Question 17.SP.2: 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; 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. Neglecting friction, determine (a) the number of revolutions executed by gear B before its angular velocity reaches 600 rpm, (b) the tangential force that gear B exerts on gear A.
STRATEGY: You are given a couple and are asked to determine the position at a given angular velocity, so use the principle of work and energy.
MODELING: For part (a), choose the system to be both gears and model each as a rigid body. In part (b), you are asked to determine an internal force, so you need to choose gear A as your system.
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.
