# How to Calculate the Torque of a Rotating Cylinder?

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1. Jun 5, 2017

### rwmiller4

Assuming we have a cylinder that is rotating on a single axis, how would I go about calculating the torque knowing the following information?

m = 2000 kg
w = 60 rpm
r = 5 m

2. Jun 5, 2017

### Staff: Mentor

Welcome to the PF.

You can calculate the energy in the rotating cylinder based on it's Moment of Inertia (which is based on it's weight distribution) and its rotational velocity.

You can then calculate the torque required to stop it, based on how long you have to stop it.

3. Jun 5, 2017

### Dr.D

It is important to specify just what axis is the axis of rotation; there are multiple possibilities. I saw nothing in the original problem statement about stopping the rotation, so I must ask, is that really the case of interest here? We need a much more complete problem statement.

4. Jun 6, 2017

### rwmiller4

Assuming the cylinder is rotating on one axis the long ways (flipping over itself). I calculated the time required to stop the rotation based on a given force value.

Rotation Like: https://pasteboard.co/fjvq76T45.png

Correct me if I am wrong:

Given Parameters:
m = 3000 kg
L = 3 m
r = 2 m
w = 6 rpm
F = 0.75 N

Inertia for Cylinder:
I = (mass * L^2) / 12 = 2250 kg m^2
w = (6 rpm)(2pi / 60) = 0.6283 rad/s
Torque = force * L = (0.75)(3) = 2.25 N*m
Angular Momentum = I*w = (2250)(0.6283) = 1413.7

time = (Angular Momentum) / Torque = (1413.7) / (2.25) = 628.32 seconds

Is this correct?

Also how would I change this up to incorporate two axis of rotation or if it were rotating on a different axis?

Edit:
- Or should I use I = (m*L^2)/4 + (m*L^2)/12 ?
- Also what if I wanted to say it was hollow? I know it would change the inertia for rotating radially, but would this change if the rotation was flipping over itself?

Last edited: Jun 6, 2017
5. Jun 6, 2017

### JBA

If the load is fixed and equally distributed in the cylinder then once it is in rotation the only resisting torque is from the shaft bearings and the air resistance from the moving barrel based upon its effective frontal area and speed of rotation.

The required torque to start the barrel rotating is based on the combination of the polar moment of inertial of the barrel with its load and the desired time to accelerate the rotation to the desired speed.

6. Jun 6, 2017

### rwmiller4

After reviewing more, torque resistance and drag are negligible.
I would want to use inertia, I = (m*r^2)/4 + (m*L^2)/12? Is that correct?

7. Jun 6, 2017

### JBA

That will only apply if the mass of the barrel and it contents are equally distributed throughout the length of the barrel (ie, a totally full barrel of a homogeneous contents); otherwise, as if a mass or weight is located at or toward the ends of the barrel then that must be treated as an I x r object relative to the axis of rotation.

8. Jun 7, 2017

### tygerdawg

Dunno, seems simple to my simple mind.

Torque = MassMomentOfInertia X AngularAcceleration

For your case of cylinder rotating about transverse axis, Wikipedia (and many other sources) show the MomentOfInertia as this (I-sub-x or I-sub-y)

AngularAcceleration =~ DeltaAcceleration / DeltaTime = [ CurrentRotationalVelocity - FinalRotationalVelocity ] / [ TimeRequired ]

CurrentRotationalVelocity = whatever is rotation speed
FinalRotationalVelocity = 0 (zero)

In what time duration do you want to reduce the rotational velocity to zero? That = TimeRequired

Plug & Chug to get Torque required to spin that cylinder down to zero in TimeRequired.

Add fudge factors for bearing friction, windage, anything else you can think of.