Physics | Mechanics | -askIITians

nikhil arora Grade: 12

7 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

7 years ago

Approved

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

7 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

7 years ago

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