Torque required from powered roller to rotate cylinder

[biggercheese]

Hi all. Let me preface this by saying that I have no formal education in physics so I apologize if there is any information that I left out. Please comment with questions and I will try to answer them to the best of my ability.

The cylinder has a mass of 30 lb, is 6 feet long, and has a diameter of 12 inches. It's weight is distributed across two rollers, a powered roller and an idle roller. Each roller has two wheels of 3-inch diameter. One of the wheels on the powered roller will be responsible for transmitting the torque required to rotate the pole.

I began by assuming that the weight was evenly distributed across the two rollers. This allowed me to simplify my free body diagram to just the powered roller:

The forces present are the normal forces of the wheels acting on the cylinder and the friction between the wheels and the cylinder. Here is where I begin to lose confidence in my approach. My initial thought is that since Friction Force= static friction coefficient * normal force I could omit the frictional component during the static analysis. Once i find my normal forces I can use it to find the friction force. I would then use my friction force to find the torque: Torque at wheel A = Friction Force at A * radius of wheel A
I am just looking for a ballpark torque value for this project but I am not sure that my train of thought is the correct approach

 

[erobz]

Hi all. Let me preface this by saying that I have no formal education in physics so I apologize if there is any information that I left out. Please comment with questions and I will try to answer them to the best of my ability.

The cylinder has a mass of 30 lb, is 6 feet long, and has a diameter of 12 inches. It's weight is distributed across two rollers, a powered roller and an idle roller. Each roller has two wheels of 3-inch diameter. One of the wheels on the powered roller will be responsible for transmitting the torque required to rotate the pole.

I began by assuming that the weight was evenly distributed across the two rollers. This allowed me to simplify my free body diagram to just the powered roller:
View attachment 354180
View attachment 354181

The forces present are the normal forces of the wheels acting on the cylinder and the friction between the wheels and the cylinder. Here is where I begin to lose confidence in my approach. My initial thought is that since Friction Force= static friction coefficient * normal force I could omit the frictional component during the static analysis. Once i find my normal forces I can use it to find the friction force. I would then use my friction force to find the torque: Torque at wheel A = Friction Force at A * radius of wheel A
I am just looking for a ballpark torque value for this project but I am not sure that my train of thought is the correct approach

If you don't want the driver to slip, then the frictional force between the driver and the rotating body must be whatever it takes to accelerate (angular) the body which you are trying to rotate.

So we go from rest to some fixed angular velocity and how quicky (time or angle) is that to be accomplished? You figure out what the minimum value of friction coefficient needs to be after that.

 

[biggercheese]

If you don't want the driver to slip, then the frictional force between the driver and the rotating body must be whatever it takes to accelerate (angular) the body which you are trying to rotate.

So we go from rest to some fixed angular velocity and how quicky (time or angle) is that to be accomplished? You figure out what the minimum value of friction coefficient needs to be after that.

Hmm for my application it isn't necessarily important but just for the sake of understanding the math say the cylinder should start from rest and accelerate to 0.5 rad/s in 3 seconds. Could you provide direction on how to calculate the friction coefficient to achieve that amount of acceleration?

 

[Baluncore]

Since the cylinder sits between the two rollers at 90°, the contact force will be increased by √2. Half the mass of the cylinder is multiplied by √2, then by g=9.8 , to get the normal force against the driving roller.

Three further advantages would be gained if the idle roller position was raised above the driving roller.
First, a greater proportion of the cylinder mass would rest on the driving roller.
Second, the pinch between the rollers would be increased.
Third, the frictional drive force, would be added to the weight of the cylinder, for the sense of rotation shown in the diagram.

 

[erobz]

Hmm for my application it isn't necessarily important but just for the sake of understanding the math say the cylinder should start from rest and accelerate to 0.5 rad/s in 3 seconds. Could you provide direction on how to calculate the friction coefficient to achieve that amount of acceleration?

Start by drawing free body diagrams of the body being rolled and the driven pinch roller. On the large body there are two forces that produce opposing torques on it (ignoring other possible sources for the time). There is friction between the driver and the roll $f_d$ that is responsible for rotating the roll and friction $f_i$ between the roll and the idler that drives the idler. Pay attention to the relative angular acceleration of each roller in the system. Calculate $\ddot \theta_r$ ( the constant angular acceleration of the roll) under the assumption of constant angular acceleration using the kinematic equations and your parameters for the desired angular velocity $\omega_f$ and time $t_f$. The friction that is accelerating the idler $f_i$ is fixed by noting that the idler is demanded to spin at a greater rate than your roll (think of them as meshing gears). If the angle the roll turns through is $\theta_r$, then the angle the idler turns through is $\beta_i$, and hence angular acceleration $\ddot \beta_i$ (related by the radii's of the meshing rollers). You get $f_i$ by analyzing the idler first in this way. $\sum \tau = I \alpha$

I see you wrote no formal education in physics, you obviously have some understanding though based off what you have already shown. Anyhow, if you get stuck please ask.

 

[biggercheese]

Start by drawing free body diagrams of the body being rolled and the driven pinch roller. On the large body there are two forces that produce opposing torques on it (ignoring other possible sources for the time). There is friction between the driver and the roll $f_d$ that is responsible for rotating the roll and friction $f_i$ between the roll and the idler that drives the idler. Pay attention to the relative angular acceleration of each roller in the system. Calculate $\ddot \theta_r$ ( the constant angular acceleration of the roll) under the assumption of constant angular acceleration using the kinematic equations and your parameters for the desired angular velocity $\omega_f$ and time $t_f$. The friction that is accelerating the idler $f_i$ is fixed by noting that the idler is demanded to spin at a greater rate than your roll (think of them as meshing gears). If the angle the roll turns through is $\theta_r$, then the angle the idler turns through is $\beta_i$, and hence angular acceleration $\ddot \beta_i$ (related by the radii's of the meshing rollers). You get $f_i$ by analyzing the idler first in this way. $\sum \tau = I \alpha$

I see you wrote no formal education in physics, you obviously have some understanding though based off what you have already shown. Anyhow, if you get stuck please ask.

Here is my attempt at the static analysis. I am not sure how to determine the angles that the friction force of the rollers acts on the cylinder. Would it also just be 45 degrees?

 

[erobz]

View attachment 354206
Here is my attempt at the static analysis. I am not sure how to determine the angles that the friction force of the rollers acts on the cylinder. Would it also just be 45 degrees?

We are just worried about the rotation aspect here. You should be applying sum of $\sum \tau = I \alpha$

The frictional forces are tangent to the cylinders.

Another thing...pick a rotation direction as positive ( counter clockwise for the big cylinder for instance ). Then free the bodies from contact with each other, but maintain consistent rotation direction as if they were meshing gears). Use Newtons 3rd law so you can look at each body as its own problem.

 

[biggercheese]

We are just worried about the rotation aspect here. You should be applying sum of $\sum \tau = I \alpha$

The frictional forces are tangent to the cylinders.

Another thing...pick a rotation direction as positive ( counter clockwise for the big cylinder for instance ). Then free the bodies from contact with each other, but maintain consistent rotation direction as if they were meshing gears). Use Newtons 3rd law so you can look at each body as its own problem.

Ok based on what you said I believe this is correct. So now I know the frictional force at the idler. I can now sum the forces on the idler to find the normal force of the idler on the cylinder?

I could then find each force and thus find the torque required at the driving wheel. Is my thought process correct?

Also I appreciate you taking the time to help me :)

 

[Lnewqban]

I am just looking for a ballpark torque value for this project but I am not sure that my train of thought is the correct approach

Welcome!

The long cylinder must be supported by something else than those two small rollers.
Could you provide a whole description of the device?

Also, the input torque can only be equal to the resistive torque coming from the cylinder (rotational inertia during start-ups plus torque required to overcome friction and any load).

Do you know the actual coefficient of static friction for the materials combination of the rollers and cylinder?

 

[erobz]

View attachment 354213
Ok based on what you said I believe this is correct. So now I know the frictional force at the idler. I can now sum the forces on the idler to find the normal force of the idler on the cylinder?

I could then find each force and thus find the torque required at the driving wheel. Is my thought process correct?

Also I appreciate you taking the time to help me :)

Just use all variables, it will be easier for me to follow, and better to make adjustments later on.

I think its ok.

$$ f_i r = \frac{1}{2}mr^2 \frac{\omega_f}{t_f} \frac{R}{r}$$
$$ \implies f_i = \frac{1}{2}mR \alpha \tag{idler}$$

Then you apply $f_i$ at $R$ on the large cylinder (of mass $M$) as a force creating a torque about the mass center of the large cylinder. You then solve that for $f_d$. ( don't forget about $f_d$ acting at $R$ as another external torque).

Once you do that we can talk about the minimum $\mu$ ( in relation to $f_d$). The normal forces $N_i,N_d$ really aren't relevant until we get to this step.

Then perhaps after we get that we can add some constant resistive toques from bearings into the analysis without much difficulty.

 

[biggercheese]

Just use all variables, it will be easier for me to follow, and better to make adjustments later on.

I think its ok.

$$ f_i r = \frac{1}{2}mr^2 \frac{\omega_f}{t_f} \frac{R}{r}$$
$$ \implies f_i = \frac{1}{2}mR \alpha \tag{idler}$$

Then you apply $f_i$ at $R$ on the large cylinder (of mass $M$) as a force creating a torque about the mass center of the large cylinder. You then solve that for $f_d$. ( don't forget about $f_d$ acting at $R$ as another external torque).

Once you do that we can talk about the minimum $\mu$ ( in relation to $f_d$). The normal forces $N_i,N_d$ really aren't relevant until we get to this step.

Then perhaps after we get that we can add some constant resistive toques from bearings into the analysis without much difficulty.

Ok I am tracking.
$$\sum_{}^{}\tau=\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}^{}+\mathrm{f}_{i}^{}*R+\mathrm{f}_{d}^{}*R=0$$
Then I can solve for $f_d$ :
$$\mathrm{f}_{d}^{} = \frac{\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}-\mathrm{f}_{i}*R}{R}$$
$$ \implies\mathrm{f}_{d}=0.036 lb$$

I would now calculate the normal forces produced by the idler and driver wheels? Would I need to find the friction coefficient of the wheel materials?

 

[biggercheese]

Welcome!

The long cylinder must be supported by something else than those two small rollers.
Could you provide a whole description of the device?

Also, the input torque can only be equal to the resistive torque coming from the cylinder (rotational inertia during start-ups plus torque required to overcome friction and any load).

Do you know the actual coefficient of static friction for the materials combination of the rollers and cylinder?

When I say roller I'm referring to the devices that are supporting the cylinder. Each roller has two wheels and there would be two or three rollers to support the cylinder. My idea is to have only one wheel on one roller powered to produce rotation on the cylinder

The materials for the wheels have not yet been chosen, I am here in an attempt to understand the math required for a project like this.

 

[Lnewqban]

Please, note that the driving wheel has mechanical advantage respect to any load torque imposed onto the cylinder:

https://www.engineeringtoolbox.com/gears-d_1307.html

 

[erobz]

Ok I am tracking.
$$\sum_{}^{}\tau=\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}^{}+\mathrm{f}_{i}^{}*R+\mathrm{f}_{d}^{}*R=0$$
Then I can solve for $f_d$ :
$$\mathrm{f}_{d}^{} = \frac{\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}-\mathrm{f}_{i}*R}{R}$$
$$ \implies\mathrm{f}_{d}=0.036 lb$$

I would now calculate the normal forces produced by the idler and driver wheels? Would I need to find the friction coefficient of the wheel materials?

The torques from the frictional forces should not be of the same sign(sense). They are in opposition.

Anyhow, after that uses $\sum F_x, \sum F_y = 0$ to get the Normal forces $N_d,N_i$.

Then whatever material you chose would have to have a $\mu$ such that at least $\mu \geq \frac{f_d}{N_d}$ is satisfied.

 

[biggercheese]

The torques from the frictional forces should not be of the same sign. They are in opposition.

Anyhow, after that uses $\sum F_x, \sum F_y = 0$ to get the Normal forces $N_d,N_i$.

Thank you for pointing that out, I corrected my free body diagram.

$$\sum_{}^{}\tau=\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}^{}-\mathrm{f}_{i}^{}*R-\mathrm{f}_{d}^{}*R=0$$
$$\implies\mathrm{f}_{d}=0.036 lbf$$
Now the static analysis:
$$\sum \mathrm{F}_{y}=-30lb-\mathrm{f}_{i}*sin(45)+\mathrm{f}_{d}*sin(45)+\mathrm{N}_{d}*sin(45)+\mathrm{N}_{i}*sin(45)=0$$
$$\sum \mathrm{F}_{x}=\mathrm{N}_{d}*cos(45)-\mathrm{f}_{d}*cos(45)-\mathrm{f}_{i}*cos(45)-\mathrm{N}_{i}*cos(45)=0$$
$$\implies\mathrm{N}_{d}=17.631lb $$
$$\implies\mathrm{N}_{i}=17.592lb $$

I am assuming we will use this information to find the friction coefficient that is required.

 

[erobz]

Thank you for pointing that out, I corrected my free body diagram.

View attachment 354241

$$\sum_{}^{}\tau=\mathrm{I}_{cylinder}^{}*\mathrm{\alpha}_{cylinder}^{}-\mathrm{f}_{i}^{}*R-\mathrm{f}_{d}^{}*R=0$$
$$\implies\mathrm{f}_{d}=0.036 lbf$$
Now the static analysis:
$$\sum \mathrm{F}_{y}=-30lb-\mathrm{f}_{i}*sin(45)+\mathrm{f}_{d}*sin(45)+\mathrm{N}_{d}*sin(45)+\mathrm{N}_{i}*sin(45)=0$$
$$\sum \mathrm{F}_{x}=\mathrm{N}_{d}*cos(45)-\mathrm{f}_{d}*cos(45)-\mathrm{f}_{i}*cos(45)-\mathrm{N}_{i}*cos(45)=0$$
$$\implies\mathrm{N}_{d}=17.631lb $$
$$\implies\mathrm{N}_{i}=17.592lb $$

I am assuming we will use this information to find the friction coefficient that is required.

You're still messing up the torques. Draw a fbd of the idler. The friction force spins it a certain direction. On the big cyliner ithat friction force acts opposite that convention. You are to be doing a FBD of the large cylinder. That means you isolate the large cylinder.

 

[biggercheese]

You're still messing up the torques. Draw a fbd of the idler. The friction force spins it a certain direction. On the big cyliner ithat friction force acts opposite that convention. You are to be doing a FBD of the large cylinder. That means you isolate the large cylinder.

Separating the wheel from the cylinder in the free body diagrams makes so much sense. I can see how the idler reacts against the cylinders motion

Redoing the previous calculation:

$$\sum \mathrm{F}_{y}=-30lb-\mathrm{f}_{i}*sin(45)-\mathrm{f}_{d}*sin(45)+\mathrm{N}_{d}*sin(45)+\mathrm{N}_{i}*sin(45)=0$$
$$\sum \mathrm{F}_{x}=\mathrm{N}_{d}*cos(45)+\mathrm{f}_{d}*cos(45)-\mathrm{f}_{i}*cos(45)-\mathrm{N}_{i}*cos(45)=0$$
$$\implies\mathrm{N}_{d}=17.631lb $$
$$\implies\mathrm{N}_{i}=17.665lb $$

As you said with your previous reply, I have everything I need to find $\mu$. This is important in material selection.

To find the torque requirement at the driver, say for a motor, would it be as simple as:
$$\tau=\mathrm{f}_{d}*r+\mathrm{I}_{d}*\mathrm{\alpha}_{d}$$
plus any friction in the bearings?

 

[erobz]

Separating the wheel from the cylinder in the free body diagrams makes so much sense. I can see how the idler reacts against the cylinders motion
View attachment 354242
Redoing the previous calculation:

$$\sum \mathrm{F}_{y}=-30lb-\mathrm{f}_{i}*sin(45)-\mathrm{f}_{d}*sin(45)+\mathrm{N}_{d}*sin(45)+\mathrm{N}_{i}*sin(45)=0$$
$$\sum \mathrm{F}_{x}=\mathrm{N}_{d}*cos(45)+\mathrm{f}_{d}*cos(45)-\mathrm{f}_{i}*cos(45)-\mathrm{N}_{i}*cos(45)=0$$
$$\implies\mathrm{N}_{d}=17.631lb $$
$$\implies\mathrm{N}_{i}=17.665lb $$

As you said with your previous reply, I have everything I need to find $\mu$. This is important in material selection.

To find the torque requirement at the driver, say for a motor, would it be as simple as:
$$\tau=\mathrm{f}_{d}*r+\mathrm{I}_{d}*\mathrm{\alpha}_{d}$$
plus any friction in the bearings?

A bearing torque from the idler works its way into the equation via $f_d$. So you need to account for it through the equations that way...( at least I believe there isn't a shortcut- but I don't really care if there is). If there was a torque from the driver bearings, then yeah it adds linearly into the equation for the driver.

 

[biggercheese]

A bearing torque from the idler works its way into the equation via $f_d$. So you need to account for it through the equations that way...( at least I believe there isn't a shortcut- but I don't really care if there is). If there was a torque from the driver bearings, then yeah it adds linearly into the equation for the driver.

I think that's everything I needed. Thank you for your time

 

[erobz]

I think that's everything I needed. Thank you for your time

You're welcome.