- #1
H_man
- 145
- 0
Hi,
I am rather perplexed by the solution to the following problem.
Particles of mass M1 and M3 are fastened to the ends of a light rod having a length l. A bead of mass M2 is free to slide along the rod between M1 and M3. Point p is the center of mass of M1 and M3, not including M2.
I is the moment of inertia of the M1, M3 rod arrangement about an axis perpendicular to the rod and passing through p . All motion is considered in a plane.
Now the translational aspect of the KE is very simple. But it was the rotational part that had me confused.
In my humble opinion we cannot use I and p coz as we add M2 the center of mass changes.
However in the answer for the rot. KE they have
[tex]\frac{1}{2}[/tex] I @theta/@dt + [tex]\frac{1}{2}[/tex] M2[.........]
The thing that bothers me is that surely with the addition of M2, the value of p and hence I would change. So how can they treat the other masses independently from the third?
I hope I have been fairly clear and am not being too silly.
Thanks guys!
I am rather perplexed by the solution to the following problem.
Particles of mass M1 and M3 are fastened to the ends of a light rod having a length l. A bead of mass M2 is free to slide along the rod between M1 and M3. Point p is the center of mass of M1 and M3, not including M2.
I is the moment of inertia of the M1, M3 rod arrangement about an axis perpendicular to the rod and passing through p . All motion is considered in a plane.
Now the translational aspect of the KE is very simple. But it was the rotational part that had me confused.
In my humble opinion we cannot use I and p coz as we add M2 the center of mass changes.
However in the answer for the rot. KE they have
[tex]\frac{1}{2}[/tex] I @theta/@dt + [tex]\frac{1}{2}[/tex] M2[.........]
The thing that bothers me is that surely with the addition of M2, the value of p and hence I would change. So how can they treat the other masses independently from the third?
I hope I have been fairly clear and am not being too silly.
Thanks guys!