Is rotational Work the same as linear Work?

AI Thread Summary
Linear work, defined as F*D, and rotational work, expressed as 1/2Iω², can be equated under certain conditions. If linear work is known, it can be set equal to rotational work in specific scenarios. However, to find angular momentum, additional information such as radius or angular speed is necessary; work and force alone are insufficient. The discussion emphasizes the need for clarity in questions regarding these concepts. Understanding the relationship between linear and rotational work is crucial in physics.
Just_enough
Messages
52
Reaction score
0
Linear work is F*D and rotation is 1/2Iω2, but if a problem as me rotational energy (Rotation worl = KE?) of a wheel and I have the linear work, can I just set Wrotation=Wlinear?
 
Physics news on Phys.org
Please correct the errors in your post, your question is not understandable as written.
 
marcusl said:
Please correct the errors in your post, your question is not understandable as written.
If I have Work1=F*D, can I use that answer to be equal to W2(=1/2Iω2)?
 
Linear and rotational work can be equated, yes.
 
  • Like
Likes Just_enough
russ_watters said:
Linear and rotational work can be equated, yes.
Thanks. Now is it possible to find angular momentum from just work and force? no radius, speed, nor mass are given. If so, how?
 
Just_enough said:
Thanks. Now is it possible to find angular momentum from just work and force? no radius, speed, nor mass are given.
No, you need at least radius or angular speed.
 
  • Like
Likes Just_enough
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