33Let it rotates by θ,
then by energy conservation:
mgR(1-cosθ)=(1/2)*(3/2)mR2ω2...
for it to just leave surface....
mω2R=mg.cosθ
solving these two equations we get θ.
33
Abhishek Priyam
·2008-11-08 02:45:16
b)
ωR is the velocity of COM before leaving contact...
33
Abhishek Priyam
·2008-11-08 02:48:25
c)
Rot K.E.=(1/2)(1/2)mR2ω2
Translational K.E. = (1/2)m(ωR)2+mg.R.cosθ
After leaving contact Rot K.E. Does not changes
1a) mgR(1-cosθ)=(1/2)(1/2 + 1)mR^2ω^2 = 3/4 mR^2ω^2
=> ω2R2 = 4/3 gR(1-cosθ)
=> v2 = 4/3 gR(1-cosθ) -----------------(1)
at any angle θ, force balancing,,,, N+mv2/R= mgcosθ
wen cylinder leaves contact,,, N=0.... so...
mv2/R= mgcosθc --------------------(2)
from eqn (1) and (2)...
cosθc = 4/7 => θc = cos-14/7.
b) velocity of center of mass while leaving contact ... from prev. part...
v = √[4/3 gR(1-cosθc)]
= √[4/3X 3gr/7]
= 2√(gr/7)
c) in third part,,, rot. K.E. will be same ... 3/4mR2ω2
=3/4 X mgr4/3 X 3/7
=3/7 mgr.
trans K.E. = 1/2 m 4gr/7 + mg r4/7 = 6mgr/7
therefore
ratio of trans/ rot =2. (i am doubtful abt this ans... plz correct if wrong...)
33
Abhishek Priyam
·2008-11-08 08:55:27
That rot K.E. u wrote: 3/4mR2ω2
thats total energy when it just looses contact... not rot K.E.
rot K.E. = (1/2)mR2ω2
translational K.E. (when center is at horizontal)is correct....
33
Abhishek Priyam
·2008-11-08 08:58:18
3rd question naa? i also thought it has mistake, i pointed that in above post....
1
skygirl
·2008-11-08 09:02:31
yup my mistake... :(
c) in third part,,, rot. K.E. will be same ... 1/4 mR2ω2
=1/4 X mgr4/3 X 3/7
=1/7 mgr.
trans K.E. = 1/2 m 4gr/7 + mg r4/7 = 6mgr/7
therefore
ratio of trans/ rot =6.
1in the above how the trans KE comes to sum of two qtys to give 6mgr/7
62i din get u ...
is it that we have ke in X component and y component?
1
skygirl
·2008-11-08 18:12:15
trans K.E. = 1/2 m 4gr/7 + mg r4/7 = 6mgr/7
first part is the KE of the body at the time of leaving contact... the second part is due to loss in height...
33
Abhishek Priyam
·2008-11-08 22:56:08
After loosing contact the rot K.E does not changes as no torque acts on it after(mg passes thorough center).
But that mg changes transl. K.E.........
so transl K.E. change......
1
Sarath George
·2008-11-09 09:13:39
i think the cylinder has acquired velocity due to loss in PE. SO we dont want to cosider both (while calculating translatiomal KE)
1
skygirl
·2008-11-09 17:40:31
'both' means? after leaving point of contact,,,, how can the cylinder at all rotate? ... it will only fall down naa ? did u get....
