|
GR0177 #91
|
|
|
|
|
Alternate Solutions |
Ning Bao
2008-02-01 08:11:02 | Let R, m,h,g=1. Then you just have
1=1/2*8/7+1/2*I*8/7
1=4/7+4/7*I
I=3/4. |  |
|
|
Comments |
Ning Bao
2008-02-01 08:11:02 | Let R, m,h,g=1. Then you just have
1=1/2*8/7+1/2*I*8/7
1=4/7+4/7*I
I=3/4.
smokwzbroiplytowej
2008-10-21 11:27:20 |
Nice - sure helps save time! Caveat: you really can't set  and  simultaneously, since they are both measures of length.
Here it works rather nicely, because H cancels out as you could tell from the answers.
|
| | neutrino
2007-10-31 03:28:11 | You can make the calculation more easy by noting that the rotational inertia will look like
 ,
with  some constant. Now, you just have to solve:
)
This might save you some time. Because, 1.7 min..., man, that is not much time. |  | Richard
2007-10-30 18:38:31 | A potentially useless note:
The solution given in (E) is found by NOT accounting for the translational kinetic energy.
| | Andresito
2006-03-05 00:00:42 | Yosun, there is a typo on your solution. When you express I, the term MV^2 has a factor of 2 implicitly.
Recall the conservation of energy equation, multiply both terms by 2 and the "only" term that has now a factor of 2 is MgH.
The correct expression should have I=(2 MgH - MV^2) (r/v)^2
Yosun, thank you very much for working out the problems. If you are a girl I send you a big kiss :). I obtained the pdf from the yahoo-group. It also has the typo. |  | keflavich
2005-11-10 22:48:10 | You don't need to consider  as part of the gravitational potential energy. When the cylinder reaches the bottom of the plane, its center of mass will remain at a distance R above the ground.
yosun
2005-11-10 22:54:19 |
keflavich: you're right that the center of mass stays at a height of R above the ground at both points. thanks for the correction.
|
|  |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
