Mechanics - ponder over this confirm its correctness .

1

Rohan Ghosh ·2009-04-17 07:43:08

hmm i dont think so celestine .

one thing this is a hollow sphere (solid will make things bad)

there is a familiar result -
ICR is the only sole point other than the points in the body from where the w of the body is equal to its actual omega ..

further ICR as it is at rest (further "centre" of rotation )

so we can simply write the velocities of the points as wr where r is the distance and w is the actual omega

but this will not work if we take ICR as the centre

because let us take eg. the point which is not rotating and lying on the sphere

it is that point which comes when we intersect the line perpendicular to velocity and passing through the centre of mass and parallel to the surface with the sphere

observe that according to your theory the velocity must be w(âˆš(r²+r²) = âˆš2wR

but that is not the case as it is travelling with com with velocity wr!!!

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Celestine preetham ·2009-04-22 22:36:39

yes rohan i agree !!! btw pls clear my dbt y are u usin 2 ids ???

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satan92 ·2009-04-22 19:30:21

take the hint from the instantaneous axis of rotation then you might agree with my answer :) as it is the instantaneous axis of rotation it will be able to give the velocities of each point ;

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Celestine preetham ·2009-04-22 09:47:18

yeah i wanted to post 4 wen i came back u ve already said that !!!

still feel only 4 are there

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Rohan Ghosh ·2009-04-22 09:25:32

just 2!!

by observation we straight away get 4!

(the ones i mentioned at a distance âˆš2R and taking the circle which is in the plane of motion and of largest radius we get another two)

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Celestine preetham ·2009-04-22 08:31:25

dont agree can u give a proof ?
btw is this ur own q ???

i m gettin jus 2 pts as ans ( will give mine after seeing urs i dont want to make a fool of myself :) )

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Rohan Ghosh ·2009-04-22 04:57:59

okay here goes the answers ..

1> it will have an instantaneous axis of rotation which passes from the 0 velocity point in touch with the ground and perpendicular to the objects motion

2>it will be like the outline of the wings of a butterfly when both of them are symmetrical w.r.t the straight body of butterfly

angle of their orientation = tan-1(2)
in other words it will be the projection of a cirle on to the sphere
circle has eqn x²+z²=r²

so only a small part of it will project in both directions

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Rohan Ghosh ·2009-04-17 08:12:07

what extra ?

there as the disk is travelling in one plane velocity of each point should be along the plane

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Celestine preetham ·2009-04-17 08:09:44

then also u r using extra info i think

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Rohan Ghosh ·2009-04-17 08:04:26

hmm there its a two dimensional world so only one perpendicular exists ..

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Celestine preetham ·2009-04-17 08:00:30

see # 13

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Celestine preetham ·2009-04-17 07:55:10

yes rohan i was wrong :)

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satan92 ·2009-04-16 08:55:14

why is nobody trying my problems?

this could be concept clearer guys cmon .

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Celestine preetham ·2009-04-16 20:59:16

@rohan
but that is the same case in ring also right
ie infinite vectors possible here also but we decide a unique one wrt the axis of rotation

so ICR alone in any case cant decide direction of velocity

and for ring ICR is pt of contact (must be familiar concept)

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Celestine preetham ·2009-04-16 20:53:36

ans
1 ..... point of contact ( but it changes )
2.......... part of the sphere which is at distance r from pt of contact

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satan92 ·2009-04-16 20:00:44

well dharun let me disclose my point of view ..

ICR if it exists then we should be able to define the velocities of all the points on the rigid body taking ICR the centre simply .

but observe that in case of three dimensional cases like these , there can be infinite vectors perpendicular to any vector joining the point and the ICR . thus then we WONT be able to decide the velocities of the points for which ICR is meant ..