## 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?

1. anonymous

|dw:1328884753571:dw|

2. AravindG

so?

3. AravindG

|dw:1327934516101:dw|

4. AravindG

i need that

5. anonymous

athu arayile vere endhenjil data given?

6. AravindG

?

7. AravindG

it is rolling and i think we use work energy yto get that result

8. 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

9. 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...

10. 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?

11. 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.

12. anonymous

yeah ok how did that linear acceleration be the net accelearation?

13. anonymous

what do u mean , net acc.??

14. anonymous

u wrote m*a=mgsin@-mK^2*a/r where ma represents net force right with linear acceleartion

15. anonymous

yea,by linear acc. i meant net acc..