A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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?
 2 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?

This Question is Closed

heena
 2 years ago
Best ResponseYou've already chosen the best response.2options are \[1) 1l ^{2}\omega ^{2}\div3g\] \[2) 1l \omega \div6g\] \[3) 1l ^{2}\omega ^{2}\div2g\] \[4) 1l ^{2}\omega ^{2}\div6g\]

heena
 2 years ago
Best ResponseYou've already chosen the best response.2it 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

heena
 2 years ago
Best ResponseYou've already chosen the best response.2eve i have solution too but need sm1 to explain may i post the solution?

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2i think this fig and think we use consevation of E

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2i get C.M point for source point of E

heena
 2 years ago
Best ResponseYou've already chosen the best response.2this qn is so bad i hate such type qn really :/

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2i don't find solve from options

heena
 2 years ago
Best ResponseYou've already chosen the best response.2loss in KE=gain in PE 1/2 Iw^2=mgh 1/2x Ml^3/3 x w^2=mgh h=l^2w^2/6g

heena
 2 years ago
Best ResponseYou've already chosen the best response.2i dont get the lat 2 steps

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2this formula is right but i can't understand why don't get result \[0.5mv^2+0.5I \omega^2=mgl 0.5(1\cos \theta)\]

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2ohhhhhhh thats right i did a big mistake

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2rod had a net rotational energy so the 2nd statement in left hand's of my equation is zero

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2and 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

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2I is moment of inertia around arbitrary axis.

hosein
 2 years ago
Best ResponseYou've already chosen the best response.2you'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
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.