Toppling of cylinder containing fluid

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Homework Help Overview

The problem involves a cylindrical vessel filled with water, which is placed on a rough horizontal surface. The scenario includes a hole in the cylinder through which water exits, and the discussion centers on determining the height of water at which the cylinder will topple. The subject area relates to fluid dynamics and mechanics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Bernoulli's principle to find the velocity of fluid exiting the hole and explores the conditions for toppling based on torque. Some participants question the validity of the original poster's reasoning and the formulation of the question, suggesting that it may allow for multiple correct answers.

Discussion Status

Participants are actively engaging with the original poster's approach, with some expressing agreement and others seeking clarification on the correctness of various options presented in the problem. There is a recognition of multiple interpretations of the question, but no explicit consensus has been reached.

Contextual Notes

There is mention of the problem potentially having multiple answers, which raises questions about the clarity of the problem statement and the assumptions involved in the setup.

Tanya Sharma
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Homework Statement



A light cylindrical vessel of radius r is kept on a rough horizontal surface with sufficient friction so that it cannot slide but can topple. It is filled with water up to a height 2h and a small hole of area a is punched in it so that the water coming out of it falls at the maximum distance from its wall along horizontal surface. Water comes out horizontally from the hole. The value of h for which the cylinder topples is (are)

A)πr3/2a
B)2πr3/a
C)3πr3/2a
D)4πr3/3a

Homework Equations


The Attempt at a Solution



Using Bernoulli's principle ,velocity of the fluid coming out of the hole v= √(2gy) where y be the distance of the hole from the top .

Now since water coming out of it falls at the maximum distance from the cylinder wall along horizontal surface ,we need to maximize the distance S = (√(2gy))(√(2(2h-y)/g) .

From this we get y=h

Water coming out of the hole gives an impulse Fdt = (dm)v to the rest of the fluid ,where dm = ρA(dx)

Fdt = ρA(dx)v
or F = ρAv2

If the block has to topple about the rightmost point P , then there has to be a net torque about that point.

i.e ρAv2h - ρπr2(2h)gr ≥0

or 2ρgAh2≥2ρgπr3h

or h≥πr3/A

But this is none of the given options .

I am not sure if my reasoning and approach is correct .

I would be grateful if somebody could help me with the problem .
 

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Last edited:
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Completely agree with you, but the formulation of the question apparently allows multiple answers...
(a bit corny way of detracting good students, this question!)
 
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So,are all the three (b), (c) and (d) correct ?
 
BvU said:
Completely agree with you, but the formulation of the question apparently allows multiple answers...
(a bit corny way of detracting good students, this question!)

Thanks for the input.
 
Would someone like to give opinion on my work in post#1 ?
 
Apart from that it's excellent ? I would go with XL and circle b, c and d.
If only I could regain my credibility again...
 
Thanks BvU :smile:
 

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