How to Calculate the Kinetic Energy of an Inhomogeneous Cylinder?

Click For Summary
To calculate the kinetic energy of an inhomogeneous cylinder rolling over a plane, one must consider both translational and rotational motion. The kinetic energy equation incorporates the center of mass velocity and the moment of inertia, leading to the expression T = (ω^2/2)(ma^2 + I + R^2). A key point is that the center of mass experiences circular motion, and the velocity of the center of mass must be accounted for correctly. The discussion highlights the importance of understanding the relationship between translational and rotational dynamics in rolling motion. The problem is complex and is addressed in Landau & Lifgarbagez's classical mechanics book.
fluidistic
Gold Member
Messages
3,931
Reaction score
281

Homework Statement


I think the problem was taken out from Landau&Lifgarbagez's book on classical mechanics.
An inhomogeneous cylinder of radius R rolls over a plane. The mass is distributed in such a way that a principal axis is parallel to the rotational axis of the cylinder and the center of mass is at a distance "a" from the rotational axis. The moment of inertia of the cylinder about the rotational axis is I. Calculate the kinetic energy of the cylinder.

Homework Equations


T=\frac{m v_{CM} ^2}{2}+ \frac{I \omega _c ^2 }{2}.

The Attempt at a Solution


The center of mass suffer from a circular motion of radius a and angular velocity \omega _c.
So that v_{CM}=a \omega _c and thus T=\frac{\omega ^2 _c }{2} (ma^2 + I).
This seems wrong to me because when a tends to 0, my kinetic energy equality tells me that there's only a rotational motion and no translational motion from the center of mass, which I believe it totally wrong.
I don't see what I did wrong though... I'd love some help to figure out what's wrong with what I did. Thanks in advance.
 
Physics news on Phys.org
Rolling means both translational and rotational motion. The centre of symmetry performs translation with velocity Vc. All points of the cylinder translate with Vc and rotate around the axis of symmetry with angular frequency equal Vc/R. The velocity of a point is the resultant of the translation and rotation.

This is not a simple problem...

ehild
 
You are absolutely right, I forgot about considering the translational motion.
If I consider rolling without slipping, I get that v of translation is worth -\omega _c R.
Thus T=\frac{\omega ^2 _c}{2} (ma^2+R^2+I). Is there something I'm missing?
 
this can't be solved like this...the velocity of cm keeps on changing during motion

the question should give the information of initial position of cm
 
I suggest to read Landau&Lifgarbagez's book on classical mechanics.
The problem is presented and solved in that book. The motion of the cylinder is considered pure rotation about the instantaneously fixed axis (the line where the cylinder touches the ground).

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 10 ·
Replies
10
Views
801
  • · Replies 3 ·
Replies
3
Views
2K
Replies
67
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
39
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
335
Views
16K
  • · Replies 8 ·
Replies
8
Views
1K