Question

☢☢the ball is rolling without slipping on incline plane , having the angle of inclination theta.☢☢Find :-
1). its acceleration.
2). frictional force.
3). the minimum value of Coefficient friction so the sphere rolls without slip.

Answer 1

hey there!



From conservation of mechanical energy at top and bottom point of inclined plane:

 →  mgh = 1/2 mv² ( 1+ k²/ r²)

hence, v =$\sqrt{2gh/1+ k ^{2}/r ^{2} }$

also, v²=u² +2as

a= v² /2s
a= 2gh/ (1+ k²/r²) 2s

now, h/s =sinQ (qetha)

hence, a =gsinQ/ 1+k²/r²

now, as its a sphere therefore, k²/r² = 2/5

hence, a= 5/7 gsinQ


-frictional force is given by: ma * k²/r² 

therefore, f = m* 5/7 gsinQ * 2/5

∴ f= 2/7 mg sinQ



- for  the sphere to roll without slip, limiting friction  ($f_{l}$) should be greater than the applied frictonal force :


f   ≤   $f_{l}$

2/7mgsinQ ≤ ∪R 

 2/7gsinQ ≤ ∪gcosQ                            (R=mgcosQ ; ∪ =coeffecient of friction)

2/7tanQ ≤ ∪

hence, $u_{min}$ = 2/7 tanQ


# HOPE THE ANSWER HELPS!

Answer 2

From conservation of mechanical energy at top and bottom point of inclined plane:

→ mgh = 1/2 mv² ( 1+ k²/ r²)

hence, v = $\sqrt{2gh/1+ k ^{2}/r ^{2} }$

also, v²=u² +2as

a= v² /2s

a= 2gh/ (1+ k²/r²) 2s

now, h/s =sinQ (qetha)

hence, a =gsinQ/ 1+k²/r²

now, as its a sphere therefore, k²/r² = 2/5

hence, a= 5/7 gsinQ

-frictional force is given by: ma * k²/r²

therefore, f = m* 5/7 gsinQ * 2/5

∴ f= 2/7 mg sinQ

- for the sphere to roll without slip, limiting friction $f_{ll$should be greater than the applied frictonal force :

f ≤ $f_{l}$

2/7mgsinQ ≤ ∪R

2/7gsinQ ≤ ∪gcosQ (R=mgcosQ ; ∪ =coeffecient of friction)

2/7tanQ, ≤ ∪

hence, $u_{min}$

= 2/7 tanQ