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Question
It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
- If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. ( g = 10 m/s2)
- A typical rain drop is about 4mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
- Estimate the time required to flatten the drop.
- Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
- Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm.
(Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through !!)
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Solution
Given, height (h) = 1 km = 100 m
g = 10 m/s2
a. Velocity attained by the raindrop in freely falling through a height h.
v = `sqrt(2gh)`
= `sqrt(2 xx 10 xx 1000)`
= `100sqrt(2)` m/s
= `100sqrt(2) xx (60 xx 60)/1000` km/h
= `360sqrt(2)` km/h ≈ 510 km/h
b. Diameter of the drop (d) = 2r = 4 mm
∴ Radius of the drop (r) = 2 mm = 2 × 10–3 m
Mass of a raindrop (m) = V × p
= `4/3 pir^3p`
= `4/3 xx 22/7 xx (2 xx 10^-3)^3 xx 10^3` .....(∵ Density of water = 103 kg/m3)
= 3.4 × 10–5 kg
The momentum of the raindrop (p) = mv
= `3.4 xx 10^-5 xx 100sqrt(2)`
= `4.7 xx 10^-3` kg-m/s
≈ `5 xx 10^-3` kg-m/s
c. Time required to fatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground
`t = d/v`
= `(4 xx 10^3)/(100 sqrt(2))`
= ``0.028 xx 10^-3` s
= `2.8 xx 10^-5` s ≈ 30 ms
d. Force exerted by a raindrop
F = `"Change in momentum"/"Time"`
= `(p - 0)/t`
= `(4.7 xx 10^-3)/(2.8 xx 10^-5)` ≈ 168 N
e. Radius of the umbrella (R) = `1/2` m
∴ Area of the umbrella (A) = `piR^2`
= `22/7 xx (1/2)^2`
= `22/28`
= `11/4 ≈ 0.8` m2
Number of drops striking the umbrella simultaneously with an average separation of 5 cm = `5 xx 10^-2` m
= `0.8/(5 xx 10^-2)^2` = 320
∴ Net force exerted on umbrella = 320 × 168 = 53760 N ≈ 54000 N.
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