Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Three identical thin rods each of length l and mass M one joined to form a letter H. If the H so formed is followed to fall from rest so as to allow it to rotate about one of the side, then what will be its angular velocity, when the plane of H is vertical..?

Three identical thin rods each of length l and mass M one joined to form a letter H. If the H so formed is followed to fall from rest so as to allow it to rotate about one of the side, then what will be its angular velocity, when the plane of H is vertical..?

Grade:12th pass

10 Answers

Vikas TU
14149 Points
7 years ago
mgl/2 + mgl = 1/2 x (0 + ml^2/3 + ml^2) x w^2 solve it and get w (angular velocity.
Vikas TU
14149 Points
7 years ago
plz approve!
karunyaa
10 Points
7 years ago
sorry couldnt get it can you plz explain it briefly
Harsh davda
41 Points
7 years ago
FIRST OF ALL, moment of inertia about one of its side will be the addition of moment of inertia of each rod about the side.. hence now moment of inertia about the first rod about its own axis is zero because its radius is zero.. now MOI of the second rod about its end is Ml²/3 and that of the third rod about the first is Ml² hence moment of inertia of the body is 4Ml²/3 ..... now torque of gravitational force on the body is R*F=Mgl/2..... weknow that... T=?a hence Mgl/2=4Ml²/3 a ...... which gives a=3g/8l..... now as a=vdv/ds .....,, a=?d?/d?..... hence... ?d?=(3g/8l).d? ....integrating both the sides..., we get ... ?²/2=3g?/8l.....putting limits.. ?²/2=3g?/16l ...... ?=v(3g?/8l)... whichis ur answer thanks, regards, HD.
Harsh davda
41 Points
7 years ago
FIRST OF ALL, moment of inertia about one of its side will be the addition of moment of inertia of each rod about the side.. hence now moment of inertia about the first rod about its own axis is zero because its radius is zero.. now MOI of the second rod about its end is Ml²/3 and that of the third rod about the first is Ml² hence moment of inertia of the body is 4Ml²/3 ..... now torque of gravitational force on the body is R*F=Mgl/2..... weknow that... T=Ia hence Mgl/2=4Ml²/3 a ...... which gives a=3g/8l..... now as a=vdv/ds .....,, a=wdw/d§..... hence... wdw=(3g/8l).d§ ....integrating both the sides..., we get ... w²/2=3g§/8l.....putting limits.. w²/2=3g(pi)/16l ...... w=v(3g(pi)/8l)... whichis ur answer thanks, regards, HD
karunyaa
10 Points
7 years ago
thank you so much noe i am clear
karunyaa
10 Points
7 years ago
but how come the limit is pi/2?
Harsh davda
41 Points
7 years ago
it was initially horizontal... finnaly became vertical..!
karunyaa
10 Points
7 years ago
oh thankz
Ashish Garg
34 Points
5 years ago
Three identical thin roda each of length l and mass M are joined together to form a letter H. The moment of inertia of the system about one of the outer sides of H is :-

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free