A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 rev in 44 sec . If the rim of the umbrella is a circle 1 metre in diameter and the height of the rim above the floor is 1.5 metre , find where the drops of water spun off the rim hit the floor.

Question

Arun · Answer

The angular speed of the umbrella is 21 x 2π rad / 44s = 3.0 rad/s.
The tangential speed of a drop which leaves contact with the umbrella is 1.0m x 3.0rad/s = 3.0 m/s.
Time to reach ground is sqrt(2g/h) = sqrt(2*9.8/4.9) = 2.0 s.
Distance travelled tangentially until reaching ground = 3.0m/s x 2.0s = 6.0m.
Using Pythagoras, distance from axis of umbrella is then sqrt(1.0^2+6.0^2) = sqrt(37) = 6.08m.
This is the radius of the circle which is the locus of drops from the umbrella.

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A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 rev in 44 sec . If the rim of the umbrella is a circle 1 metre in diameter and the height of the rim above the floor is 1.5 metre , find where the drops of water spun off the rim hit the floor.

A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 rev in 44 sec . If the rim of the umbrella is a circle 1 metre in diameter and the height of the rim above the floor is 1.5 metre , find where the drops of water spun off the rim hit the floor.

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A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 rev in 44 sec . If the rim of the umbrella is a circle 1 metre in diameter and the height of the rim above the floor is 1.5 metre , find where the drops of water spun off the rim hit the floor.

A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 rev in 44 sec . If the rim of the umbrella is a circle 1 metre in diameter and the height of the rim above the floor is 1.5 metre , find where the drops of water spun off the rim hit the floor.

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