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Answer to Question #40671 in Mechanics | Relativity for Sanjeev

Question #40671
A WET OPEN UMBRELLA IS HELD VERTICAL AND IT IS WHIRLED ABOUT THE HANDLE AT A UNIFORM RATE OF 21 REV. IN 44s. IF THE RIM OF THE UMBRELLA IS A CIRCLE OF 1m IN DIAMETER AND THE HEIGHT OF THE RIM ABOVE THE FLOOR IS 4.9m. THE LOCUS OF THE DROP IS A CIRCLE OF RADIUS :
1. (2.5)^1/2 m
2. 1 m
3. 3 m
4. 1.5 m
Expert's answer
Find the time of drop falling:
g* (t^2) /2 = 4.9
t = sqrt(2 *4.9 /g)
Find the normal accelration:
w = 21 * 2 * pi / 44
an = R * w^2 = 9
Find the circle radius:
R = 1 + an * (t^2) /2 = 1 + an * 4.9 /g = 5.5 (m)

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01.04.14, 15:47

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31.03.14, 20:06

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