More on Angular Kinematics

There are other ways to find angular quantities than the ones we discussed previously. Indeed, the equations I’m about to present are often much easier to work with, since it’s generally convenient to confine oneself to either linear or rotational motion, when possible.

As I mentioned, each (significant) linear quantity has a direct analogue among angular quantities: $ \boldsymbol{\theta}$ for $ \mathbf{s}$, $ \boldsymbol{\omega}$ for $ \mathbf{v}$, $ \boldsymbol{\tau}$ for $ \mathbf{F}$, $ I$ for $ m$, and so forth. Time, of course, is the same in either system. Fortunately for us, the units work out such that we can in many cases directly substitute the latter for the former in common mechanics equations. A few examples follow:

  • $ \boldsymbol{\omega}=\frac{d\boldsymbol{\theta}}{dt}$ just as $ \mathbf{v}=\frac{d\mathbf{x}}{dt}$
  • $ \boldsymbol{\alpha}=\frac{d\boldsymbol{\omega}}{dt}$ just as $ \mathbf{a}=\frac{d\mathbf{v}}{dt}$
  • $ \boldsymbol{\omega}_f = \boldsymbol{\omega}_o+\boldsymbol{\alpha}\Delta t$ just as $ \mathbf{v}_f = \mathbf{v}_o+\mathbf{a}\Delta t$
  • $ \omega_f^2 = \omega_o^2 + 2\boldsymbol{\alpha}\cdot\Delta\boldsymbol{\theta}$ just as $ v_f^2 = v_o^2 + 2\mathbf{a}\cdot\Delta\mathbf{x}$
  • $ \Delta\boldsymbol{\theta} = \frac{1}{2}\boldsymbol{\alpha}\Delta t^2 + \boldsymbol{\omega}_o\Delta t$ just as $ \Delta\mathbf{x} = \frac{1}{2}\mathbf{a}\Delta t^2 + \mathbf{v}_o\Delta t$
  • $ \mathbf{L}=I\boldsymbol{\omega}$ just as $ \mathbf{p}=m\mathbf{v}$
  • $ \boldsymbol{\tau}=I\boldsymbol{\alpha}$ just as $ \mathbf{F}=m\mathbf{a}$

Another useful analogue of a linear quantity is rotational kinetic energy, $ K_{\textrm{rot}}$, defined as $ K_{\textrm{rot}} = \frac{1}{2}I\omega^2$

This comes up canonically in the case of balls rolling. Suppose that a (solid spherical) ball with $ r=0.10\textrm{ m}$ and $ m=3.00\textrm{ kg}$ is rolling without sliding on a flat frictionless surface with $ \mathbf{v}_{\mathrm{CM}}=2.50\textrm{ m}\cdot\textrm{s}^{-1}$. It then comes to an incline. How far up the wedge will it roll?

A cursory glance at the problem might make you think about using $ K=\frac{1}{2}mv^2$ and considering that that energy must be its maximum $ U_g$, achieved at the highest point it reaches on the incline; knowing that you could easily find the height at the top. However, that approach will give you the wrong answer, since the object has kinetic energy derived not only from its linear motion ($ K=\frac{1}{2}mv^2$) but also from its rolling motion ($ K=\frac{1}{2}I\omega^2$).

First, we’ll need to find its $ I$ and its $ \omega$. $ I$ for a solid sphere is $ \frac{2}{5}mr^2$, as we showed in the previous lesson, so $ I=\frac{2}{5}(3.00\textrm{ kg})(0.10\textrm{ m})^2 = 0.012\textrm{ kg}\cdot\textrm{m}^2$ Finding the angular speed $ \omega$ is similarly easy: since $ \boldsymbol{\omega}=\mathbf{r}\times\mathbf{v}$, $ \omega = (0.10\textrm{ m})(2.50\textrm{ m}\cdot\textrm{s}^{-1}) = 0.25\textrm{ rad}\cdot\textrm{s}^{-1}$

Now we find the total $ K_i$ since $ \sum\!K = K_{\mathrm{linear}}+K_{\mathrm{rot}}$:

$ K_i$ $ =$ $ \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
  $ =$ $ \frac{1}{2}(3.00\textrm{ kg})(2.50\textrm{ m}\cdot\textrm{s}^{-1}……2}(0.012\textrm{ kg}\cdot\textrm{m}^2)(0.25\textrm{ rad}\cdot\textrm{s}^{-1})^2$
  $ =$ $ 9.38\textrm{ J}$

(Okay, it’s not that big a difference from if we’d neglected the $ K_\mathrm{rot}$, but bear with me. The concept is important.)

Now we know that $ U_{gf} = \sum\!K_i$ and that $ U_g = m\mathbf{a}_g\cdot\mathbf{y}$. So

$ 9.38\textrm{ J}$ $ =$ $ m\mathbf{a}_g\cdot\mathbf{y}$
  $ =$ $ (3.00\textrm{ kg})(9.80\textrm{ m}\cdot\textrm{s}^{-2})\cdot\mathbf{y}$

and then $ \mathbf{y} = 0.319\textrm{ m}$ which is bomb-digalicious.

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