Question 17.SP.2: Gear A has a mass of 10 kg and a radius of gyration of 200 m...

ANALYSIS:
Kinematics. The velocity of the point of contact, P, is the same for both gears (Fig. 1), so you have

$v_P=r_A \omega_A=r_B \omega_B        \quad \omega_A=\omega_B \frac{r_B}{r_A}=\omega_B \frac{100  \mathrm{~mm}}{250  \mathrm{~mm}}=0.40  \omega_B$

Calculations. For   $\omega_B=600  \mathrm{rpm}$,  you have

$\begin{aligned}\omega_B &=62.8  \mathrm{rad} / \mathrm{s}      \quad \omega_A=0.40  \omega_B=25.1  \mathrm{rad} / \mathrm{s} \\\bar{I}_A &=m_A \bar{k}_A^2=(10  \mathrm{~kg})(0.200  \mathrm{~m})^2=0.400  \mathrm{~kg} \cdot \mathrm{m}^2 \\\bar{I}_B &=m_B \bar{k}_B^2=(3  \mathrm{~kg})(0.080  \mathrm{~m})^2=0.0192  \mathrm{~kg} \cdot \mathrm{m}^2\end{aligned}$

Principle of Work and Energy: Apply the principle of work and energy

$T_1  +  U_{1 \rightarrow 2}=T_2$            (1)

You need to calculate the initial and final kinetic energy and the work.

Kinetic Energy. The system is initially at rest, so  $T_1$ = 0. Adding the kinetic energies of the two gears when  $v_B$ = 600 rpm gives

$\begin{aligned}T_2 &=\frac{1}{2} \bar{I}_A \omega_A^2  +  \frac{1}{2} \bar{I}_B \omega_B^2 \\&=\frac{1}{2}\left(0.400  \mathrm{~kg} \cdot \mathrm{m}^2\right)(25.1  \mathrm{rad} / \mathrm{s})^2  +  \frac{1}{2}\left(0.0192  \mathrm{~kg} \cdot \mathrm{m}^2\right)(62.8  \mathrm{rad} / \mathrm{s})^2 \\&=163.9  \mathrm{~J}\end{aligned}$

Work. Denote the angular displacement of gear B by  $\theta_B$.  Then

$U_{1 \rightarrow 2}=M \theta_B=(6  \mathrm{~N} \cdot \mathrm{m})\left(\theta_B \mathrm{rad}\right)=\left(6 \theta_B\right) \mathrm{J}$

Substituting these terms into Eq. (1) gives you

$\begin{aligned}0  +  \left(6 \theta_B\right) \mathrm{J} &=163.9  \mathrm{~J} \\ \theta_B &=27.32  \mathrm{rad}           \quad \theta_B=4.35  \mathrm{rev}\end{aligned}$

Motion of Gear A.

Kinetic Energy. Initially, gear A is at rest, so  $T_1=0$.  When  $\omega_B=600  \mathrm{rpm}$, the kinetic energy of gear A is

$T_2=\frac{1}{2} \bar{I}_A \omega_A^2=\frac{1}{2}\left(0.400  \mathrm{~kg} \cdot \mathrm{m}^2\right)(25.1  \mathrm{rad} / \mathrm{s})^2=126.0  \mathrm{~J}$

Work. The forces acting on gear A are shown in Fig. 2. The tangential force F does work equal to the product of its magnitude and of the length  $\theta_A r_A$  of the arc described by the point of contact. Since  $\theta_A r_A=\theta_B r_B$,  you have

$U_{1 \rightarrow 2}=F\left(\theta_B r_B\right)=F(27.3  \mathrm{rad})(0.100  \mathrm{~m})=F(2.73  \mathrm{~m})$

Substituting these values into work and energy gives

$\begin{aligned}T_1  +  U_{1 \rightarrow 2} &=T_2 \\0  +  F(2.73  \mathrm{~m}) &=126.0  \mathrm{~J} \\F &=+46.2  \mathrm{~N}        \quad           \quad \mathrm{~F}=46.2 \mathrm{~N} \swarrow\end{aligned}$

REFLECT and THINK: When the system was both gears, the tangential force between the gears did not appear in the work and energy equation, since it was internal to the system and therefore did no work. If you want to determine an internal force, you need to define a system where the force of interest is an external force. This problem, like most problems, also could have been solved using Newton’s second law and kinematic relationships.