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Thread: Interesting but difficult motion problem

  1. #1
    Oct 2010

    Angry Interesting but difficult motion problem

    Consider a wheel of radius R = 1.50m rolling with constant velocity, without slipping, v = 18.1(i) m/s (where i is the unit vector), at the top of a horizontal cliff with height h = 20.0m. Take t[0] = 0 as the time the wheel reaches the end of the cliff, i.e. the centre of the wheel has the position r(t[0]) = (h+R)j. After this, the wheel falls freely. The point that is at the top of the wheel at t[0] = 0 is denoted by P, i.e. r[p](t[0]) = (h+2R)j

    a) find the position of the centre of the wheel for t>t[0]

    b) at what time t[f] does the wheel hit the ground? Which point of the wheel hits the ground first? (I am assuming the wheel continues to rotate at constant w = v/R during the fall - w = omega)

    c)find the position and velocity of point P of the wheel for t[0]<t<t[f]. make a sketch of the position of P as a function of time

    d) find the acceleration a[p] of point P for t[0]<t<t[f]. for t[1] = 0.125s, resolve the acceleration a[p] into parallel and perpendicular components, a[p](t[1]) = a[par](t1) + a[per](t1). Make a sketch of the path of point P at time t~t[1], including v[p](t1) (the velocity of P), a[p](t1), a[par](t1) and a[per](t1)

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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Let's assume a right-ward motion off the cliff, so that the wheel is rotating in a clockwise fashion. The center of the wheel will obey the equations

    $\displaystyle \displaystyle\mathbf{r}_{\text{center}}=\left\lang le vt,h+R-\frac{gt^{2}}{2}\right\rangle.$

    This constitutes the answer to a). For the first part of b), solve the y equation for t, when y = R.

    The point P will just continue to rotate about the center with constant angular velocity $\displaystyle \omega=v/R.$ Hence, the vector describing the position of point P is given by

    $\displaystyle \displaystyle\mathbf{r}_{P}=\mathbf{r}_{\text{cent er}}+R\left\langle \sin\left(\frac{vt}{R}\right),\cos\left(\frac{vt}{ R}\right)\right\rangle=\left\langle vt+R\sin\left(\frac{vt}{R}\right),h+R-\frac{gt^{2}}{2}+R\cos\left(\frac{vt}{R}\right)\ri ght\rangle.$

    For the second part of b), plug in the time found into this equation to find the position of point P. From that you should be able to figure out, by rotations, which part of the wheel hits the ground first.

    c) and d) are plug-and-chug using the standard formulas.

    Make sense?
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