☢☢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.
1). its acceleration.
2). frictional force.
3). the minimum value of Coefficient friction so the sphere rolls without slip.
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Hey there!From conservation of mechanical energy at top and bottom point of inclined plane:
→ mgh = 1/2 mv² ( 1+ k²/ r²)
hence, v =
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 ≤
2/7mgsinQ ≤ ∪R
2/7gsinQ ≤ ∪gcosQ (R=mgcosQ ; ∪ =coeffecient of friction)
2/7tanQ ≤ ∪
hence,
# HOPE THE ANSWER HELPS!
From conservation of mechanical energy at top and bottom point of inclined plane:
→ mgh = 1/2 mv² ( 1+ k²/ r²)
hence, v =
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
should be greater than the applied frictonal force :
f ≤
2/7mgsinQ ≤ ∪R
2/7gsinQ ≤ ∪gcosQ (R=mgcosQ ; ∪ =coeffecient of friction)
2/7tanQ, ≤ ∪
hence,
= 2/7 tanQ