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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:




9 years ago

Answers : (6)

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


putting this value of f in eqn 1



put the value of I=2MR^2/5

9 years ago
snehashish bal
21 Points

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

9 years ago
vikas askiitian expert
509 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)


9 years ago
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
2 years ago
21 Points

 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?

2 years ago
15 Points
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
one year ago
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