# 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?

Fawz Naim
37 Points
13 years ago

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

snehashish bal
21 Points
13 years ago

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

509 Points
13 years ago

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)

Stella
11 Points
7 years ago
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
ayush
21 Points
7 years ago

### 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?

Nageshwar
15 Points
5 years ago
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

Krish Gupta
4 years ago

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
Yash Chourasiya
4 years ago
Dear Student

Net force on c.o.m (center of mass) = mg.sinФ

In this case friction force is acting upwards.

For linear motion, mg.sinФ - f = ma .......1

for angular motion

fR = Iλ ........2

from 1 & 2

mg.sinФ = ma + Iλ/R ...........3

the body is rolling without slipping then we apply a = λR

mgsinΦ= ma + 2ma/5 (Isphere = 2mR²/5)

a = 5gsinФ/7

a =5g/14 (sin30 = 1/2)