## anonymous 4 years ago a thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. its maximum angular speed is w(omega). its centre of mass rises to a max. height of?

1. anonymous

options are $1) 1l ^{2}\omega ^{2}\div3g$ $2) 1l \omega \div6g$ $3) 1l ^{2}\omega ^{2}\div2g$ $4) 1l ^{2}\omega ^{2}\div6g$

2. anonymous

it comes in my xam and this qn move above from my head sso i didnt solve this :P and asking u guys how to solve

3. anonymous

eve i have solution too but need sm1 to explain may i post the solution?

4. anonymous

i think this fig and think we use consevation of E

5. anonymous

yes

6. anonymous

|dw:1329508279870:dw|

7. anonymous

i get C.M point for source point of E

8. anonymous

this qn is so bad i hate such type qn really :/

9. anonymous

i don't find solve from options

10. anonymous

sory say that

11. anonymous

np :)

12. anonymous

loss in KE=gain in PE 1/2 Iw^2=mgh 1/2x Ml^3/3 x w^2=mgh h=l^2w^2/6g

13. anonymous

i dont get the lat 2 steps

14. anonymous

this formula is right but i can't understand why don't get result $0.5mv^2+0.5I \omega^2=mgl 0.5(1-\cos \theta)$

15. anonymous

ohhhhhhh thats right i did a big mistake

16. anonymous

rod had a net rotational energy so the 2nd statement in left hand's of my equation is zero

17. anonymous

and in right hand (mgl/2)cos theta=h and I =(1/3)ml^2 hence moment of inertia by this axis equal to$I=I _{C.M}+Ma ^{2}$ where "a" is perpendicular distance between that axis and C.M axis I(C.M)for rod is (1/12)ml^2

18. anonymous

do you get it?

19. anonymous

I is moment of inertia around arbitrary axis.

20. anonymous

you'r made mistake in typing last realtion,i think this:$(1/2)I \omega^2=mgh$ in gain point rod has only PE, hence in that point "omega"=0