Why is the length L divided by 2 in physical pendulum calculations?

  • Thread starter Thread starter calgal260
  • Start date Start date
  • Tags Tags
    Physical Rod
AI Thread Summary
In physical pendulum calculations, the length L is divided by 2 to account for the distance from the pivot point to the center of gravity, which is located at half the length of the rod. This adjustment is necessary because the center of mass influences the pendulum's motion, and using L/2 accurately reflects the distribution of mass. The discussion highlights that for simple systems like a single mass on a string, the full length L is appropriate, while more complex shapes require additional considerations. The confusion arises from misunderstanding how the center of gravity affects the dynamics of the pendulum. Understanding these principles is crucial for accurate calculations in pendulum physics.
calgal260
Messages
8
Reaction score
0
physical pendulum--thin rod

Why do you have to divide the length L by 2 in these cases? The explanation that's provided in my book says "the distance L between the pivot and the center of gravity is .5 the length L."

This makes no sense whatsoever. The length should be the same regardless of where you measure it from, be that from the center of the gravity or not.
 
Physics news on Phys.org


Taking L/2 accounts for where the actual mass is. If you had a single mass on a string, you would use L. Other shapes require more complicated adjustment than the L/2.
 

Thread '1:1 versus 2:1 Mechanical Advantage debate'

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...
View full post »

Thread 'Newton's first law?'

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?
View full post »
Thread 'Beam on an inclined plane'

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...
View full post »

Similar threads

I A rod that falls while rotating from the end of a table
Replies
2
Views
1K
A Modeling the damping of a physical rigid pendulum
Replies
1
Views
2K
How does the inclusion of a massive rod affect the dynamics of a pendulum?
Replies
10
Views
5K
I Lagrangian for pendulum
Replies
11
Views
2K
Compound physical pendulum - max velocity
Replies
19
Views
2K
EMF induced in moving rod in B-field, why is "L" length of wire frame?
Replies
4
Views
743
Spherical Pendulum Motion: Solve the Mystery
Replies
3
Views
2K
Question about torque and center of mass
Replies
6
Views
2K
Forces that act on a physical pendulum
Replies
7
Views
3K
Stress on a spaghetti rod balanced from its middle
Replies
6
Views
1K

Hot Threads

Recent Insights

Back
Top