Mechanics> Sphere rolling down plane...

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?

Gundoos 1234 , 14 Years ago

Grade 11

8 Answers

Fawz Naim

Last Activity: 14 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

Approved

snehashish bal

Last Activity: 14 Years ago

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

Approved

vikas askiitian expert

Last Activity: 14 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)

Approved

Stella

Last Activity: 8 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

Last Activity: 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

Last Activity: 6 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

Last Activity: 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

Last Activity: 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)

I Hope this answer will help you.
Thanks & Regards
Yash Chourasiya