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nikhil arora Grade: 12
        

1947_24875_IMG_0285.JPG

6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

total change in height = dh =H-R =50cm=0.5m


from work energy theorem,


 change in kinetic energy = total work


  kinetic energy = mgdh


 mv2(1+k2/r2)/2 = mgh


 mv2= 2mgh/(1+k2/r2)    .....................1


k is radius og gyration & for sherical ball it is Rsqrt(2/5) ...


mv2 = 10mg/7           .................2


now when theta =0 then


   N  = mv2/R


so N = 10mg/7R=10Newton


since the body is not falling off so net friction force should be equal to weight =mg


f = 0.7newton

6 years ago
nikhil arora
54 Points
										

thanks for the reply.... normal reaction is correct but frictional force is wrong by ur answer..... the right answer is 0.2 N 


 


sorry ,i forgot to tell that it is small sphere that is is rolling down the inclined plane ....


 


i did it this way (but failed)


frictional force = f , alpha=angular acceletion


at  point A


f - mg = ma..... (1)


f . r = I (alpha)....(2)


putting f = I (alpha)/r in (1)


I(alpha)/r=m(g+a)


for sphere I=2mr(square)/5 and alpha=a/r


so


2mr(sqr)/5 * a/r * 1/r=m(g+a)


2/5 a= g+a


a=-5/3g


putting this in (1)


i got f=0.46


 


plz remove my confusion


thanks

6 years ago
vikas askiitian expert
510 Points
										

 eq 1 should be


   mg - f = ma ..........1


(this is because ,direction of net force on center of mass is downward ...)


  f = Ia/r2


mg = 7f/2


f = 2mg/7 =0.2N ans


please approve if u like my ans

6 years ago
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