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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
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
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
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
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
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
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