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# 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?

10 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

10 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)

4 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
4 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?

3 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

one year 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
one year 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)