Question 17.SP.6: Gear A has a mass of 10 kg and a radius of gyration of 200 m......
Gear A has a mass of 10 kg and a radius of gyration of 200 mm, and gear B has a mass of 3 kg and a radius of gyration of 80 mm. The system is at rest when a couple M of magnitude 6 N · m is applied to gear B. (These gears were considered in Sample Prob. 17.2.) Neglecting friction, determine (a) the time required for the angular velocity of gear B to reach 600 rpm, (b) the tangential force which gear B exerts on gear A.
We apply the principle of impulse and momentum to each gear separately. Since all forces and the couple are constant, their impulses are obtained by multiplying them by the unknown time t. We recall from Sample Prob. 17.2 that the centroidal moments of inertia and the final angular velocities are
\begin{aligned}\bar{I}_A & =0.400\text{ kg}\cdot m ^2 & \bar{I}_B & =0.0192 \text{ kg} \cdot m ^2 \\\left(\omega_A\right)_2 & =25.1\text{ rad}/ s & \left(\omega_B\right)_2 & =62.8 \text{ rad} / s\end{aligned}
Principle of Impulse and Momentum for Gear A. The systems of initial momenta, impulses, and final momenta are shown in three separate sketches.
\text { Syst Momenta }{ }_1+\text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }_2
\begin{aligned}+\uparrow \text { moments about } A: & 0-F t r_A=-\bar{I}_A\left(\omega_A\right)_2 \\F t(0.250 m ) & =\left(0.400\text{ kg}\cdot m ^2\right)(25.1\text{ rad} / s ) \\F t & =40.2 N \cdot s\end{aligned}
Principle of Impulse and Momentum for Gear B.
\text { Syst Momenta }_1+\text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }_2
\begin{aligned}& +\uparrow \text { moments about } B: \quad 0+M t-F t r_B=\bar{I}_B\left(\omega_B\right)_2 \\& \quad+(6 N \cdot m ) t-(40.2 N \cdot s )(0.100 m )=\left(0.0192\text{ kg}\cdot m ^2\right)(62.8\text{ rad} / s )\end{aligned}
t = 0.871 s
Recalling that Ft = 40.2 N · s, we write
F(0.871 s) = 40.2 N · s F = 146.2 N
Thus, the force exerted by gear B on gear A is F =46.2 N \swarrow
