Rate of Increase in Water Level Due to Falling Cone in Cylinder

In summary, the conversation discusses a problem involving a cylinder partially filled with water and an inverted cone falling into it with a constant speed. The task is to find the rate of increase of water level when half the height of the cone is submerged in water. After presenting relevant equations and attempting a solution, it is found that the correct answer is 2 cm/s, not 1.875 cm/s as originally thought. The mistake was in assuming that the downward velocity of the cone was equal to dy/dt, when in fact it is (dy/dt)-(dh/dt).
  • #1
Priyadarshi Raj
8
1

Homework Statement


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A cylinder of radius ##{R}## is partially filled with water. There is an inverted cone of cone angle 90° and base radius ##{\frac{R}{2}}## which is falling in it with a constant speed ##v=30~cm/s##.
Find the rate of increase of water level (in ##cm/s##) when half the height of cone is immersed in water.

The given answer is: ##2~cm/s##

Homework Equations


Let the height of the cone be h, and height of cylinder be H
Volume of cone ## = \frac{1}{3}\pi (\frac{R}{2})^2 h = \frac{1}{12} \pi R^2 h##
Volume of cylinder = ##\pi R^2 H##

The Attempt at a Solution


Now for the cone ##\frac{dh}{dt}=v=30~cm/s##

Let the increase in cylinder's water level be ##dH## when ##dV## of cone's volume is immersed in it.
So
##dV = \pi R^2 dH##

⇒ ## \frac{dH}{dt} = \frac{\frac{dV}{dt}}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{\frac{d}{dt}(\frac{1}{12} \pi R^2 h)}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{\frac{1}{12} \pi R^2~\frac{dh}{dt}}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{1}{12} × 30 = 2.5~cm/s##

Please help.
Thank you.
 
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  • #2
Volume of the part of the cone that is immersed depends on h in a way that differs from what you use... The r of that part of the cone is dependent on h...
 
  • #3
@BvU ,@Chestermiller ,do you agree with the given answer 2 cm/s ? I believe it is incorrect . It should be 1.875 cm/s .

2cm/s would have been correct if the cone was falling at the speed of 32 cm/s .
 
Last edited:
  • #4
Vibhor said:
@BvU ,@Chestermiller ,do you agree with the given answer 2 cm/s ? I believe it is incorrect . It should be 1.875 cm/s .

2cm/s would have been correct if the cone was falling at the speed of 32 cm/s .
I get 2 cm/s.

Let z be the distance between the tip of the cone and the bottom of the cylinder, and let h(z) be the distance between the water surface and the bottom of the cylinder. Then the amount of cone submerged is h - z, and the volume of cone submerged is $$V_s=\frac{\pi (h-z)^3}{3}$$. The total volume of water and cone below the water surface is ##\pi R^2h##, and this is also equal to the volume of water Vw plus the submerged volume of cone:

$$V_w+\frac{\pi (h-z)^3}{3}=\pi R^2h$$
Taking the time derivative of this equation gives:
$$(h-z)^2\left(\frac{dh}{dt}-\frac{dz}{dt}\right)=R^2\frac{dh}{dt}$$
 
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  • #5
Hi ,

Thanks for replying.

Let y be the distance between the tip of the cone and the water surface, and let h be the distance between the water surface and the bottom of the cylinder.

$$V_w =\pi R^2h - \frac{\pi y^3}{3}$$

Taking the time derivative of this equation gives: $$R^2\frac{dh}{dt} = y^2\frac{dy}{dt}$$ .

When ##y=\frac{R}{4}## , ##\frac{dh}{dt} = \frac{30}{16}##

What is the mistake ?
 
  • #6
Vibhor said:
Hi ,

Thanks for replying.

Let y be the distance between the tip of the cone and the water surface, and let h be the distance between the water surface and the bottom of the cylinder.

$$V_w =\pi R^2h - \frac{\pi y^3}{3}$$

Taking the time derivative of this equation gives: $$R^2\frac{dh}{dt} = y^2\frac{dy}{dt}$$ .

When ##y=\frac{R}{4}## , ##\frac{dh}{dt} = \frac{30}{16}##

What is the mistake ?
The downward velocity of the cone is not dy/dt. It is (dy/dt)-(dh/dt).
 
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  • #7
Chestermiller said:
The downward velocity of the cone is not dy/dt. It is (dy/dt)-(dh/dt).
:doh:

Thanks a lot :smile:
 

Related to Rate of Increase in Water Level Due to Falling Cone in Cylinder

1. What is the rate of rise in water level?

The rate of rise in water level refers to the speed at which the water level in a body of water is increasing over time. This can be measured in various units, such as feet per hour or millimeters per day.

2. What factors can affect the rate of rise in water level?

There are several factors that can influence the rate of rise in water level, including precipitation, evaporation, inflow from surrounding bodies of water, and human activities such as damming or water extraction.

3. How is the rate of rise in water level measured?

The rate of rise in water level is typically measured using water level gauges and sensors that record changes in water level over time. These measurements can then be used to calculate the rate of rise in various units.

4. Can the rate of rise in water level be predicted?

Yes, the rate of rise in water level can be predicted using various models and data analysis techniques. However, it is important to note that natural phenomena and human activities can cause unexpected changes in the rate of rise.

5. Why is the rate of rise in water level important?

The rate of rise in water level is important for various reasons, including monitoring potential flooding or drought conditions, managing water resources, and understanding the impacts of climate change on bodies of water.

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