My question is pretty fundamental but has me stumped. Long story short I can't seem to calculate the correct required power to accelerate a mass to a set speed in a set distance. Every time I calculate my equations I end up with a power value that is double what it should be or a mismatch between the two ways that I am using to calculate it.


Picture a point mass being accelerated down a cylinder in a helical spiral pattern (think threaded hole). I am trying to calculate the necessary power it would take to accelerate this mass to a certain speed before the end of the cylinder. The cylinder is stationary and can not move.

Known Variables:

$\omega_f$ [radians] = final velocity
$\omega_0$ [radians] = starting velocity = 0
m [kg] = mass of projectile
r [meters] = radius of projectile
Q [$\frac{rev.}{m}$] = thread revolutions per meter
L [m] = length of cylinder

$\theta_f$ [rad] = final position = $2 \pi Q L$
$\theta_0$ [rad] = initial position = 0


[Eq. 1] $\omega_f^2 = \omega_0^2 + 2\alpha(\theta_f-\theta_0)$
[Eq. 2] $\omega_f = \omega_0 + \alpha t$
[Eq. 3] $I_p = m r^2$
[Eq. 4] $T = I \alpha$
[Eq. 5] $P = T \omega_f$
[Eq. 6] $E_\textrm{torque} = T \Delta\theta = T \theta_f $
[Eq. 7] $E_\textrm{power} = P t$
[Eq. 8] $E_\textrm{inertia} = \frac12 I \omega_f^2$

Attempt and Problem:

Given that I know $\omega_f$ and $\theta_f$, and initial values are all zero, I can rearrange Eq. 1 and calculate $\alpha$: $$\alpha=\frac{\omega_f^2}{2 \theta_f}$$ Now that I know $\alpha$ and I already knew m and r I can calculate the $T$: $$T = I \alpha = \left(mr^2\right)\left(\frac{\omega_f^2}{2 \theta_f}\right) = \frac{m r^2 \omega_f^2}{2 \theta_f}$$ Now I have torque. This is where things get confusing for me. If I calculate energy directly using Eq. 6 and Eq. 8 I get the same answer, but if I calculate Power directly using Eq. 5 and then energy using Eq. 7 I get a different answer from Eq. 6 and Eq. 8.
Method Using Eq. 6: $$E_\textrm{torque} = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)\left(\theta_f\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 8: $$E_\textrm{inertia} = \frac12\left(mr^2\right)\left(\omega_f^2\right) = \frac{m r^2 \omega_f^2}{2}$$ Method Using Eq. 2, 5 and 7: $$t = \frac{\omega_f}{\alpha}$$ $$P = T \omega_f = \left(\frac{m r^2 \omega_f^2}{2 \theta_f}\right)(\omega_f) = \frac{m r^2 \omega_f^3}{2 \theta_f}$$ $$E_\textrm{power} = P t = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{\omega_f}{\alpha}\right) = \left(\frac{m r^2 \omega_f^3}{2 \theta_f}\right)\left(\frac{2 \theta_f}{\omega_f}\right) = \frac{m r^2 \omega_f^2}{1}$$


Why does $E_\textrm{power}$ not equal the other two energy calculations and where did I go wrong? Ultimately I need the power, but I don't trust my power value in this calculation because it gives the wrong final energy value.
I hope everything was clear if not I will gladly attempt to explain anything further.

  • $\begingroup$ The power is not constant (assuming constant torques and increasing angular speed). Multiplying the final power by the time will not give you the total kinetic energy. You eq 7 is not right. 6 and 8 give the same result. $\endgroup$
    – nasu
    Commented Oct 26, 2016 at 20:09
  • $\begingroup$ 1. Is this mass accelerating due to gravity? 2. Is it a point mass? 3. Is $r$ the radius of mass or the radius of ths cylinder? 4. Do you need to find the average power over the whole journey? $\endgroup$
    – Farcher
    Commented Oct 26, 2016 at 20:33
  • $\begingroup$ @nasu Thanks I had forgotten that. What would be the proper way then to calculate the peak power required to accelerate the point mass to the final speed? Farcher, Assume no gravity in this scenario, I'm trying to calculate peak power required to accelerate the point mass up final speed. r in this case is the radius of the cylinder. $\endgroup$
    – Wired365
    Commented Oct 27, 2016 at 13:33
  • $\begingroup$ @nasu wait I fully understand it now. The average power is found by dividing the energy by time. The peak power is found by multiplying torque times max speed, as I did in Eq. 5. Then finally in order to get the energy to match up for all three equations I would have had to integrate the changing power over time. $\endgroup$
    – Wired365
    Commented Oct 27, 2016 at 13:50
  • $\begingroup$ @nasu could you make your comment and answer so I can give you credit for answering my question? $\endgroup$
    – Wired365
    Commented Oct 27, 2016 at 16:45

1 Answer 1


As explained by @nasu, the discrepancy between the results arises because in calculating $P=T\omega_f$ you have used final angular velocity $\omega_f$ instead of average angular velocity $\frac12 \omega_f$. This is similar to calculating distance = final velocity x time, instead of distance = average velocity x time. If you include a factor of $\frac12$ this will make $E_{power}$ the same as $E_{inertia}$.

I think you are also missing the fact that the particle has translational (axial) KE as well as rotational (circular) KE. As well as rotating around the cylinder axis it also moves along the axis.

No work is done in constraining the particle to move in a circle, so the problem is equivalent to linear motion, which is much easier to handle. This avoids the complication of splitting the motion into rotational and axial components. Assuming there is no friction, all of the energy supplied is transformed into translational kinetic energy along the helical curve.

In the equivalent linear case, the particle is accelerated from rest up to speed $v$ in a straight line over distance $s$ which is the distance around the helix.

The acceleration $a$ is given by $v^2=2as$. The force accelerating the particle is $F=ma$. The instantaneous power delivered is $P=Fv=mav=m\frac{v^3}{2s}$. As noted already, the power delivered is not constant, but increases linearly, because $a$ is constant while $v$ increases linearly. Peak power is the final power $P=mav$. Average power is $\frac12mav$.

The only problem remaining is to relate the curvi-linear variables $s,v$ (ie along the helix) to rotational variables $L, Q, \theta, \omega$. I doubt whether this is worthwhile, because it makes the formulae unnecessarily complicated. It depends what variables you can or must measure.

When the particle has made one revolution it has moved forward a distance $1/Q$ along the axis and $2\pi R$ around the circumference of a circle of radius $R$, so the pitch angle $\phi$ is given by $\tan\phi=\frac{1}{2\pi RQ}$. When the distance moved along the helix is $s$, the axial distance is $L=s\sin\phi$; the number of revolutions is $LQ$ and the distance moved around a circle is $s\cos\phi=2\pi RLQ=R\theta$ where $\theta=2\pi LQ$$ is the final angular position.

When the particle has reached speed $v=\dot s$ along the helix, the angular velocity (measured around a plane circle, perpendicular to the axis) is $\omega=\frac{v\cos\phi}{R}$.

Substitute into the eqn for power in the linear case :
$$P=m(\frac{R\omega}{\cos\phi})^3 \frac{\cos\phi}{2R\theta}=m(\frac{R}{\cos\phi})^2 \frac{\omega^3}{2\theta}$$

  • $\begingroup$ Sam, the power equation that you list at the end of your explanation and derivation. That would be the speed dependent total power when combining both the transnational and rotational energies correct? If I wanted to get peak power required to reach my desired speed I would use the final rotational velocity and if I wanted the average power then I would use $\frac{1}{2} \omega$? $\endgroup$
    – Wired365
    Commented Nov 1, 2016 at 14:20
  • $\begingroup$ Also could you clarify your equation for relating pitch angle to Q. Is it $\frac{1}{2\pi}RQ$ or something else? $\endgroup$
    – Wired365
    Commented Nov 1, 2016 at 14:27
  • $\begingroup$ @Wired365 : I have revised my answer in the light of your comments. There was an error in relating pitch angle to Q. $\endgroup$ Commented Nov 2, 2016 at 1:06

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