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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? 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?
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?
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
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
there is a nice derivation of this formula in resnick halliday and krane......this is a proved result.....
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)
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)
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
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?
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 solving10/√2 - 10/√2×1/√3 =a So answer is a ~= 1.732
Dear StudentNet force on c.o.m (center of mass) = mg.sinФIn this case friction force is acting upwards.For linear motion, mg.sinФ - f = ma .......1for angular motion fR = Iλ ........2 from 1 & 2mg.sinФ = ma + Iλ/R ...........3the 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 & RegardsYash Chourasiya
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