# Torque for rotating mass inside cylincder

What I have:

• A hollow cylinder filled 1/3 its dia with slurry.
• Wt of cylinder is 987251 pounds
• Wt of slurry is 358157 pounds
• Radius of cylinder = 8.5 ft.
• RPM = 7.5
• Time taken = 20 seconds

What I need is :
• Torque required to roate the cylinder
• Torque required to rotate the slurry inside along with the cylinder

My try:
• Mass = Wt / accn due to gravity = 987251 / 32.2 = 30659.9 (where accn.due to gravity is in ft/sec^2)
• Torque required to rotate cylinder is Tcyl= (30659.9 * 8.5^2 * 2* Pi*RPM)/(60*20)

How to arrive for both combined..
• Is it the same procedure too? That is calculate torque reqd for the slurry too and then add both ?
Instead, can we use Torque = Force*Radius principle to arrive at the Torque required for rotating the slurry
• How to calculate the start up torque
• Remember having learnt some perpendicular axes theorem or parallel axes thoerom., do those apply ????
Any other method towards solving this ??

Any help is much appreciated

Last edited:

Simon Bridge
Homework Helper
The slurry will redistribute itself under rotation, changing the calculation.
You do not need any torque to keep something rotating at a constant rate.
Any torque greater than the static losses will rotate the cylinder. Any torque equal to the kinetic losses will keep it rotating.

Wt/g is not moment of inertia. Wt = weight = mg after all. But you did m/g which is not right either.

Look up "moment of inertia hollow cylinder".

You want: $$\sum \tau = I \frac{\Delta\omega}{\Delta t}$$

Last edited:
The slurry will redistribute itself under rotation, changing the calculation.
You do not need any torque to keep something rotating at a constant rate.
Any torque greater than the static losses will rotate the cylinder. Any torque equal to the kinetic losses will keep it rotating.

Okie how do you calculate the start up torque for the slurry. assuming it to be stationary. What is the torque required to bring the entire system (hollow cylinder with slurry ) to 7.5 RPM

Thats Mass= WT / Accn. due to gravity ... my BAD

jbriggs444
Homework Helper
The required startup torque will depend on two important details:

1. How good your bearings are.

2. How long you are prepared to wait for the assembly to reach the 7.5 rpm operating speed.

If your bearings are perfect and you are prepared to wait arbitrarily long then the required startup torque can be arbitrarily small.

change in angular momentum = net torque multiplied by elapsed time

So compute the angular momentum of the assembly rotating at 7.5 rpm and divide by the time you are willing to wait for spin-up.

sophiecentaur
Gold Member
2020 Award
Just checking here. Is the axis vertical or horizontal?

Just checking here. Is the axis vertical or horizontal?

horizontal

Tried another way, dunno if its ryt :

Rotational Moment of Inertia = I= m*r^2

in my case, I(slurry)= (Wt/32.2)*8.5^2

Torque , Tslurry = I*ω*(1/time)=I*(2*∏*rpm)*(1/time*60)

... is this correct ?

The motion of the fluid will depends on its viscosity, density, compressibility, as well as the dynamic friction with the walls.

Simon Bridge
Homework Helper
Your cylinder is only open on one end right?

Under ideal conditions, if the slurry is sticky enough to keep it's shape on the first half-turn, then you need to provide enough torque to lift the mass of the slurry a bit under the diameter of the cylinder. So $\pi \tau = 2mgr$ + losses etc. This will put the slurry momentarily at rest at the top of the cylinder.

Gravity helps you on the second half of the first turn.

You may prefer to modify this so that the centrifugal reaction is strong enough to hold the clump of slurry in place too. In which case you need the speed at the top.

If, when spun up, the slurry is basically evenly spread around the walls, then you can model it as another cylinder. In between the math gets messy and depends a lot on the properties of the slurry. I suspect the actual slurry kinda rolls around the bottom for some of the motion, gradually climbing the wall, until the rotation is fast enough for it to stick and it smooths out.

If all axis are along the same line, then the moments of inertia just add up.

This is the kind of thing we'd normally do empirically because of the variables ... to get a back-of-envelope prediction (so we don't go into the experiment blind) we'd need to know the specifics like: what do you need to know for etc. This sort of thing sets the restrictions on how good the approximation needs to be.

sophiecentaur