|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 23:21:06 IST
|
|
|
a uniform rod of lenght l,hinged at the loswer end is free to rotate in the vertical plane.if the rod is held vertically in the beginning and thn released , the angular accelaration of the rod whn it makes an angle of 45 wid the horizontal is: rates assured.....
|
|
|
|
28 Nov 2007 23:28:00 IST
|
|
|
well MI of rod abt hinge = mL^2/3
T =Ia
mgLsin(theta)/2 = mL^2/3 * a { a---> ang acceleration, sin component acts at com of rod so put L/2 instead of L}
a = 3gsin(theta)/2L
theta =45
a = 3g/2L*sqrt(2)
|
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: 2 points
(with 0 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Nov 2007 23:30:53 IST
|
|
|
but the answer given is 3g/2sqrt 2 l
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Nov 2007 23:37:48 IST
|
|
|
T=Ia
I=mL^2/3
T=mg*L/2*cos45
=mgL/2root(2)
a=T/l
=3g/2sqrt 2 l
|
science-
the most fundamental
the most eternal
|
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Nov 2007 23:44:53 IST
|
|
|
the mean kinetic energy of a particle of mass m with constant force in any interval of time will be......whr w1 and w2 are the initial and final velocities
|
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}
