If friction is present between the surface of ball and
bowl then find out the range of co for which ball does
not slip (u is the friction coefficient)
Friction develop a range of co for which the particle will
be at rest.
Answers
Explanation:
When the bowl rotates, there will be a range of angular speed for which the block will not move.
When the angular speed of the bowl is minimum in this range, the block will have a tendency to roll down. So, the frictional force will act in the upward direction along the surface of the bowl.
When the angular speed of the bowl is maximum in this range, the block will have a tendency to roll up. So, the frictional force will act in the downward direction along the surface of the bowl.
We only need to consider the second case here.
Lets say the block moves in a horizontal circle with the centre at C as shown in diagram so that the radius is PC=OPsin=rsinθ
Its acceleration is therefore ω
2
rsinθ. Resolving the forces along PC and applying Newton's second law,
Nsinθ+μNcosθ=mω
2
rsinθ --------(1)
As there is no vertical acceleration,
Ncosθ−μNsinθ=mg
⇒N=
(cosθ−sinθ)
mg
---------(2)
Replacing the value of N from equation (2) in (1)
(cosθ−μsinθ)
mg(sinθ+μcosθ)
=mω
2
rsinθ
⇒ω=[
rsinθ(cosθ−μsinθ)
g(sinθ+μcosθ)
]
1/2
solution
Answer:
sorry I don't know the answer