Mechanical Advantage

Power proportional to the magnitude of the angular velocity

Moment of Inertia

• $I=MR^{2}$: Point mass $M$. Radius of path curvature $R$.

Integrate Moment of Inertia with respect to Angular Velocity ($\omega$): Angular Momentum

• $L=I\omega +c_{1}$: Magnitude of angular velocity $\omega$.
• $L=I*(V/R)+c_{1}$: Velocity $V$.

Derivative of Angular Momentum with respect to time (t): Torque

• $\tau ={\frac {dL}{dt}}$

Integrate Torque with respect to angular velocity: Power proportional to the angular velocity

• $P=\tau \omega +c_{2}$: Magnitude of angular velocity $\omega$.

Therefore, for power to increase directly rpm, the torque curve must be flat.

In electric vehicles

In electric vehicles, electric motors with flat torque curves are often used. The electric motors provide greater torque at starting speeds when compared to internal combustion engines.

Power proportional to the magnitude of the angular velocity cubed

Start with "Power proportional to the magnitude of the angular velocity"

• $P=\tau \omega$: Magnitude of angular velocity $\omega$.

Now assume that torque is no longer constant.

• $P={\frac {\tau }{\omega ^{2}}}\omega ^{3}$

Determine the constant ${\frac {\tau }{\omega ^{2}}}$

• $P={\frac {dL/dt}{\left(d\theta /dt\right)^{2}}}\omega ^{3}$
• $P={\frac {\frac {dL}{d\theta }}{\omega }}\omega ^{3}$
• $d\theta =d/r$ Distance at radius r $d$.
• $P={\frac {\frac {mvr}{d/r}}{\omega }}\omega ^{3}$
• $P={\frac {mr^{2}/dt}{\omega }}\omega ^{3}$
• $P={\frac {mr^{2}}{d\theta }}\omega ^{3}$

Apparently, if $d\theta$ increases with time, then surely the power being used to drive mass $m$ would not necessarily decrease. If $r$ is constant, then ${\frac {m}{d\theta }}$ is constant. This implies that change in angle would increase the $m$ involved. Therefore $m$ should be replaced by the derivative of m $dm$.

• $P={\frac {dm}{d\theta }}r^{2}\omega ^{3}$

Thus, for power to increase with the cube of the rpm, a spinning mass must cause a flow of matter.

In fans

For a fan, power output is proportional to mass flow rate times static pressure. We can separate the mass flow rate in the above equation as such:

• $P={\frac {dm}{dt}}{\frac {r^{2}}{d\theta }}\omega ^{3}dt$
• $P={\frac {dm}{dt}}r^{2}\omega ^{2}$

Where $\frac {dm} {dt}$ is the mass flow rate and $r^{2}\omega ^{2}$ is proportional to static pressure, according to fan laws. The output torque of a fan is:

• $\tau ={\frac {dm}{dt}}r^{2}\omega$

Since $v = r\omega$:

• $\tau ={\frac {vr}{dt}}dm$
• $\tau ={\frac {dL}{dt}}$

This result vindicates the calculations above.

Power proportional to the magnitude of the angular velocity squared

From the section above:

• $P={\frac {dm}{dt}}r^{2}\omega ^{2}$

If the power in question is proportional to the square of the rpm, then:

• ${\frac {dm}{dt}}r^{2}$ must be a constant.

Since $r={\frac {v}{\omega }}$:

• ${\frac {dm}{dt}}{\frac {vr}{\omega }}$ must be a constant.
• ${\frac {dL}{dt}}{\frac {1}{\omega }}$

This would mean that the torque would have to be directly proportional to the rpm.

From the first section:

• $P=\tau \omega$

Thus:

• $P={\frac {\tau }{\omega }}\omega ^{2}$

If ${\frac {\tau }{\omega }}$ were constant such that power increases $\omega^2$, we realize that the torque would have to increase in proportion to the rpm as well, and that is how power proportional to the rpm squared is attained.

In electrical circuits

In an electrical circuit, power increases with the square of the voltage, which in turn, increases with the square of the rate of rotation. To visualize this, consider the power induced into a coil as it rotates inside a magnetic field, or vise versa. The voltage induced in the coil increases with the strength of the magnetic field, the rate of rotation of the coil, and the number of turns in the coil. The amperage induced in the coil increases with the induced voltage and decreases with the electrical impedance of the coil. As the rate of rotation is increased, the induced voltages and currents increase in proportion, such that power is also proportional to the square of the current.

Conservation of Energy and Angular Momentum on a lever

Length of Input Arm 1 $L_1$

Length of Output Arm 2 $L_2$

$L_{1}||L_{2}$

Force on Arm 1 $|F_{1}|=-|F_{2}|$

Force on Arm 2 $|F_{2}|=-|F_{1}|$

$|L_{1}XF_{1}|=|L_{2}XF_{2}|$

Velocity of Arm 1 $V_1$

Velocity of Arm 2 $V_2$

Power in=Power out $|F_{1}|V_{1}=|F_{1}|V_{2}$

Conservation of momentum and implications on a lever

Force = change in momentum / change in time

Impulse = force * change in time

Per section above:

Length of Input Arm 1 $L_1$

Length of Output Arm 2 $L_2$

$L_{1}||L_{2}$

Force on Arm 1 $|F_{1}|=-|F_{2}|$

Force on Arm 2 $|F_{2}|=-|F_{1}|$

$|L_{1}XF_{1}|=-|L_{2}XF_{2}|$

Velocity of Arm 1 $V_1$

Velocity of Arm 2 $V_2$

Power in=Power out $|F_{1}|V_{1}=|F_{1}|V_{2}$

However: $|F_{1}|$ may be unequal to $|F_{2}|$ if $L_1$ is unequal to $L_2$. If $|F_{1}|<|F_{2}|$, then $v_{1}>v_{2}$.

Phenomenon of underutilized torque

Say one has two gears connected to each other. The larger diameter gear weighs less and the smaller diameter gear weighs more. A motor is attached to the smaller diameter gear and the teeth of the smaller diameter gear apply a force at right angles to the axis of the large diameter gear. Because of the larger diameter of the lighter gear, it recieves more torque than the heavier, smaller gear. Let's say that the moment of inertia of the larger diameter gear is "x" times as much as of that of the smaller diameter gear. The mass of the larger diameter gear is "x" times less and its radius is "x" times as much. Every "x" turns of the smaller gear will cause the larger gear to turn one time. The torque on the larger gear is "x" times greater than the torque on the smaller gear. The velocity of the outer perimeter of each gear is the same. Therefore, the power output of the small diameter gear is matched by the power input of the big diameter gear. However, the larger diameter gear receives a greater torque than torque applied by the smaller gear, and should gain angular momentum at a faster rate, inspite of its mass being "x" times less, its radius being "x" times greater, and its outer velocity being the same. This setup suggests that not all the torque of the motor is being used to turn that larger gear meaning that there is a significant loss of utility if nothing else is connected. However, if the teeth of the smaller gear were connected to the teeth a massless gear that is sqrt(x) times wider connected 1:1 with the larger gear, all of the torque would be used in order to spin that larger wheel. The angular velocity of the larger gear would be sqrt(x) times greater in the second case though its torque would be sqrt(x) times less. This would make up for the lost potential to do work that we found earlier.