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I'm a 57 yr old EE, but never had any mechanical physics classes.

If we have a vertically oriented disc, affixed to an axle in bearings, 1m in diameter, with two 1kg weights attached at the outermost edges (directly opposite each other), rotating at 30 RPM, how do I calculate the torque required to turn the disc?

I know the one rising will have a higher effective weight, due to moving against gravity, and the one "falling" will have an equal weight loss... but I'm at a loss myself as to how to solve this.

I want to know how to solve it, not just the answer.

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closed as off-topic by CuriousOne, ACuriousMind, user36790, Gert, honeste_vivere May 12 '16 at 15:22

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    $\begingroup$ You specify the disc is rotating at 30 RPM. Are you asking about torque required to keep it rotating at the rate, or to speed it up, or stop it? Torque changes the rate of rotation in an analogous way that force changes velocity. $\endgroup$ – M. Enns May 11 '16 at 1:56
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    $\begingroup$ The masses are always balanced, are they not? Then you only have to overcome the moments of inertial of this rotating rigid body. $\endgroup$ – CuriousOne May 11 '16 at 2:06
  • $\begingroup$ I was asking about the torque required to maintain the 30 RPM rate. $\endgroup$ – ronbot May 11 '16 at 2:30
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    $\begingroup$ The torque required to maintain the 30 RPM rate the depends on the friction of the bearing (and whatever other resisting forces there might be) if you've got a perfect bearing then no torque is required. $\endgroup$ – M. Enns May 11 '16 at 2:38
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The gravitational force of the first mass is

$$F_1 = m_1\cdot g/ r\cdot sin(\varphi)$$

and on the second mass

$$F_2 = m_2\cdot g/ r\cdot sin(\varphi+\pi)$$

which is in total

$$F_g=F_1+F_2=m_1\cdot g/ r\cdot sin(\varphi)-m_2\cdot g/ r\cdot sin(\varphi)$$

which cancels to $0$ if $m_1=m_2$.

So if your disk is aligned vertically and the two masses are equal and aligned opposite each other you can neglect gravity.

Therefore if you want to stop the disk from rotating you need to find the rotational energy, which is in your case with ω = 30 rounds/minute = π/sec

$$E = \omega\cdot r\cdot (m_1+m_2)/2 =\pi \frac{\text{ kg m}^2}{\text{s}^2}$$

If you apply your torque over φ = 180° which is again π radians you need a torque of

$$\tau = E/\varphi = 1\frac{\text{ kg m}^2}{\text{s}^2}$$

If you want to reverse the motion of your disk you of course have to add a factor of 2 to the calculation.

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  • $\begingroup$ Thanks much... I don't know why I was in such a fog over this... why I didn't see that they exactly cancel out... duh... $\endgroup$ – ronbot May 11 '16 at 2:22

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