Question

vikas askiitian expert · Answer

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

nikhil arora · Answer

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

vikas askiitian expert · Answer

eq 1 should be mg - f = ma ..........1 (this is because ,direction of net force on center of mass is downward ...) f = Ia/r 2 mg = 7f/2 f = 2mg/7 =0.2N ans please approve if u like my ans

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