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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?

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9 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
```
9 years ago
```							there is a nice  derivation of this formula in resnick halliday and krane......this is a proved result.....
```
9 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)

```
9 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
```
3 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 solving10/√2 - 10/√2×1/√3 =a So answer is a ~= 1.732
```
2 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 solving10/√2 - 10/√2×1/√3 =a So answer is a ~= 1.732
```
5 months ago Yash Chourasiya
246 Points
```							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
```
4 months ago
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