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.

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

vP=rAωA=rBωB    ωA=ωBrBrA=ωB100  mm250  mm=0.40 ωBv_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   ωB=600 rpm\omega_B=600  \mathrm{rpm},  you have

ωB=62.8 rad/s    ωA=0.40 ωB=25.1 rad/sIˉA=mAkˉA2=(10  kg)(0.200  m)2=0.400  kgm2IˉB=mBkˉB2=(3  kg)(0.080  m)2=0.0192  kgm2\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

T1 + U12=T2T_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  T1T_1 = 0. Adding the kinetic energies of the two gears when  vBv_B = 600 rpm gives

T2=12IˉAωA2 + 12IˉBωB2=12(0.400  kgm2)(25.1 rad/s)2 + 12(0.0192  kgm2)(62.8 rad/s)2=163.9  J\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  θB\theta_B.  Then

U12=MθB=(6  Nm)(θBrad)=(6θB)JU_{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

0 + (6θB)J=163.9  JθB=27.32 rad     θB=4.35 rev\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  T1=0T_1=0.  When  ωB=600 rpm\omega_B=600  \mathrm{rpm}, the kinetic energy of gear A is

T2=12IˉAωA2=12(0.400  kgm2)(25.1 rad/s)2=126.0  JT_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  θArA\theta_A r_A  of the arc described by the point of contact. Since  θArA=θBrB\theta_A r_A=\theta_B r_B,  you have

U12=F(θBrB)=F(27.3 rad)(0.100  m)=F(2.73  m)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

T1 + U12=T20 + F(2.73  m)=126.0  JF=+46.2  N                  F=46.2 N\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.

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