A hemispherical bowl of radius 0.1m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity omega. A particle of mass m = 10^(-2) kgplaced inside the bowl also the particle from the bottom of the bowl is h, find the relation between h and omega.

A hemispherical bowl of radius 0.1m is rotated about a vertical axis p

1.	A hemispherical bowl of radius 0.1m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity omega. A particle of mass m = 10^(-2) kgplaced inside the bowl also the particle from the bottom of the bowl is h, find the relation between h and omega.
Answer» SOLUTION :The particle is height h from the bottom of the bowl. With the rotation of the bowl, the particle also rotates about the vertical axis along a circular path of radius r In this situation, the weight mg of the particle of mass m, the CENTRIFUGAL force `m OMEGA^(2)`r and the normal force (R) on the particle by the surface of the bowl KEEP the particle in equilibrium. So,Rsin`theta = m omega^(2) r "" and " " Rcos theta = ` mg `therefore"" tan theta = (omega^(2) r)/(g)` If the radius of the bowl is a , then from we get tan`theta = (r )/(a- h)" " therefore (r)/(a - h)= (omega^(2)r)/(g) " " or, " " a - h = (g)/(omega^(2))` or, `"" h = a - (g)/(omega^(2)) = 0.1 - (9.8)/(omega^(2)) = 0.1 (1 - (98)/(omega^(2)))`m.