Question 17.SP.7: 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 with a magnitude of 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 that gear B exerts on gear A.
STRATEGY: Since you are given an angular velocity and are asked for time, use the principle of impulse and momentum.
MODELING: You are asked to find the internal tangential force, so you need two systems for this problem; that is, gear A and gear B. Model the gears as rigid bodies. Since all forces and couples are constant, you can obtain the impulses by multiplying the forces and moments by the unknown time t.
ANALYSIS: 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 \mathrm{~kg} \cdot \mathrm{m}^2 & \bar{I}_B &=0.0192 \mathrm{~kg} \cdot \mathrm{m}^2 \\\left(\omega_A\right)_2 &=25.1 \mathrm{rad} / \mathrm{s} &\left(\omega_B\right)_2 &=62.8 \mathrm{rad} / \mathrm{s}\end{aligned}
Principle of Impulse and Momentum for Gear A. The impulse-momentum diagram (Fig. 1) for gear A shows the initial momenta, impulses, and final momenta.
\text { Syst Momenta }{ }_1 + \text { Syst Ext } \operatorname{Imp}_{1 \rightarrow 2}=\text { Syst Momenta }{ }_2
\begin{array}{rlr}+\circlearrowleft \text {moments about } A: & 0 – F t r_A=-\bar{I}_A\left(\omega_A\right)_2 \\F t(0.250 \mathrm{~m}) & =\left(0.400 \mathrm{~kg} \cdot \mathrm{m}^2\right)(25.1 \mathrm{rad} / \mathrm{s}) \\F t & =40.2 \mathrm{~N} \cdot \mathrm{s}\end{array}
Principle of Impulse and Momentum for Gear B. Draw a separate impulse–momentum diagram for gear B (Fig. 2).
\text { Syst Momenta }{ }_1 + \text { Syst Ext Imp } \operatorname{Im}_{1 \rightarrow 2}=\text { Syst Momenta }{ }_2
\begin{aligned}&+\circlearrowleft \text { moments about } B: \quad 0 + M t – Ft r_{B}=\bar{I}_B\left(\omega_B\right)_2 \\&+(6 \mathrm{~N} \cdot \mathrm{m}) t – (40.2 \mathrm{~N} \cdot \mathrm{s})(0.100 \mathrm{~m})=\left(0.0192 \mathrm{~kg} \cdot \mathrm{m}^2\right)(62.8 \mathrm{rad} / \mathrm{s})\end{aligned}
t=0.871 \mathrm{~s}
Recall that Ft = 40.2 N·m, so you have
F(0.871 \mathrm{~s})=40.2 \mathrm{~N} \cdot \mathrm{s} \quad F=+46.2 \mathrm{~N}
Thus, the force exerted by gear B on gear A is
\mathbf{F}=46.2 \mathrm{~N} \swarrow
REFLECT and THINK: This is the same answer obtained in Sample Prob. 17.2 by the method of work and energy, as you would expect. The difference is that in Sample Prob. 17.2, you were asked to find the number of revolutions, and in this problem, you were asked to find the time. What you are asked to find will often determine the best approach to use when solving a problem.
