Torque requirements to roll vertically?

In summary: It does make more sense. Thanks.In summary, to model a problem about rolling up a wire, you need to calculate the Torque required to climb at a given speed.
  • #1
imathieson
3
0
Hi,

I'm trying to model a problem about rolling up a wire. I don't know how to go about setting it up to calculate the Torque required to climb at a given speed. Obviously coefficients of friction, normal forces, and weight will all play a roll but am hoping I can get a hand setting up the problem.

Any help would be greatly appreciated.

Thanks

Ian
 
Engineering news on Phys.org
  • #2
I would start with an energy approach. Let’s assume that the wheel is rolling up the wire without slipping at constant speed. That means constant linear and angular velocity. So no kinetic energy is being added or removed. No slipping means no loss of energy to friction or heat. (that’s an idealization of course) However, as the wheel rises against the force of gravity, it gains potential energy. At a constant speed, the wheel requires an input of potential energy every second equal to mgv where v is the velocity of the center of the wheel. The units are of power (Newton-meters per second). The torque you apply must provide this power. But torque is not power. Its Newton-meters. However, if constantly applied, torque is units of power. (Applying a torque of 10 Newton-meters for a full second is exactly what you get from an engine that delivers 10 Newton-meters per second power.) So you need a constant torque of magnitude mgv.

But you’re not finished. You need to assure the non-slip condition that was previously assumed. To get a non-slip condition, you need to generate friction between the wheel and wire. That is, you need to press the wheel to the wire. Given a wheel radius, torque and static coefficient of friction, are you comfortable calculating the normal force?
 
Last edited:
  • #3
Hi everyone, I'm just a new member and have been trying to post a question in the forum but couldn't find no link to do this. Can anyone tell me how to do this please... Any direct link or instructions to the post page will be ok...

Thanks
 
  • #4
Hi Mike,

Thanks for your reply. I'm comfortable getting the normal force give that data, however I have a question on the initial assumptions. With the kinetic energy being constant, it is the potential energy that is increasing correct. Therefore the equation essentially becomes kinetic energy+torque input=potential energy correct? If this is correct that is it correct to take the potential energy value at it's peak height?

Thanks for the reply!
 
  • #5
No, I don't think you phrased that last equation right.

Kinetic Energy + Energy provided by Torque = Kinetic Energy + Potential Energy.

Or, Energy provided by Torque = Potential Energy

The potential energy comes from the constantly applied torque.
You can really leave Kinetic Energy out of the equation.

As for peak height, it does not come into play either. You are thinking of potential energy as mgh, right? But I described a steady state problem. When, or if, the wheel stop does not matter. What matters is power while the wheel is moving. ***Energy per unit time*** And the amount of that power depends upon the amount of mass you are raising and how fast you are raising it. Its mgv where v is the velocity of the center of the wheel.
 
Last edited:
  • #6
us001007 said:
Hi everyone, I'm just a new member and have been trying to post a question in the forum but couldn't find no link to do this. Can anyone tell me how to do this please... Any direct link or instructions to the post page will be ok...

Thanks

Welcome us001007. I suppose the options for posting a question is different for different browsers. But on my machine there is a button at the top of the list of topics for Mechanical Engineering forum that says "New Topic". That's what I use to post a question.
 
  • #7
To imathieson,

As I reviewed my posting, I’m thinking it is not clear. I started out talking about energy but moved to talking about power. The determining equation really is,

Power delivered by the Torque = Power required to raise the wheel

Potential Energy, as you know, is mgh (units Newton-meters)
Power is Energy per unit time, in this case, mgv (units Newton-meters/second)

where m is the mass of the wheel,
g is gravity
and v is the velocity of the center of the wheel.

The power delivered by any torque (in units of Newton-meters) is the magnitude of the torque as long as that torque is being applied continuously (Newton-meters/second). That’s what you have to do to maintain a constant velocity for the wheel. You have to apply a torque of mgv continuously.

Does that make more sense? Or is it more confusing?
 

1. What is torque?

Torque is a measure of the amount of rotational force applied to an object, often measured in units of newton-meters (Nm).

2. How is torque related to rolling vertically?

When an object is rolling vertically, its torque is directly related to its angular acceleration and the moment of inertia of the object.

3. What factors affect the torque required to roll vertically?

The torque required to roll vertically is affected by the weight and shape of the object, as well as the surface on which it is rolling and any external forces acting on the object.

4. How can torque requirements be calculated for rolling vertically?

To calculate torque requirements for rolling vertically, the moment of inertia of the object must be determined, as well as the angular acceleration needed to achieve the desired rolling motion. These values can then be used in the equation: torque = moment of inertia x angular acceleration.

5. How can torque requirements be adjusted for different objects or surfaces?

To adjust torque requirements for different objects or surfaces, the moment of inertia and angular acceleration must be recalculated for the specific object and surface conditions. Additionally, external forces such as friction or air resistance may need to be taken into account.

Similar threads

  • Mechanical Engineering
Replies
2
Views
1K
Replies
8
Views
2K
Replies
2
Views
2K
  • Mechanical Engineering
Replies
8
Views
3K
  • Mechanical Engineering
Replies
17
Views
2K
  • Mechanical Engineering
Replies
5
Views
3K
  • Mechanical Engineering
Replies
3
Views
1K
Replies
3
Views
2K
Replies
5
Views
1K
  • Mechanical Engineering
Replies
2
Views
1K
Back
Top