Kinetic energy and rigid bodies

AI Thread Summary
Kinetic energy in rigid bodies must specify both the axis of rotation and the point about which the body rotates, as the moment of inertia varies with these parameters. The formula K = ½ * I * w^2 requires clarity on the axis to accurately calculate kinetic energy. Without this specification, the kinetic energy value may be misleading or incorrect. Understanding these details is crucial for proper analysis in physics. Therefore, it is essential to define the axis and point of rotation when discussing kinetic energy.
Niles
Messages
1,834
Reaction score
0

Homework Statement


Does it make sense to talk about kinetic energy without specifying, which axis the body is rotation about? I mean, since K = ½ * I * w^2, I have to say that " ... the kinetic energy of the body around the x-axis is ...", right?
 
Physics news on Phys.org
Niles said:

Homework Statement


Does it make sense to talk about kinetic energy without specifying, which axis the body is rotation about? I mean, since K = ½ * I * w^2, I have to say that " ... the kinetic energy of the body around the x-axis is ...", right?

Not only you have to say around what axis it is rotating but in addition you have to say around what point! In other words, you have to specify completely the axis around which the object rotates (and notice that the value of the moment of inertia depends on what is the axis around which the object revolves).
 
Ok, good - then I have understood it correctly :-)

Thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top