Question

A Hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without sleeping on its surface. If the surface of bowl is smooth and the angle made by the radius through the ball with the vertical is alpha. Find the angular speed at which the ball is rotating.

Answer 1

Answer:

The angular speed at which the ball is rotating is √g/Rcosθ

Explanation:

A Hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without sleeping on its surface. If the surface of bowl is smooth and the angle made by the radius through the ball with the vertical is alpha.

We know,

N cosθ=mg  

N= mg/cosθ...........(i)

N sinθ=mω^2R sinθ ...... (ii)

where

ω=Angular speed

N=Normal reaction

and R=Radius of the Hemispherical bowl.

putting the value of N from (i) at (ii)

mg/cosθ x [sinθ]=mω^2R sinθ

from here we need to find the angular speed ω

mg/cosθ=mω^2R

removing m from both side,

g/cosθ=ω^2R

ω^2=g/Rcosθ

ω=√g/Rcosθ