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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?

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  1. anonymous
    • 4 years ago
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    |dw:1328884753571:dw|

  2. AravindG
    • 4 years ago
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    so?

  3. AravindG
    • 4 years ago
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    |dw:1327934516101:dw|

  4. AravindG
    • 4 years ago
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    i need that

  5. anonymous
    • 4 years ago
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    athu arayile vere endhenjil data given?

  6. AravindG
    • 4 years ago
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    ?

  7. AravindG
    • 4 years ago
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    it is rolling and i think we use work energy yto get that result

  8. anonymous
    • 4 years ago
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    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

  9. anonymous
    • 4 years ago
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    |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...

  10. anonymous
    • 4 years ago
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    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?

  11. anonymous
    • 4 years ago
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    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.

  12. anonymous
    • 4 years ago
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    yeah ok how did that linear acceleration be the net accelearation?

  13. anonymous
    • 4 years ago
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    what do u mean , net acc.??

  14. anonymous
    • 4 years ago
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    u wrote m*a=mgsin@-mK^2*a/r where ma represents net force right with linear acceleartion

  15. anonymous
    • 4 years ago
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    yea,by linear acc. i meant net acc..

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