|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 19:52:38 IST
|
|
|
L be length of rod
two force acting on it ((dm)g downwards
nd (dm)v^2/r , radially outwards,
by energy conservation
P.E of dm at 60 degree angle = KE at 37 degree + PE at 37 degree
dmgL(1-cos60) = dm v^2/2 + dmgL(1-cos37)
dmv^2/2=dmgL(cos37-cos60)
dmV^2/L = 2dmg( 4/5 - 1/2)
dmV^2/L = 3dmg/5 this is radially outward
now angle between dmg and dmV^2/L is 37 degree
F = sqrt( ((dmg)^2 + (3dmg/5)^2 + 2*(3dmg/5)^dmg*cos37 ))
solve it to get answer
you shud get 1/5* dmg*sqrt(58)
check for calculation mistakes i might have committed a few, :),
|
The heights which great men reached and kept, Were not attained by sudden flight, They, whilst their companions slept, Were toiling upwards in the night.... |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
