Exploring Torque and Angular Acceleration in Nonuniform Free Fall

In summary, objects with a nonuniform mass density experience a net torque, but it is not always responsible for an angular acceleration.
  • #1
UMath1
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9
I am confused about whether objects when a nonuniform mass density experience net torque during free fall. If you take a rod that is heavier on one end and lighter on the other, should it not experience a net torque and angularly accelerate? But at the same time, the acceleration by gravity is uniform for the whole rod. And usually in the absence of air resistance it wouldn't rotate. It seems it only rotate if there is a pivot point. But I don't understand why.
 
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  • #2
UMath1 said:
I am confused about whether objects when a nonuniform mass density experience net torque during free fall. If you take a rod that is heavier on one end and lighter on the other, should it not experience a net torque and angularly accelerate? But at the same time, the acceleration by gravity is uniform for the whole rod. And usually in the absence of air resistance it wouldn't rotate. It seems it only rotate if there is a pivot point. But I don't understand why.
Just because there's a net torque about a point does not necessarily mean that there will be an angular acceleration about that point. That "rule" applies for non-accelerating points or for the center of mass. I'm sure you'll agree that there's no net torque about the center of mass. And if you take torques about one end of the non-uniform rod, realize that that point is accelerating--it will turn out that the angular acceleration about that point will be zero.
 
  • #3
When the CM and the centre of pressure lie on a vertical line then there will be no torque. If the CM is vertically below the CP then the rod will be in stable equilibrium (there is a restoring torque for any displacement) and if the CM is above the CP, the equilibrium is unstable and the rod will soon start to rotate.
 
  • #4
Why bring up center of pressure? The rod's in free fall.
 
  • #5
I was considering the case where the center of mass is the axis of rotation. Then if the left side were denser, there would be a net counterclockwise torque? But it seems the axis of rotation must be at fulcrum for that to apply...but I can't explain to myself why that is.
 
  • #6
UMath1 said:
I was considering the case where the center of mass is the axis of rotation. Then if the left side were denser, there would be a net counterclockwise torque?
No, there would be no net torque about the center of mass. If the left side of the rod were denser, the center of mass would be closer to that end.

Consider the definition of center of mass and how to calculate it. Compare that to calculating the net torque.
 
  • #7
Doc Al said:
Why bring up center of pressure? The rod's in free fall.
I was throwing the rod out of an aeroplane (Woops!). That's the kind of 'free fall' that I thought of at first.
 
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  • #8
My bad...I mean the center of length of the rod.
 
  • #9
UMath1 said:
Then if the left side were denser, there would be a net counterclockwise torque?
But if everything on the object is accelerating at g, there is no moment of torque of any point about any fulcrum you may choose.
accelerationrelative X mass = forcerelative
No relative acceleration produces no force.
 
  • #10
UMath1 said:
My bad...I mean the center of length of the rod.
You cannot take just any point as your axis, if that point is accelerating. (Unless it is the center of mass.) Is there a net torque about the center of the rod? Sure. Does that net torque lead to an angular acceleration? No.
 
  • #11
so why is that you need a pivot or fixed point for angular acceleration to occur? Or do you? In the case of a ball rolling, its axis of rotation is not fixed.
 
  • #12
UMath1 said:
so why is that you need a pivot or fixed point for angular acceleration to occur? Or do you?
You are making the assumption that torque about some axis always equals ##I\alpha##. That's a reasonable assumption in intro physics, but it turns out not to be true except under special conditions. It's a topic covered in intermediate classical mechanics. (In intro physics just about all of the problems meet those special conditions, so you don't have to worry about it. Unless you're thinking ahead!)

UMath1 said:
In the case of a ball rolling, its axis of rotation is not fixed.
One of those 'special conditions' is when the chosen point is the center of mass. In that case you can deduce that a net torque about such an axis does lead to an angular acceleration.
 
  • #13
That actually helps answer another one of my questions. I was wondering why a cylinder wouldn't roll on a frictionless incline if gravity provided a torque if the axis were considered the point of contact.

Can you send me material that goes over this topic?
 
  • #14
UMath1 said:
Can you send me material that goes over this topic?
This is the sort of thing covered in just about any classical mechanics text. (If I find something online, I'll post a link.)
 
  • #15
UMath1 said:
I was wondering why a cylinder wouldn't roll on a frictionless incline
this is a bit off topic, but short answer - without friction there is no torque on the cylinder. it would slide down. the orientation doesn't matter.
 
  • #16
No but if you consider the axis of rotation to be the point of contact with the ground, gravity does produce torque.
 
  • #17
UMath1 said:
No but if you consider the axis of rotation to be the point of contact with the ground, gravity does produce torque.
And that point of contact is accelerating, so things are a bit different when using that axis. The end result is the same, of course: No rotational acceleration about the center of mass.
 
  • #18
The center of mass also accelerates
 
  • #19
UMath1 said:
The center of mass also accelerates
The center of mass has special properties. It doesn't matter if that point accelerates or not.
 
  • #20
Right..thats what I want to know about. A text that goes over these special properties and the requirements for rotation.
 
  • #22
UMath1, the CM coordinate is the result of an averaging. Ie for the gorisontal located linear body:
[itex]x_c =\frac{1}{L} \int_0^L \rho _x (x)dx[/itex]
Suppose we have an inhomogeneous gorisontal rod. Obviously its center mass is not located in the middle. Let us mentally divide the rod through the center of mass by the the vertical plane. We will get the unequal geometrical parts whith an equal masses. Where will be the center mass each of them situated?
As I can see, your avatar tells that math is the thing you understand well.
So You can easily prove that the centers of mass of two geometricaly unequal parts obtained by the such dividing are situated at equal distances from the general CM (where the dividing plane located). So why have the torque be when we have two an equal masses located on equal distances from CM? That looks like balanced weighing-machine does not it?
 
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Related to Exploring Torque and Angular Acceleration in Nonuniform Free Fall

1. What is torque?

Torque is a rotational force that causes an object to rotate around an axis. It is a vector quantity, meaning it has both magnitude and direction.

2. How is torque different from force?

While force is a linear or straight-line push or pull on an object, torque is a rotational force. It is the product of force and the distance between the force and the axis of rotation.

3. How does torque affect an object's angular acceleration in nonuniform free fall?

In nonuniform free fall, an object experiences a constant acceleration as it falls due to the force of gravity. Torque acts on the object to cause it to rotate as it falls, and this rotation affects the object's angular acceleration.

4. What factors affect the torque experienced by an object in nonuniform free fall?

The magnitude of the force, the distance between the force and the axis of rotation, and the angle at which the force is applied all affect the torque experienced by an object in nonuniform free fall.

5. How can we calculate torque in nonuniform free fall?

To calculate torque in nonuniform free fall, we can use the equation T = F * r * sin(theta), where T is torque, F is the force applied, r is the distance between the force and the axis of rotation, and theta is the angle between the force and the lever arm.

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