33b)
ωR is the velocity of COM before leaving contact...
This is the IIT-JEE 1995 question. So U guys must have seen it. Not a very tough question. But it gives a lot of clarity in concepts and idea of how to apporoach such questions.
A rectangular rigid fixed block has a long horizontal edge. A solid homogenous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in the figure. There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine:
a) The angle θc through which the cylinder rotates before it leaves contact with the edge,
b) The speed of the centre of mass of the cylinder before leaving contact with the edge
c) The ratio of the translation to rotational kinetic energies of the cylinder with its centre of mass is in horizontal liine with the edge.
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16 Answers
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.
33c)
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...)
33That 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....
333rd question naa? i also thought it has mistake, i pointed that in above post....
1yup 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
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...
33After 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......
1i think the cylinder has acquired velocity due to loss in PE. SO we dont want to cosider both (while calculating translatiomal KE)
1'both' means? after leaving point of contact,,,, how can the cylinder at all rotate? ... it will only fall down naa ? did u get....