A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is µ. Find the range of the angular speed for which the block will not slip.
Answers
Refers to the attachment for the answer.
We know for the maximum velocity, the block will try to move upwards as a result friction will be acting downwards.
From attachment,
N Cosθ = F Sinθ + mg.
N Cosθ = μN Sinθ + mg.
N (Cosθ - μSin θ) = mg.
⇒ N = mg/(Cosθ - μSin θ)
Second equation,
N Sinθ + μNCosθ = mω²r Sinθ
N (Sin θ + μCosθ) = mω²r Sinθ
Substituting the above values of N in this,
We will get,
mg/(Cosθ - μSin θ) × (Sin θ + μCosθ) = mω²r Sinθ
⇒ ω² = g(Sin θ + μCosθ)/rSinθ (Cosθ - μSin θ)
∴ ωmax. = √[g(Sin θ + μCosθ)/rSinθ (Cosθ - μSin θ)]
Now, Similarly, ωmin. can be find. For easy calculations, Substitute the Plus sign instead of minus sign and vice versa in max. so that we can get the min. angular velocity.
∴ ωmin. = √[g(Sin θ - μCosθ)/rSinθ (Cosθ + μSin θ)]
Note, This method of sign change in not the good method, if you are studying in 11th or 12th. But will be good for competitions. I means that you need to learn the concept. I have already done one part, you can follow the same steps, you will get the same result.
Hope it helps.
Answer. I hope it helps you
