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Gundoos 1234 Grade: 11
        

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?

6 years ago

Answers : (4)

Fawz Naim
37 Points
										

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

6 years ago
snehashish bal
21 Points
										

there is a nice  derivation of this formula in resnick halliday and krane......this is a proved result.....

6 years ago
vikas askiitian expert
510 Points
										

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)


 

6 years ago
Stella
11 Points
										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
										
3 months ago
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