Question

find the minimum velocity at which water inside the container came out
while rotating it

iit(ppt) physics ​

Answer 1

Answer:

Your potential energy function U(z)U(z) doesn't show at all that the water surface is parabolic. What you need to find is the functional form of the rotating water surface, i.e. the surface height zz as a function of rr in cylindrical coordinates rr and zz. (Because of rotational symmetry ϕϕ is not necessary.) The centrifugal force is

Fcf=mω2r

Fcf=mω2r

and the gravitational force is

Fgrav=−mg

Fgrav=−mg

The water surface is orthogonal to the direction of the resultant force

F⃗ =F⃗ cf+F⃗ grav

F→=F→cf+F→grav

Thus the slope of the water surface is

dz(r)dr=|Fcf||Fgrav|=mω2rmg

dz(r)dr=|Fcf||Fgrav|=mω2rmg

From this we get by integration

z−z0=12gω2r2

z−z0=12gω2r2

Thus we get indeed a parabolic surface in the rotating water in the bucket.

Answer 2

Answer:

Angular speed=√2gh/r, Min. Velocity=√2gh

Explanation:

For water to spill out, y=h,x=r,

h=(Angular speed)^2×r^2/2g

2gh=(Angular speed)^2×r^2

Angular speed=√2gh/r

We know, Velocity=Angular speed×radius

Velocity=[√2gh/r]×r=√2gh

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