Rotation of free to rotate rigid bodies I recently came across this statement in a book that the line joining the point of suspension and center of mass of a body should be parallel to the acceleration due to gravity(g or more precisely, g effective). Why is this so? Can anyone please explain this to me?

Emma Hobbs

Emma Hobbs

Answered question

2022-11-18

Rotation of free to rotate rigid bodies
I recently came across this statement in a book that the line joining the point of suspension and center of mass of a body should be parallel to the acceleration due to gravity(g or more precisely, g effective). Why is this so? Can anyone please explain this to me?

Answer & Explanation

Antwan Wiley

Antwan Wiley

Beginner2022-11-19Added 13 answers

Recall that we adjust other Forces or acceleration effect in g effective. So there are only 2 External Forces on the rigid body, in the frame of body.
1.Tension due to string
2.Effective Gravitational force
Now since tension always applies along the string, it will align with effective g during equilibrium.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Force, Motion and Energy

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?