# Maximum angular velocity to stop in one rotation with a known torque

I have an object I can rotate with a given torque. I would like to stop applying torque once I've reached a defined maximum rotational speed. The maximum rotational speed should be defined so that applying maximum torque will stop the rotation of the object within one rotation. If I know my torque and moment of inertia, how can I find the maximum rotational velocity to allow me to stop the object in one rotation?

Time is whatever is needed.

I've tried finding the angular acceleration required to stop the object, but that leaves me with the time variable. Of the equations I've tried, I'm left with a time variable as well as the maximum angular velocity.

• What approaches have you tried? Where are you hung up with this question? – Jerry Schirmer Jun 2 '13 at 22:04
• I edited the question to give a little details about what I've tried. And it's not homework. More like hobby work. – MichaelHouse Jun 2 '13 at 22:09
• For people with similar homework questions: try the equivalent exercise in linear motion. E.g. the direct equivalent here is, given a stopping length L and a force F, what is the maximum velocity v ? You'll quickly spot that you need the second derivative a which you get from F=m*a. – MSalters Jun 3 '13 at 0:00

To stop the object you must do work. For a constant torque perpendicular to the moment arm, the work it does is equal to $\tau\cdot\Delta\theta$, and you want $\Delta\theta\leq2\pi$.

It should be obvious that the greatest angular velocity that a torque $\tau$ can stop will take it the full $2\pi$ radians to stop. In a rotating system, the rotational kinetic energy is given by $E_r=\frac12I\omega^2$ (a direct analogue of $E_K=\frac12mv^2$ ). Now consider work-energy equivalence.

• Great, I should have taken the energy approach. Been too long since physics. Thanks. – MichaelHouse Jun 2 '13 at 22:29
• No problem. J. Lowney has a full-detail solution above - you should check your algebra. A word of advice: on a problem like this, omit your direction signs and only consider magnitudes. Signs will only confuse you and freak you out when you end up with an imaginary angular velocity. – Zen Jun 2 '13 at 22:31

Building off of Zen's response, the energy will be $E_r = \frac{1}{2}I\omega^2$. The work done in one rotation is $\tau\Delta\theta$. These two terms are equivalent in your case. I.e. you will have the following expression

$$E_r = \frac{1}{2}I\omega^2 = \tau_\text{max} \Delta\theta$$ $$\omega_\text{max} = \sqrt{\frac{2\tau_\text{max}\Delta\theta}{I}}$$

You're treating your $\Delta\theta$ as $2\pi$, for one full rotation, hence:

$$\omega_\text{max} = \sqrt{\frac{4\pi\tau_\text{max}}{I}}$$

Where $I$ is the moment of inertia of your object. $\omega$ is the angular velocity. $\tau$ is your torque.

• How would I go about expanding this to use multiple torque values at different positions around the object? – MichaelHouse Jun 2 '13 at 22:43
• If you have a function $\tau(\theta)$ that gives you torque at every position around the object then simply replace the naive expression for work $W=\tau\Delta\theta$ by $$W=\int_0^{2\pi}\tau(\theta)d\theta$$ – Zen Jun 2 '13 at 22:56

Alternative solution:

I see you tried to do it by first finding angular acceleration $\alpha=\frac\tau I$ (where tau is your applied torque). This also works! However I suspect you got stuck at $$\alpha t=-\omega_{max}$$ This is perfectly understandable because t is somewhere between 0 and a full period (under the original angular velocity $\omega$ that is), but you don't know what t is.

You could try to set up some equations to solve simultaneously for $t$, but that's not necessary because you don't care about $t$, you only care about $\omega_{max}$ There is another equation you can use, the so-called "timeless equation": $$\omega_f^2=\omega_o^2+2\alpha\Delta\theta$$ where $\omega_f$ is final angular velocity, $\omega_o$ is initial angular velocity, $\alpha$ is angular acceleration, and $\Delta\theta$ is angular displacement (this again is a direct analogue of the linear kinematic equation $v^2=v_0^2+2a\Delta x$).

Let $\omega_f=0$ and $\omega_0=\omega_{max}$ (i.e., you start out at maximum velocity). By the same reasoning as above $\Delta\theta=2\pi$ and if you know $\tau$ and $I$ you can find $\alpha$. Then you have: $$0=\omega_{max}^2+2\alpha\cdot2\pi$$ $$\omega_{max}^2=2\alpha\cdot2\pi$$ (again, ignore signs) $$\omega_{max}=\sqrt{4\alpha\pi}$$ which is the same as above, since $\alpha=\frac \tau I$