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AravindG
 4 years ago
3:hw do we get acceleration=g*sin theta/(1+k^2+r^2) in the case of a body rolling down an inclined plane?
AravindG
 4 years ago
3:hw do we get acceleration=g*sin theta/(1+k^2+r^2) in the case of a body rolling down an inclined plane?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328884753571:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0athu arayile vere endhenjil data given?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1it is rolling and i think we use work energy yto get that result

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0acceleration of rolling body is equal to acceleration of com and tangential acceleration ( http://cnx.org/content/m14384/latest/) then accelerartion of com= gsin@ provided by gravitational force along the slope so net acceleration of the body=I*angular acceleration angular acc*r=tangentail acceleration there fore we get ang. acc=gsin@/r i dont know how to proceedd after this arvind

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328886048559:dw \[mg \sin \theta  F_{friction} = ma\] \[F = I \alpha\] (let I = m\[K^{2}\] ) also for pure rolling, \[\alpha = a/R\] puttin above eqn in first one \[mg \sin \theta  mK^{2} \times a/r = ma\] solve for a...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did u bring in frictional force and how is it equal to the angular acceleration of the body also (tangential acceleration of sphere is given as the net accelrartion of sphere how?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0arey, frictional force will be thr to increase the angular acc.of the rolling object to sustain pure rolling... n i didn't equate frictional force to angular acc.? i wrote F equals I (moment f inertia) times alpha.. where alpha equals linear acc divided by radius.. BTW its nt sphere.. this is a general case.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ok how did that linear acceleration be the net accelearation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do u mean , net acc.??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u wrote m*a=mgsin@mK^2*a/r where ma represents net force right with linear acceleartion

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea,by linear acc. i meant net acc..
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