Rotational Equilibrium and Dynamics

In summary: In summary, Dr. Donald Luttermoser of East Tennessee State University does a grievous error here, in conflating centre of gravity with centre of mass. The centre of gravity is just where the force of gravity can be considered to ACT, while the centre of mass is the mass of an object, and is different from the definition given by Luttermoser. Additionally, in order to calculate torques, in 3-D, engineers must assume that the net force and torque are perpendicular vectors, and not necessarily coplanar.
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Author: Dr. Donald Luttermoser of East Tennessee State University
 

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Lutermoser does a grievous error here, in conflating centre of gravity with centre of mass.

Although he has the correct definition of centre of gravity (i.e, as the point where we might consider the weight concentrated (if such a point exists)), he sets it equivalent to the centre of mass, which is totally differently defined! :grumpy:

The centre of gravity is just where the force of gravity can be considered to ACT.
Where a force acts, is of course, mainly of importance when computing torques, and for a 2-D situation, in which direction vectors and forces are coplanar, net torque [itex]\tau[/itex] and net force [itex]\vec{F}=(F_{x},F_{y})[/itex] we define the centre of action [itex]\vec{r}_{c.a}=(x_{c.a},y_{c.a})[/itex] to be the that point with least magnitude that satisfies:
[tex]x_{c.a}F_{y}-y_{c.a}F_{x}=\tau[/tex]
(The origin being the point we compute the torque with respect to, say C.M)

This yields:
[tex]\vec{r}_{c.a}=\frac{\tau}{F_{x}^{2}+F_{y}^{2}}(F_{y},-F_{x})[/tex]

For a uniform gravitational field, C.M and C.G. (that is, c.a.) coincides, but not necessarily with a varying field.


Finally, in the 3-D case, we must assume that the net force&torque are perpendicular vectors in order to be able to define a common centre of action.
If that is the case, we have, as above:
[tex]\vec{r}_{c.a}=\frac{\vec{F}\times\vec{\tau}}{||\vec{F}||^{2}}[/tex]

I'd like to close with saying that I don't regard concepts like "centre of gravity" to be particularly useful, in that the positin of C.G. depends on such factors as the orientation of the object and which point we happen to compute the torque with respect to.

In engineering, particularly in STATICS, the concept of "centre of pressure" has proven useful, though.
 
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  • #3
Hyperphysics has a good discussion on moments of inertia.

Rotational-Linear Parallels
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html


And these might be useful

Area Moment of inertia
http://em-ntserver.unl.edu/NEGAHBAN/EM223/note18/note18.htm

Mass moment of inertia
http://em-ntserver.unl.edu/NEGAHBAN/EM223/note19/note19.htm

Proofs of moment of inertia equations
http://homepages.which.net/~paul.hills/Spinningdisks/MOI/MOIproofs.html
 
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What is rotational equilibrium?

Rotational equilibrium is a state where an object is not rotating or changing its angular velocity. This means that the net torque acting on the object is equal to zero.

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied.

How is rotational equilibrium different from translational equilibrium?

Rotational equilibrium refers to the balance of forces that cause an object to rotate, while translational equilibrium refers to the balance of forces that cause an object to move in a straight line at a constant velocity.

What is the role of the center of mass in rotational equilibrium?

The center of mass is the point where an object's mass is evenly distributed. In rotational equilibrium, the center of mass is the point where the net force and net torque act, allowing for a stable and balanced rotational motion.

How is angular momentum related to rotational equilibrium?

Angular momentum is a measure of an object's tendency to continue rotating. In rotational equilibrium, the angular momentum of an object does not change since there is no net torque acting on the object.

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