AravindG
  • AravindG
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?
Physics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1328884753571:dw|
AravindG
  • AravindG
so?
AravindG
  • AravindG
|dw:1327934516101:dw|

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AravindG
  • AravindG
i need that
anonymous
  • anonymous
athu arayile vere endhenjil data given?
AravindG
  • AravindG
?
AravindG
  • AravindG
it is rolling and i think we use work energy yto get that result
anonymous
  • anonymous
acceleration 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
  • anonymous
|dw: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
  • anonymous
how 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
  • anonymous
arey, 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
  • anonymous
yeah ok how did that linear acceleration be the net accelearation?
anonymous
  • anonymous
what do u mean , net acc.??
anonymous
  • anonymous
u wrote m*a=mgsin@-mK^2*a/r where ma represents net force right with linear acceleartion
anonymous
  • anonymous
yea,by linear acc. i meant net acc..

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