# 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

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)

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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## Comments

Assignment Expert01.04.14, 14:47Dear Sanjeev,

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