Variable torque due to rotation

In summary, Naty1 has hit a snag in his physics project and is looking for help. He has calculated the system's moment of inertia and the magnitude of the three torques, but is not sure how to relate the varying quantities to time. He is looking for ideas on how to solve this differential equation.
  • #1
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Hi, all! I just registered here, and hope to enjoy enlightening discussions about the subject of physics in the future.
The spesific reason for the post is a project I am currently concerning myself with. I hit a snag recently, and hope to resolve it with help from others here. Here goes:

Background

I'll ignore friction and air resistance in the following.
I'm making an apparatus consisting of a beam (B) rotating about a pivot (A), loaded with two masses M and m; one attached to each end. It looks like a sort of see-saw.
Now, there are three forces in play, all due to gravity:
-The force on the largest mass M, GM
-The force on the smaller mass m, Gm
-The force on the beam itself, GB.

Please note that A may be different from the beam's centre of gravity.
The system is designed so that of the three resulting torques (only two if the beam's weight doesn't set up a torque about A), TM will always be bigger than the other(s).
The result is that the system will be set in rotational motion if held in an initial position, and then released. The mass M will move in an arc downwards, and then hit some kind of blockage (e.g. the floor).

I have calculated the system's moment of inertia (I), and the magnitude of the three torques are easily found. The angle between the beam and the horizontal axis will be denoted as θ.

Problem:

Once the system is released, the total torque will accelerate the system from rest because of ΣT = Iα.
In so doing, the mass M will move through a small angle dθ, and the lever arm of TM will be shortened since the arm is equal to R*cos(θ). Thus the torque decreases, and also the acceleration. The same is true for the mass m on the other side of the pivot, only that its torque in the other (negative) direction is decreased.
The net result I'm guessing will be a steadily decreasing acceleration and an angular speed that keeps growing throughout the motion, but less and less with time.

How do I model this change in acceleration and speed mathematically?
My goal is to calculate the final speed of the masses M and m, but I can't quite relate the varying quantities to time. My guess is that this will lead to some differential equations, and I'm not steady enough with my math to crack this one.
Any help will be greatly appreciated!
 
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  • #2
I don't know how to answer your question, but I'd guess a first year college text in dynamics (as opposed to statics) would provide some formulas and application ideas. I guess that would be part of a mechanical engineering curriculum? Wikipedia might also be a source with some formulas to use...
 
  • #3
Thank you for your answer, Naty1!
I have a book on university physics, but sadly mechanics are not my field, and the theory on rotational dynamics covered doesn't quite help me figuring it out.
Maybe this should have been posted on the mathematics subforums instead? It seems like this calls for more calculus than I can muster at the moment.

I will continue my investigations, and appreciate all additional thoughts on the matter! =)
 
  • #4
Okay, I think I made some headway, so I'll share my thoughts to see whether I'm on the right track.

My new approach is to use energy considerations. I'll treat the two halves of the system as separate, that is, each mass and the part of the beam that is on its side of the pivot is one entity.
Then, by treating both the masses and the beam parts as point masses, and setting a sensible reference level for the gravitational potential energy (the floor), the entire rotational motion amounts to moving four point masses up or down in the gravitational field. Thus

EP1 + EK1 = EP2 + EK2,

and since the system is at rest at startup, it reduces to

EP1 = EP2 + EK2.

Now, as the half with the heavier mass M is going to fall downwards, M's potential energy and its beam part's PE is going to decrease.
The lighter mass m and its beam part, however, will rise up, and gain PE.
Calculating the net resulting PE and comparing with the initial PE gives a difference which must be the system's final kinetic energy.
And finally, given this kinetic energy, one can use

EK = (1/2)*I*ω2

yielding

ω = sqrt(2*I*EK).

Since this last calculation considers the moment of inertia, I don't see that my simplification in the energy part should influence the final speed here.

That's it, any thoughts?
 
  • #5


Hello and welcome! It's great to have a fellow physicist here. Your project sounds very interesting and I'm happy to help you solve this problem.

You are correct in your assumption that this situation will require the use of differential equations. Specifically, you will need to use Newton's Second Law for rotational motion, which states that the sum of the torques acting on an object is equal to the moment of inertia multiplied by the angular acceleration: ΣT = Iα.

In this case, the total torque will be the sum of the three torques you mentioned: TM, Gm, and GB. As you noted, TM will decrease as the lever arm shortens, while the other two torques will remain constant. This means that the net torque will also decrease with time, leading to a decrease in angular acceleration.

To model this change in acceleration and speed mathematically, you will need to set up a differential equation that relates the angular acceleration (α) to time (t). This can be done by using the fact that acceleration is the derivative of velocity, and velocity is the derivative of position. In other words, you can use the equations α = dω/dt and ω = dθ/dt, where ω is the angular velocity and θ is the angular position.

Using these equations and plugging in the values for the torques and moment of inertia, you should be able to solve for the final angular velocity and position of the masses M and m. Just be sure to keep track of the signs of the torques and angular acceleration, as they will determine the direction of the motion.

I hope this helps you with your project! If you have any further questions, please don't hesitate to ask. Good luck!
 

1. What is variable torque due to rotation?

Variable torque due to rotation is the force that causes an object to rotate around an axis. It is dependent on several factors such as the rotation speed, the distance from the axis of rotation, and the mass of the object.

2. What causes variable torque due to rotation?

Variable torque due to rotation is caused by an unbalanced force acting on an object that is free to rotate. This force can be applied by an external source, such as a motor or a person, or it can be a result of the distribution of mass within the object.

3. How is variable torque due to rotation measured?

Variable torque due to rotation is typically measured in units of Newton-meters (Nm) or foot-pounds (ft-lb). This measurement is calculated by multiplying the force applied to the object by the distance from the axis of rotation at which the force is applied.

4. What are some real-life examples of variable torque due to rotation?

One common example of variable torque due to rotation is seen in a bicycle wheel. The force applied by the cyclist to the pedals causes the wheel to rotate around its axis. Another example is a door handle, where the force applied by the hand causes the door to rotate around its hinges.

5. How does variable torque due to rotation affect the stability of an object?

The variable torque due to rotation can affect the stability of an object by either increasing or decreasing its rotational speed. If the torque is strong enough, it can cause the object to rotate at a faster rate, making it more difficult to control. On the other hand, a decrease in torque can result in a decrease in rotational speed, making the object more stable and easier to control.

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