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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:




8 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

8 years ago
snehashish bal
21 Points

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

8 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)


8 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
one year 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?

one year 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
4 months ago
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