Rotational force caused by a rod through the radius of a cylinder

In summary, a rod through a cylinder's radius causes rotational force equal to the difference in foot poundsof the two sides, but what happens once the cylinder has rotated enough so that the end of the rod is touching the same surface that the cylinder is resting on? To calculate the rotational force, you must first calculate the weight of the rod at either end, and then apply that weight to the rod to find the rotational force.
  • #1
jehubuddaka
5
0
I am playing with math and physics during my off-time at work, and I am wondering how to calculate a specific force, or if I am even looking at it correctly. Here it is.

A rod through a cylinder's radius so that if it is off to one side causes rotational force equal to the difference in foot poundsof the two sides seems simple enough. But what happens once the cylinder has rotated enough so that the end of the rod is touching the same surface that the cylinder is resting on? How do I calculate that force. I am assuming that the cylinder has a hole through it just larger than the rod to allow motion. Where is the force applied? Thanks in advance for the answers :)
 
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  • #2
I find your description a bit unclear, so let me ask. You have this cylinder lying down on a surface, a rod is inserted "in the side of it", ie. in radial direction so that it's not symmetrical but only points out on one side of the cylinder. And then you rotate it from one side til it hits the surface on the other drawing out the path of a half circle?
 
  • #3
This problem was posted before, but I don't recall the exact wording required to find that thread or the answer.

Assume a rod goes through a pair of frictionless holes or perhaps even better, a frictionless tube across the diameter of the cylinder. Assume that the initial condtion is that the end of the rod rests on the same surface as the cylinder. What are the forces, and could any movement result from these forces?

In the previous case, there wasn't any maximum length given to the rod, so a very long rod could be well extended, resting on the surface at a near horizontal angle. Another critical point would be when the rod was vertical (assume the rod has a rounded end and ther's no rolling resistance). I'm not sure if the rod was considered to have any mass at all in the previous thread.
 
  • #4
Yes Jeff has my basic description covered quite well. Frictionless hole through the diameter with a rod inserted into that hole. basically tip the cylinder onto one of its flats, then insert a rod through it from left to right, directly through the middle of the cylinder. then tip it back up onto its round side and you've got a good idea of how the rod placement should be. The only thing other than that to remember is that the rod is free to slip out of the hole to a certain distance. Now, with this information, how would I calculate what kind of resistance to continued rotation would be caused when the protruding end of the rod strikes the surface that the cylinder is rolling on?

Add: I am trying to calculate as if the rod is weightless, but with 1lb of weight at either end located exactly on the ends of the rod. My theory is that the weight of the rod is pushing down and sideways with a downward force equal to a larger amount for angles closer to horizontal, and vice versa for the sideways push. Am I anywhere near How I should be thinking on this? And how does this linear force translate to rotational force?
 
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1. What is rotational force caused by a rod through the radius of a cylinder?

Rotational force, also known as torque, is the measure of the force that can cause an object to rotate around an axis. When a rod is inserted through the radius of a cylinder, it creates a lever arm which can apply a rotational force on the cylinder.

2. How is rotational force calculated?

Rotational force is calculated by multiplying the force applied to the lever arm by the length of the lever arm. This can be expressed as T = F x r, where T is the torque, F is the applied force, and r is the length of the lever arm.

3. What factors affect the strength of the rotational force?

The strength of rotational force is affected by several factors, including the magnitude of the applied force, the length of the lever arm, the angle at which the force is applied, and the mass of the object.

4. How does the position of the rod affect the rotational force on the cylinder?

The position of the rod along the radius of the cylinder will affect the distance from the axis of rotation, and therefore the length of the lever arm. The longer the lever arm, the greater the torque applied to the cylinder.

5. What is the difference between rotational force and linear force?

Rotational force, or torque, causes an object to rotate around an axis, while linear force causes an object to move in a straight line. Rotational force is measured in units of Newton-meters (Nm), while linear force is measured in units of Newtons (N).

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