Subscripts
Unknown macro: {latex}
\large L, R, M
correspond the left half-shaft, right half-shaft, and middle gear, respectively.Three observations:
- When we hold the right axle fixed and turn the left axle one revolution, the middle sprocket makes half a revolution in the same angular direction.
- Holding the middle sprocket in place and turning the left axle one revolution causes the right axle to make one revolution in the opposite direction.
- The above observations are the same if the left and right axle are switched.
The results can be summarized as such:
Unknown macro: {latex}
\large\begin
Unknown macro: {equation*}
\theta_{\text{M}} = \frac
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\theta_{\text{L}} + \frac
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\theta_{\text{R}}
\end
where
Unknown macro: {latex}
\large $\theta$
represents angular position. In other words, the position of the middle gear is half the sum of the left and right gears. Say that
Unknown macro: {latex}
\large $\omega$
represents angular velocity,
Unknown macro: {latex}
\large $\alpha$
represents angular acceleration, and
Unknown macro: {latex}
\large $\tau$
represents torque.
Unknown macro: {latex}
\large $r$
is the radius of the gear, and
Unknown macro: {latex}
\large $v$
is the linear velocity at the edge of a gear. From the first equation, we can find the relationships between the angular and linear velocities of the gears.
Unknown macro: {latex}
\large
\begin
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\omega_{\text{M}} =& \frac
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\left( \omega_{\text{L}} + \omega_{\text{R}} \right)
\omega_{\text{M}} r_{\text{M}} =& \frac
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( \omega_{\text{L}} r_{\text{M}} +
\omega_{\text{R}} r_{\text{M}})
v_{\text{M}} =& \frac
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\left( \omega_{\text{L}} r_{\text{M}} \frac{r_{\text
Unknown macro: {L}
}}{r_{\text
}} +
\omega_{\text{R}} r_{\text{M}} \frac{r_{\text
Unknown macro: {R}
}}{r_{\text
}} \right)
v_{\text{M}} =& \frac
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\left( v_{\text{L}} \frac{r_{\text
Unknown macro: {M}
}}{r_{\text
Unknown macro: {L}
}} +
v_{\text{R}} \frac{r_{\text
}}{r_{\text
Unknown macro: {R}
}} \right)
\end
We assume that these ideal gears are frictionless
and massless. Therefore, we can use conservation of energy to say that input power equals output power. Say
Unknown macro: {latex}
\large $P$
represents power as a function of time.
Unknown macro: {latex}
\large
\begin
Unknown macro: {align*}
&P_{\text{M}} = P_{\text{L}} + P_{\text{R}}
&\tau_{\text{M}} \cdot \omega_{\text{M}} = \tau_{\text{L}} \cdot \omega_{\text{L}} +
\tau_{\text{R}} \cdot \omega_{\text{R}}
&\tau_{\text{M}} \cdot \left[ \frac
Unknown macro: {1}
Unknown macro: {2}
\left( \omega_{\text{L}} + \omega_{\text{R}} \right) \right]
= \tau_{\text{L}} \cdot \omega_{\text{L}} +
\tau_{\text{R}} \cdot \omega_{\text{R}}
\end
If we consider
Unknown macro: {latex}
\large $\omega_{\text{L}}$
and
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$\omega_{\text{R}}$
separately, we find that
Unknown macro: {latex}
\large
\begin
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\frac
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\tau_{\text{M}} \omega_{\text{L}} &= \tau_{\text{L}} \omega_{\text{L}}, \;\;\;
\frac
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\tau_{\text{M}} \omega_{\text{R}} = \tau_{\text{R}} \omega_{\text{R}}
\frac
Unknown macro: {1}
\tau_{\text{M}} &= \tau_{\text{L}}, \;\;\;\;\;\;\; \;\; \; \;\; \frac
Unknown macro: {1}
Unknown macro: {2}
\tau_{\text{M}} = \tau_{\text{R}}
\end
Therefore, we can conclude that the relationships between torques is the same as the relationships between angular position; a torque on the middle gear is evenly divided between the torque on the left gear and the torque on the right gear.
Unknown macro: {latex}
\large
\begin
Unknown macro: {align*}
\tau_{\text{M}} &= \tau_{\text{L}} + \tau_{\text{R}}
\tau_{\text{L}} &= \tau_{\text{R}} = \frac
Unknown macro: {1}
Unknown macro: {2}
\tau_{\text{M}}
\end