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An inclined plane makes angle 30 with horizontal. a solid sphere rolling down without slipping has linear acceleration equal to:
ans: 5g/14.
They used:
a=Mgsinθ/(M+Icm/R^2)
How?
when a sphere rolls on an inclined plane then force of friction opposes its motion and acts in the direction opposite to the direction of motion. radius of sphere is R and mass of sphere is M. alpha is the angular acc. of the sphere
applying newtons second law of motion
Mgsintheta-f=Ma ................ 1
the net torque produced by the force of frction f*R=I*alpha ............2
I is the moment of inertia of sphere which is equal to 2MR^2/
a=alpha*R=>alpha=a/R .........3
putting value of alpha from eqn 3 in 2
f=Ia/R^2
putting this value of f in eqn 1
Mgsintheta-Ia/R^2=Ma
a=Mgsintheta/(M+I/R^2)
put the value of I=2MR^2/5
there is a nice derivation of this formula in resnick halliday and krane......this is a proved result.....
net force on center of mass = mgsin@
friction force = f (acting upwards)
for linear motion ,
mgsin@ - f = ma ..............1
for angular motion
fR = I(alfa) ........2 (alfa is angular accleration)
from 1 & 2
mgsin@ = ma + I(alfa)/R ...............3
now, if it is given that the body is rolling without slipping then we apply a = (alfa)R
now eq 3 becomes
mgsin@ = ma + 2ma/5 (Isphere = 2mR2/5)
a = 5gsin@/7
=5g/14 (sin30 = 1/2)
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