# Torque required to turn a drum/barrel

I need to spec a motor to turn a mixing barrel. The barrel contains loose earth and can be filled to a maximum of 50% of its interior volume. It is a horizontal cylinder, and will rotate through its central axis. What effort/torque is required to start the barrel turning?

In my case, the mass of the barrel when filled to its maximum is 50 kg, its length is 1.2m, diameter is 700mm. Friction in the bearings and the drive is negligible.

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it actually makes a more useful question if you don't include the specific numbers, or at least leave them out of the main part of the question. Rather than closing the question I edited it for you. – David Z May 25 '11 at 1:17
What kind of motor do You look for? Electric? Combustion? – Georg May 25 '11 at 8:34
@David Zaslavsky Ok but I'll add back in the part about keeping the barrel turning – Craig May 25 '11 at 12:44
@Georg An electric DC motor. I have my eye one one with a 516:1 gearbox giving a torque of 12,000 g.cm – Craig May 25 '11 at 12:45
sure, that's fine. – David Z May 25 '11 at 18:14

## BoTE

Assume the pile will, on average have a slope of 45 degrees. Take $\theta = 0$ to be the horizontal on the side where the pile accumulates.

Then integrate:

$$\bar\tau = \int_0^R dr \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} d\theta \rho l (-g) r^2 \cos(\theta)$$

$$\bar\tau = g \rho l \frac{R^3}{3} \left[ \sin(\theta) \right]_{\theta=-\frac{3\pi}{4}}^{\frac{\pi}{4}}$$

$$\bar\tau = g \rho l \frac{R^3}{3} [\frac{\sqrt{2}}{2} - -\frac{\sqrt{2}}{2} ]$$

$$\bar\tau = g \rho l \frac{\sqrt{2}}{3} R^3$$

Where $\rho$ represents the mass density of the mixture, $l$ the depth of the drum, and $R$ the radius of the drum.

Now $\rho l \frac{\pi R^2}{2} = 50\text{ kg}$ so

$$\bar\tau = g \frac{2\sqrt{2}}{3 \pi} (50\text{ kg}) R$$

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that torque has units of force, i think – Mark Eichenlaub May 25 '11 at 1:06
@Mark. I think you are right. That's what I get for composing latex and integrating at the same time. The top line has the right units...no it didn't, but now it does, and so does the result. – dmckee May 25 '11 at 1:11
@dmckee Ah, good. The integration looks right, but the substitution doesn't seem to follow. $2\sqrt{2}/3$ is $\approx .95$, which must be too high. I think your result ought to be $\sqrt{2}/2$ times my result. – Mark Eichenlaub May 25 '11 at 1:42
@Mark: Yep. That looks right, and now I've found all the loose factors. – dmckee May 25 '11 at 1:52
@dmckee Looks right - I just cheated and looked the integral up! – Mark Eichenlaub May 25 '11 at 1:55

Theoretically, zero torque is required to start the barrel turning as long as it starts out in equilibrium. To keep it turning at a constant speed, the required torque varies periodically with a time-average of zero. This assumes that there is no friction in the barrel and that the dirt doesn't roll around inside.

Realistically, it depends on a number of factors, including the friction in the bearings and other sources of friction, the size of the barrel, the speed at which it rotates, and the extent to which the dirt tumbles down on itself inside the barrel. If the barrel has a large radius and small thickness, more torque is required. If the barrel has low friction, less torque is required. If the dirt all tumbles inside, the effect is complicated, probably increasing the average power input required but decreasing the peak torque. There is simply a lot of information missing, without which the torque can't be calculated.

One possible estimate is to find the maximum torque required to counter gravity.

The mass of the dirt is 50 kg, so its weight is 500 N. If the 50%-full cylinder is split vertically, with the left half full and the right half empty, we would maximize the torque from gravity, which is 500 N multiplied by the distance of the center of mass of the barrel from the central axis.

The center of mass of the cylinder is $4/3\pi$ the way to the edge of the barrel, so multiplying this length by the 500 N gives the maximum torque that can be exerted by gravity.

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The barrel length is 1.2m, diameter is 700mm. So 500*3/4*pi*0.7=824Nm? – Craig May 25 '11 at 0:22
@Craig Yes, that's it. – Mark Eichenlaub May 25 '11 at 1:43

The answer is: Torque = Force x Distance = (Mg) x ((4Pi/3)R sin(Theta)).

M = Mass. For simplicity, assume the barrel weight is 0 and the CONTENTS are 50kgs.

g = acceleration due to gravity = 9.8 m/(s^2)

Force = Mg = 490N.

R = max radius of the INSIDE of the barrel, which I assume to be 350mm = 0.35m.

'Theta' is angle of rotation. Theta = Angular displacement from load level in the barrel. For the puposes of this problem, we constrain Theta to the interval 0 to 90 degrees.

Distance = horizontal displacement of the Center of Gravity (CG) of the load from the center of the barrel.

The TOTAL distance of the CG of the load from the axis of barrel rotation is (4/3Pi)(R).

The HORIZONTAL distance of the CG of the load from the axis of barrel rotation is the TOTAL distance (which is (4/3Pi)(R)) times COS(90-Theta). On our 0 to 90 degree interval, COS(90-Theta) is the same as SIN(Theta). So the horizontal torque arm length is ((4/3Pi)(R))(SIN(Theta)).

Mark is correct: the theoretical torque to start the barrel moving (disregarding friction) is zero. The max torque to keep it moving is given by using the formula above (first line of this post) and choosing the appropriate Theta. I would pick Theta = 90 degrees, which makes SIN(Theta) = 1, and collapses the formula to:

Max torque = (Mg)(4/3Pi)R = 490N x (4/3Pi)(0.35m) = 72.8 (Nm) of torque. This is roughly 53.6 foot-lbs.

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## protected by Qmechanic♦Apr 28 '13 at 21:22

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