Rotational Equilibrium and Dynamics

AI Thread Summary
Dr. Donald Luttermoser incorrectly equates the center of gravity (C.G.) with the center of mass (C.M.), despite their distinct definitions. The center of gravity is the point where gravitational force acts, which is crucial for calculating torques, especially in 2-D scenarios. While C.G. and C.M. coincide in a uniform gravitational field, they can differ in varying fields. In 3-D cases, a common center of action can be defined only if the net force and torque are perpendicular. The concept of C.G. is deemed less useful in practical applications compared to the center of pressure, particularly in engineering statics.
Messages
19,773
Reaction score
10,726
Author: Dr. Donald Luttermoser of East Tennessee State University
 

Attachments

Physics news on Phys.org
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!

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 \tau and net force \vec{F}=(F_{x},F_{y}) we define the centre of action \vec{r}_{c.a}=(x_{c.a},y_{c.a}) to be the that point with least magnitude that satisfies:
x_{c.a}F_{y}-y_{c.a}F_{x}=\tau
(The origin being the point we compute the torque with respect to, say C.M)

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

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:
\vec{r}_{c.a}=\frac{\vec{F}\times\vec{\tau}}{||\vec{F}||^{2}}

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.
 
Last edited:
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
 
Last edited by a moderator:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top