\beta)\sin(\theta_m) - \left(g + a\sin(\beta)\right)


      0 = (g + a \sin(\beta))(\cos(\theta_m)-1) + a\cos(\beta)\sin(\theta_m)


      0 = -(g + a \sin(\beta))(1-\cos(\theta_m)) + a\cos(\beta)\sin(\theta_m)


      -a\cos(\beta)\sin(\theta_m) = -(g + a \sin(\beta))(1-\cos(\theta_m))

      Now we proceed into the murky depths of trigonometric properties. If we suppose \theta_m = 2 \alpha then we can use double-angle and half-angle properties on the left and right respectively:

      -a\cos(\beta)\sin(2\alpha) = -(g + a \sin(\beta))\frac{2}{2}(1-\cos(2\alpha))


      a\cos(\beta)(2\sin(\alpha)\cos(\alpha)) = 2(g + a \sin(\beta))\sin^2(\alpha)


      \frac{2a\cos(\beta)}{2(g + a \sin(\beta))}  = \frac{\sin^2(\alpha)}{\sin(\alpha)\cos(\alpha)}


      \frac{a\cos(\beta)}{g + a \sin(\beta)}  = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)


      \frac{a\cos(\beta)}{g + a \sin(\beta)}  = \tan\left(\frac{\theta_m}{2}\right)

      Thus:

       \theta_m = 2\arctan\left(\frac{a\cos(\beta)}{g + a \sin(\beta)}\right)

      That was a big error!

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