# Torque required to rotate a plate in a fluid

• LT72884
In summary: My understanding is that the reason that we use a differential equation is to model the real world. In the real world, there is no such thing as a 'point mass'. Everything occupies space and has mass.But, even in the real world, we can often model things using a point mass – the mathematics often works fine.However, when we are considering a body that is not a point mass but occupies space, we need to take account of the body’s shape.In this case the body is a disc, and when the disc rotates, the outer parts of the disc have a larger radius and therefore move faster than the inner parts.This means that the outer parts of the disc experience a larger force than the inner parts.In

#### LT72884

Homework Statement
A 12-in.-diameter circular plate is placed over a fixed bottom
plate with a 0.1-in. gap between the two plates filled with glycerin
as shown in Fig. P1.89. Determine the torque required to rotate the
circular plate slowly at 2 rpm. Assume that the velocity distribution
in the gap is linear and that the shear stress on the edge of the
rotating plate is negligible.
Relevant Equations
Torque=radius*(Viscous shear force) where viscous shear force is Fs
Fs = (shear stress)(area)
Torque = r(t)(a)
Hello all:) thanks for the help. This is for my fluids class. This is an example problem and I am stuck on it.

ok, so i found a video on youtube which walks through the entire process, but I am confused at a few sections. Here is the link to the video so you can follow:

at time stamp 2:37, he puts T=rta in differential form dT=rtdA

i don't understand why we need to do that? and why was area chosen to be the variable differentiated? why not r?

My biggest question though is why do we even need to make a diffeq? I am not seeing the reason why.

thanks much. its been a few years since i had diffeq

LT72884 said:
Homework Statement:: A 12-in.-diameter circular plate is placed over a fixed bottom
plate with a 0.1-in. gap between the two plates filled with glycerin
as shown in Fig. P1.89. Determine the torque required to rotate the
circular plate slowly at 2 rpm. Assume that the velocity distribution
in the gap is linear and that the shear stress on the edge of the
rotating plate is negligible.
Relevant Equations:: Torque=radius*(Viscous shear force) where viscous shear force is Fs
Fs = (shear stress)(area)
Torque = r(t)(a)

Hello all:) thanks for the help. This is for my fluids class. This is an example problem and I am stuck on it.

ok, so i found a video on youtube which walks through the entire process, but I am confused at a few sections. Here is the link to the video so you can follow:

at time stamp 2:37, he puts T=rta in differential form dT=rtdA

i don't understand why we need to do that? and why was area chosen to be the variable differentiated? why not r?

My biggest question though is why do we even need to make a diffeq? I am not seeing the reason why.

thanks much. its been a few years since i had diffeq

The shear force between 2 surfaces (separated by viscous fluid) is proportional to the speed of one surface relative to the other and proportional to the area of overlap. That’s where area comes in.

We can't just use the area of the whole disc becase different parts of the disc have different speeds (v = ωr). So we have to be clever.

If we consider the disc’s surface as being lots (infinity!) of concentric rings, we can find the total torque by adding up the torques produce by all the rings.

Each ring has an area 2πr.dr and a speed, ωr. The force on the ring can be found in terms of 2πr.dr and ωr (and other parameters). Then multiplying by r gives the torque produced by that ring.

Finally we need to add up all the torques from the many rings. That’s the integration.

• Lnewqban and LT72884
Steve4Physics said:
The shear force between 2 surfaces (separated by viscous fluid) is proportional to the speed of one surface relative to the other and proportional to the area of overlap. That’s where area comes in.

We can't just use the area of the whole disc becase different parts of the disc have different speeds (v = ωr). So we have to be clever.

If we consider the disc’s surface as being lots (infinity!) of concentric rings, we can find the total torque by adding up the torques produce by all the rings.

Each ring has an area 2πr.dr and a speed, ωr. The force on the ring can be found in terms of 2πr.dr and ωr (and other parameters). Then multiplying by r gives the torque produced by that ring.

Finally we need to add up all the torques from the many rings. That’s the integration.
im still a little confused, how come we can't use the T=Fr/j where j is mr^2/2 for a disk?

thanks for the explanation from that view. I am going to try and understand it

LT72884 said:
im still a little confused, how come we can't use the T=Fr/j where j is mr^2/2 for a disk?
I don't recognise your equation but the moment of inertia of a disc is ½MR².

It is typically used when calculating the angular acceleration of a disc, or a disc’s kinetic energy.

For example, you would use the moment of inertia if you wanted to calculate how quickly a disc would speed-up or slow-down when a torque is applied.

But that’s not what this question is about. The question is about calculating the torque produced by a viscous fluid. The disc has a constant angular speed (no acceleration) and the moment of inertia (½MR²) is not relevant.

When you use a formula, it is important to know in what circumstances it applies.

• Lnewqban and LT72884
Steve4Physics said:
I don't recognise your equation but the moment of inertia of a disc is ½MR².

It is typically used when calculating the angular acceleration of a disc, or a disc’s kinetic energy.

For example, you would use the moment of inertia if you wanted to calculate how quickly a disc would speed-up or slow-down when a torque is applied.

But that’s not what this question is about. The question is about calculating the torque produced by a viscous fluid. The disc has a constant angular speed (no acceleration) and the moment of inertia (½MR²) is not relevant.

When you use a formula, it is important to know in what circumstances it applies.
the above formula is for torque of a disk. came from my machine design course.

i still do not see why it is even relevant to use a differential equation to solve this. Just find the torque required to move the disk... i don't know why they have to make it so complicated and use a diff eq? that's what i do not understand at all.

thanks for taking the time to reply:)

Cosider a di
LT72884 said:
i don't know why they have to make it so complicated and use a diff eq?

There is no way I know of to find the answer without using calculus!

But, if you are not familiar/comfortable with calculus, I can see that it would appear complicated.

LT72884 said:
the above formula is for torque of a disk. came from my machine design course.

i still do not see why it is even relevant to use a differential equation to solve this. Just find the torque required to move the disk... i don't know why they have to make it so complicated and use a diff eq? that's what i do not understand at all.

thanks for taking the time to reply:)
Is the fluid exerting a torque on the disk or not? This problem focuses on determining the torque the fluid exerts on the disk. It is not concerned with the torque balance on the disk, since the angular acceleration of the disk is zero anyway, and the only other torque on the disk is one that has to be applied to maintain a constant angular velocity against the torque of the fluid on the disk.

• Lnewqban
Steve4Physics said:
Cosider a di

There is no way I know of to find the answer without using calculus!

But, if you are not familiar/comfortable with calculus, I can see that it would appear complicated.
I am very familiar with calculus and and diff eq, but that does not answer my question as to WHY WHY WHY do they use a diff eq to answer the question. Has nothing to do if one is familiar with calc. I am looking for why do WE HAVE TO USE diff eq? i can solve the diff eq all do with no issues, but WHY do we have to use it?

in machine design, strengths of materials, and dynamics, we never used diff eq at all to solve problems, we had a list of formulas and equations to use. Rotating disks and torque were most of my machine design class i just finished. I tried using those equations on this problem with failure, but i wnat to know why they don't work and why we have to use diff eq?

thanks

Last edited:
Chestermiller said:
Is the fluid exerting a torque on the disk or not? This problem focuses on determining the torque the fluid exerts on the disk. It is not concerned with the torque balance on the disk, since the angular acceleration of the disk is zero anyway, and the only other torque on the disk is one that has to be applied to maintain a constant angular velocity against the torque of the fluid on the disk.
ok, so if i follow, since the fluid is causing a torque on the disk, and at the center of the disk, shear stress and velocity would be zero, and as we move out towards the edge of the disk, the shear stress and velocity increase meaning, in order to find the torque the fluid is causing over the entire disk, we would have to use diff eq because it is incremental?

• Steve4Physics, berkeman and Chestermiller
LT72884 said:
in machine design, strengths of materials, and dynamics, we never used diff eq at all to solve problems, we had a list of formulas and equations to use. Rotating disks and torque were most of my machine design class i just finished. I tried using those equations on this problem with failure, but i wnat to know why they don't work and why we have to use diff eq?
I expect many of the 'formulas and equations' you are given have themselves been derived using calculus. For example, the moment of inertia of a disc (½MR²) is derived using calculus. Your machine design class would use these standard formula when finding acceleration of a disc say, without requiring any calculus.

But in this problem you don't already have a formula relating to viscous torque acting on a disc. You have to be prepared to derive a formula (which must contain the viscosity of the fluid, μ) for yourself. The only way I can think of is to use calculus. (I think you have understood this because of what you said in Post #9.)

• LT72884
Steve4Physics said:
I expect many of the 'formulas and equations' you are given have themselves been derived using calculus. For example, the moment of inertia of a disc (½MR²) is derived using calculus. Your machine design class would use these standard formula when finding acceleration of a disc say, without requiring any calculus.

But in this problem you don't already have a formula relating to viscous torque acting on a disc. You have to be prepared to derive a formula (which must contain the viscosity of the fluid, μ) for yourself. The only way I can think of is to use calculus. (I think you have understood this because of what you said in Post #9.)
Yes, i think i am beginning to see why i need to derive the formulas. In machine design, we only had to worry about the solid body or just a pulley and a rod, but now we have a fluid interreacting with the rotating disk which is now causing shear stress and velocity changes along the entire disk at intervals rather than constant.

for example, pressure at a certain depth of water, i have always been taught it is P=pgh, but now in fluids class, water density is not constant so we have to derive a diff eq...

• Steve4Physics