In Fig. Q9.4, how are the radial accelerations of points at the teeth of the two sprockets related? Explain.

The expression for the radial acceleration is as follows:

${\mathit{a}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here, v is the linear velocity, and is the radius.

Consider the following figure.

The linear velocity(v) is same.

The radial accelerations in the case of rear and front sprocket will be given by the following equations.

$a_{rad,rear}=\frac{v^2}{r_{rear}}$

And,

$a_{rad,front}=\frac{v^2}{r_{front}}$

Take the ratio of both accelerations as follows.

$\frac{a_{rad,front}}{a_{rad,rear}}=\frac{r_{rear}}{r_{front}}.........\left(1\right)$

From the given figure, it can be observed that $r_{front}>r_{rear}$. So, it can be concluded that localid="1667978783875" $a_{rad,front}<a_{rad,rear}$.

The radial speed and the radius are related as follows.

$$\begin{gathered} r_{rear}{\omega }_{rear}=r_{front}{\omega }_{front} \\ \frac{r_{rear}}{r_{front}}=\frac{{\omega }_{rear}}{{\omega }_{front}} \end{gathered}$$

From equation (1),

$\frac{{\alpha }_{rad,front}}{{\alpha }_{rad,rear}}=\frac{{\omega }_{front}}{{\omega }_{rear}}$

Therefore, the radial acceleration of two sprockets is related as $\frac{{\alpha }_{rad,front}}{{\alpha }_{rad,rear}}=\frac{{\omega }_{front}}{{\omega }_{rear}}$.