Circular motion formulas, what does the angle without the function mean?

AI Thread Summary
The discussion centers on the interpretation of an angle without a trigonometric function in circular motion formulas, specifically referring to delta theta. Users clarify that this angle can be expressed in either degrees or radians, but consistency in units is crucial throughout calculations. For example, if the angle represents a quarter of a circle, it can be denoted as Δθ = π/2 in radians. While radians are typically preferred, degrees can also be used as long as all related measurements, like angular acceleration, are compatible. Understanding this flexibility helps in applying circular motion formulas correctly.
Femme_physics
Gold Member
Messages
2,548
Reaction score
1
http://img864.imageshack.us/img864/9467/tnua.jpg

I see an angle without a trigonometric function (delta theta - at the left side equations). I've never seen it before in a formula, and am not sure how to use it. Can anyone tell me how? Do I just translate degrees to radians and just plug the number I get in radians?
 
Last edited by a moderator:
Physics news on Phys.org
Yes, typically you see everything in radians. So if you went through a quarter of a circle, you'd have \Delta \theta = {{\pi} \over{2}}

You could use degrees as long as everything is consistent (such as the angular acceleration being in degrees/sec for example).
 
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