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Mg(2R)=(1/2)(MR2/2)ω2+(1/2)M(ωR)2
solving this we get
ω28g/3R
ω=√8g/3R
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Or You can find moment of inertia of system about hinge and then apply (1/2)Ipω2
Ip=(MR2/2+MR2) = 3MR2/2
then Mg(2R)=(1/2)Ipω2
2MgR=3/4MR2ω2
ω=√8g/3R
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1Abhishek my problem is that in the first method you gave , why we have taken the angular velocity of mass M and the disc to be same
11Why aren't they the same, karan?
33
Abhishek Priyam
·2009-02-03 23:43:46
if ω is angular speed of disc then velocity of any particle on its circumference..(mass M is also there..) is ωR..
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karan9989 gupta
·2009-02-04 16:37:14
please can you explain giving some example or some concept that why the angular velocities are same
33
Abhishek Priyam
·2009-02-04 21:00:05
Consider a disc rotating with ang speed ω about com.. vith vcom=0
velocity of a particle at a distance R/2 from center is............
velocity of a particle at a distance R from center is............
or consider a rod rotating with ang speed ω then speed of particle halfway the rod is....... (if rod of length l is hinged at one end...)
If U know answers to these u know the one u asked...