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)
 
						

							a=((1+h/R)F)/mβPut h=0       F=mg sinθ       β=(1+ (k^2/R^2))where, k=radius of gyration.. Here solid sphere=> (k^2/r^2)=2/5=>β=7/5So finally eqn stands as:a=[ {(1+(0/R)}*mg sinθ]/[m*(7/5)] = (5g sin 30)/ 7 =5g/14
						

							
 an object of mass 5 kg is sliding down on an inclient plane at an angle of 30 degree without an acceration now if the inclination is increased to 45 degree then findd acceration in the object?
						

							
5gsin30°- ∆5gcos30°=0  (as ,accln =0 & ∆= friction)
 So ∆= 1/√3----- by solving
 
       (OR as  accln =0  tan30° = ∆)
 
 Now as theta = 45° 
 
Then acceleration is
 
5gsin45°- ∆5gcos45°=5a
 By solving
10/√2 - 10/√2×1/√3 =a
 
So answer is a ~= 1.732