Question

A pinion with radius $r_1,$ and inertia $I_1$ is driving a gear with radius $r_2$ and inertia $I_2.$ Torque ${\tau }_1$ is applied on pinion. The following are free body diagrams of pinion and gear showing important forces $\left(F_{1}and{F}_2\right)$ of interaction. Which of the following relations hold true?

• A. $F_1\ne F_2;{\tau }_1=I_1\ddot{\theta };F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$
• B. $F_1=F_2;{\tau }_1=I_1[I_1+I_2{\left(\frac{r_1}{r_2}\right)}^2]\ddot{\theta };F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$
• C. $F_1=F_2;{\tau }_1=I_1{\ddot{\theta }}_1;F_2=I_2\frac{1}{r_2}{\ddot{\theta }}_2$
• D. $F_1\ne F_2;{\tau }_1=I_1[I_1+I_2{\left(\frac{r_1}{r_2}\right)}^2]{\ddot{\theta }}_1;F_2=I_2\frac{1}{r_2}{\ddot{\theta }}_2$

Answer 1

The correct option is B $F_1=F_2;{\tau }_1=I_1[I_1+I_2{\left(\frac{r_1}{r_2}\right)}^2]\ddot{\theta };F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$

${\tau }_1-F_1{r}_1=I_1{\ddot{\theta }}_1...\left(i\right)$

$F_1=F_2..\left(ii\right)$

[Action and reaction are equal by Newton's 3rd law]

$F_2{r}_2=\tau =I_2{\ddot{\theta }}_2...\left(iii\right)$

From equations (ii) and (ii)

$F_2=F_1=\frac{I_2{\ddot{\theta }}_2}{r_2}...\left(iv\right)$

From equations (i) and (iv)

${\tau }_1-\left(\frac{I_2{\ddot{\theta }}_2}{r_2}\right)r_1=I_1{\ddot{\theta }}_1...\left(v\right)$

We know, $r_1{\omega }_1=r_2{\omega }_2$

$r_1{\dot{\theta }}_1=r_2{\dot{\theta }}_2$

Differentiating with respect to time

$r_1{\ddot{\theta }}_1=r_2{\ddot{\theta }}_2$

${\ddot{\theta }}_2=\frac{r_1}{r_2}{\ddot{\theta }}_1$

${\tau }_1=I_1{\ddot{\theta }}_1+\frac{I_2}{r_2}\left[\frac{r_1}{r_2}{\ddot{\theta }}_1\right]r_1$

${\tau }_1=[I_1+I_2{\left[\frac{r_1}{r_2}\right]}^2]{\ddot{\theta }}_1$

${\tau }_2=I_2{\ddot{\theta }}_2=F_2{r}_2$

$F_2=\frac{I_2}{r_2}{\ddot{\theta }}_2=\frac{I_2}{r_2}\left[\frac{r_1}{r_2}{\ddot{\theta }}_1\right]$

$F_2=I_2\frac{r_1}{r_2^2}{\ddot{\theta }}_1$