Newton's Second Law for Rotation — Question 9

Question 9

A driven gear must also accelerate the gear it meshes with

Two externally meshing gears have radii $R_1=0.10$, $R_2=0.30\,\mathrm m$ and inertias $I_1=0.020$, $I_2=0.180\,\mathrm{kg\,m^2}$. A torque $0.120\,\mathrm{N\,m}$ drives gear 1. There is no friction or slip.

1. Find both angular accelerations.

2. Find the tangential contact force and explain why using $\tau/I_1$ alone overestimates acceleration.

Question 9 - Solution

A driven gear must also accelerate the gear it meshes with

(a) Derivation and calculation

Use opposite gear rotation senses and a shared tangential acceleration: $$\alpha_2=-\frac{R_1}{R_2}\alpha_1,\quad \tau-FR_1=I_1\alpha_1,\quad FR_2=I_2|\alpha_2|.$$ Eliminating the contact force gives $$\alpha_1=\frac{\tau}{I_1+I_2(R_1/R_2)^2} =\frac{0.120}{0.020+0.180/9}=\boxed{3.0\ \mathrm{rad/s^2}}.$$ $$\boxed{\alpha_2=-\frac13(3)=-1.0\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

The contact force and its resisting moment on gear 1 are $$\boxed{F=\frac{I_2|\alpha_2|}{R_2}=\frac{0.180(1)}{0.30}=0.60\ \mathrm N},\quad FR_1=0.060\ \mathrm{N\,m}.$$ Only $0.120-0.060=0.060$ N m accelerates gear 1. Using $\tau/I_1=6$ rad/s$^2$ ignores the torque required to accelerate gear 2.

Physical check

The no-slip accelerations match in magnitude: $R_1\alpha_1=R_2|\alpha_2|=0.30$ m/s$^2$.