Old time mills

Old time mills for grinding wheat or corn into flour were often powered by a water wheel. Some water wheels had water falling over paddles, other water wheels dipped their paddles in a moving river, which as it flowed past turned the wheel. One particular mill has a water wheel with radius 3 m 3~\mbox{m} suspended above a river such that the bottom edge of the paddles are just in contact with and move with the river water. If the river flows at 0.5 m/s 0.5~\mbox{m/s} then what is the period of the spinning water wheel?


The answer is 37.7.

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6 solutions

Dipu Joy
Aug 26, 2013

The radius of the wheel ( r ) is 3 m.

The period of the water wheel is the time required to cover the entire circumference of the wheel.

The circumference of the wheel = 2 * pi * r

                                                   = 2 * 3.14  * 3

                                                   = 18.84 m

Speed of the river = .5 m/s

Period of the wheel = time required to cover the cirumference at .5 m/s

                         = 18.81 / .5

                         = 37.68 seconds
Michael Tang
Aug 26, 2013

As the water wheel spins, the bottoms of the paddles trace out a circle with radius 3 , 3, so every paddle travels exactly 2 π r = 2 π 3 = 6 π m 2\pi r = 2\pi \cdot 3 = 6\pi \; \text{m} in one period. The speed of the river (and so the speed of the paddles) is 0.5 m/s , 0.5 \; \text{m/s}, so the period of the wheel is 6 π m 0.5 m/s = 12 π sec , \dfrac{6\pi \; \text{m}}{0.5 \; \text{m/s}} = 12\pi \; \text{sec}, which is about 37.7 sec . \boxed{37.7}\; \text{sec}.

The speed of water flowing is will be used to rotate the wheel. So liner speed of Wheel = Speed of Water flowing

Here, liner speed of Wheel = Vc, Speed of Water flowing = Vw

So Vc = Vw

In one revolution The Wheel will rotate 360 degree or 2 pi radian.

From the defination of radian angle, angle = s/r

or, s = angle x r , angle = 2 pi , or, s = 2 x pi x r
,pi = 3.1416 , r = 3, s = 2 x 3.1416 x 3 = 18.85 Now, Vc = Vw = 0.5 vc = s/t t = s/ vc t = 18.8 / 0.5= 37.6 seconds

Snehdeep Arora
Sep 1, 2013

omega(angular velocity)=tangential velocity/radius=0.5/3=0.1667 omega * time=theta(in radians) here,theta = 2 * pi time =2 * pi/0.1667=37.7

Michael Tong
Aug 25, 2013

The key to this problem is knowing that the rate at which the wheel turns is equal to the rate of the flowing water, namely 0.5 m / s 0.5 m/s . Thus, the time it takes for the water wheel is equal to the length of its circumference divided by 0.5 m / s 0.5 m/s . So, all we need to do now is plug and chug: C = 2 π r = 6 π P = 6 π 0.5 = 12 π = 37.7 C = 2 \pi r= 6\pi \rightarrow P = \frac {6 \pi}{0.5} = 12 \pi = 37.7 reported with 3 3 significant figures.

Kyle Coughlin
Aug 25, 2013

The radius of the wheel is 3m, so the circumfrence is 6π. You multiply the circumfrence by the reciprocal of the river speed to find the time it takes to complete one rotation. So 12π = 37.699

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