9.3 Dynamics of rotational motion: rotational inertia  (Page 4/8)

Section summary

• The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
• If we exert a force $F$ on a point mass $m$ that is at a distance $r$ from a pivot point and because the force is perpendicular to $r$ , an acceleration $\text{a = F/m}$ is obtained in the direction of $F$ . We can rearrange this equation such that
  $\mathrm{F\; =\; ma}\text{,}$

  and then look for ways to relate this expression to expressions for rotational quantities. We note that $\mathrm{a\; =\; r\alpha }$ , and we substitute this expression into $\mathrm{F=ma}$ , yielding

  $\mathrm{F=mr\alpha }$
• Torque is the turning effectiveness of a force. In this case, because $F$ is perpendicular to $r$ , torque is simply $\tau =\mathit{rF}$ . If we multiply both sides of the equation above by $r$ , we get torque on the left-hand side. That is,
  $\text{rF}={\text{mr}}^2\alpha$

  or

  $\tau ={\text{mr}}^2\alpha \text{.}$
• The moment of inertia $I$ of an object is the sum of ${\text{MR}}^2$ for all the point masses of which it is composed. That is,
  $I=\sum {\text{mr}}^2\text{.}$
• The general relationship among torque, moment of inertia, and angular acceleration is
  $\tau =\mathrm{I\alpha }$

  or

  $\begin{array}{}\alpha =\frac{\text{net τ}}{I}\cdot \\ \end{array}$

Conceptual questions

While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?

Problems&Exercises