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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.`
11 months ago

Arun
22613 Points
```							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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11 months ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions