| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Mar 2008 20:18:09 IST
|
|
|
|
|
|
|
26 Mar 2008 20:44:05 IST
|
|
|
in 37..
T = ma
and T/2.R= I *a/R
solve these to get ans...
I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
Q 50 torQe=t t=I  F*r=(mr^2+mr^2) 5*0.25=2mr^2* =1.25/2*0.5*0.025*0.25  =10 rad/s time=10sec =10+0.10*20 =10+2 =12 rad/sec answer
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
26 Mar 2008 21:32:33 IST
|
|
|
Q 69 let l be the lenght of rod m be mass of rod by energy conservation 1/2I 2-0=mg/2(cos 37- cos 60) 1/2m 2/3=mg1/2(4/5-1/2) 2=9g/10l m =mg(1.2 sin 37) =mg(1.2)*3/5 =0.9(g/l) is angular accleleration therefore to find the force acting on tip of particle Fc=centrifugal force =(dm)l 2=0.9(dm)g Ft= tangential force =(dm) l=0.9(dm)g so,total force will be F=  (Fc )2+(Ft) 2 =0.9 ]2(dm)g answer
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Mar 2008 12:20:23 IST
|
|
|
unable to understand this step m=mg(1.2 sin 37) =mg(1.2)*3/5 =0.9(g/l) instead it should be I*=MgSIN37*L WHERE L=1m ND I=ML^2/3 =9/5g NOT WT U HAD WRITTEN
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Mar 2008 12:23:09 IST
|
|
|
|
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}
