The Ball in Liquid columns Coulumns
A heavy cylinder of radius R lies in equilibrium on a smooth surface, separating two liquids of densities ρ l on the left and ρ r on the right, with ρ r > ρ l .
If the height of the liquid on the right is R , what is the height h of the liquid on the left in terms of R ?
Neglect the surface tensions of the liquids.
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2 solutions
With your solution, are you essentially encapsulating the observation from @Rohit Gupta , given above?
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I had not read his observation, but yes, I believe my solution exemplifies his point.
Thanks for providing a solution. However, I just want to point out that I believe I see an obvious flaw. Applying a purely vertical force to the cylinder at any off-center point would cause the cylinder to roll. It seems the problem is actually much more complicated.
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That is true. However, there are three forces acting on the cylinder:
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the gravitational force, which acts (on average) in the center and produces no torque
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the normal force from the supporting surfaces, which acts on the bottom with a line of force through the center, and produces no torque
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the force from water pressure, which acts perpendicular to the cylinder surface, with a line of force through the center, and produces not torque
For this reason, there is no reason a priori to believe that the cylinder should rotate. If there was a net force in the horizontal direction, static friction might act horizontally on the bottom of the cylinder and cause a torque, so that we get rolling motion. However, the problem states that there is equilibrium, so that there is not net force in horizontal direction, and no static friction is needed to prevent slipping.
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Oh ok, right! It's a "smooth surface." So, the horizontal forces must balance out. So, I guess we don't really care whether it rotates or not. Considering its rotation for a second though... Wouldn't the cylinder be subject to a rotational acceleration?
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@James Wilson – Specifically, it would be driven to rotate counterclockwise, if I'm not mistaken.
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@James Wilson – No, if the net force is zero, the net torque is also zero because all forces (except possibly static friction) act along a line through the central axis of the cylinder. As you said before, you'd have to push "off-center" to cause rotation!
Now, if we'd add too much liquid on the left, the increased water pressure results in a net force to the right. If the bottom of the cylinder and the floor are rough enough, static friction from the ground would prevent slipping by pushing to the left; the cylinder would roll to the right while rotate clockwise.
Or if we'd have too little liquid on the left, the water pressure on the right "wins", causing a force to the left, and static friction would result in rolling motion to the left while rotating counterclockwise.
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@Arjen Vreugdenhil – With all due respect, I believe you are wrong, Sir. The fact/assumption that the forces act along a line through the central axis of the cylinder leads to the conclusion that there are off-center vertical forces acting on the cylinder. If you reflect the liquid on the left that is above the line y = R/2 across the line y = R/2, it is easily seen by cancellation (by symmetry) that the net vertical force on the left side is equal to the force of a lower density liquid (compared with the right side) applying an upward force over a smaller area (than the right side). Thus, the net upward force on the right side is greater than that of the left.
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@James Wilson – Don't forget that (on the left side) the force pushing from above is less than the force pushing from below.
Today I will post a more detailed calculation of the torques due to the vertical force component.
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@Arjen Vreugdenhil – I'm not sure you understood what I was saying. I took that into account already.
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@James Wilson – That's pretty ambitious of you. I would love to see your result.
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@James Wilson – I realize now that you are right: the vertical force components result in a counterclockwise torque on the cylinder. However, the horizontal force components also result in a clockwise torque, and the two balance!
Detailed analysis
Take a coordinate system with ( 0 , 0 ) at the bottom of the cylinder. The height of the liquid (density ρ ) on one side is h . Consider an infinitesimal area element at height y . Then p = ρ g ( h − y ) , x = R 2 − ( y − R ) 2 , d x = x R − y d y . The vertical force component due to the liquid on the area is d F y = p d x d z = ρ g ( h − y ) ( R − y ) / x d y d z , which contributes a torque d τ y = x d F y = ρ g ( h − y ) ( R − y ) d y d z . Integrate this to find τ y = ρ g z ∫ 0 h ( h − y ) ( R − y ) d y = ρ g z [ R h y − 2 1 ( h + R ) y 2 + 3 1 y 3 ] 0 h = 6 1 ρ g z h 2 ( 3 R + h ) .
Now consider the horizontal force component: d F x = p d y d z = ρ g ( h − y ) d y d z . This contributes a torque d τ x = y d F x = ρ g y ( h − y ) d y d z . Integration results in τ x = ρ g z [ 2 1 h y 2 − 3 1 y 3 ] 0 h = 6 1 ρ g z h 3 .
These torques τ x and τ y act in opposite directions, so that the total torque due to the liquid on one side is equal to τ = 6 1 ρ g z h 2 ( 3 R + h ) − 6 1 ρ g z h 3 = 2 1 ρ g z h 2 R . In the analysis of forces we concluded that ρ h 2 has the same value on either side of the cylinder. Therefore the two liquids exert exact opposite amounts of torque, resulting in no rotation.
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@Arjen Vreugdenhil – That is really interesting! I am reading through your solution right now! I guess I was wrong about the cylinder rotating!
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@James Wilson – I just realized that the torques I calculated are about the point ( 0 , 0 ) , i.e. the bottom of the cylinder. If instead we consider the torque about the central axis of the cylinder, the last part becomes d τ x = ( R − y ) d F x = ρ g ( R − y ) ( h − y ) d y d z , which is equal to the d τ y , showing immediately that the torque contributions due to horizontal and vertical force components cancel on every part of the cylinder. This is to be expected, because the pressure acts perpendicular to the cylinder, creating a force through the cylinder's axis, and therefore no torque about that axis.
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@Arjen Vreugdenhil – Yes, indeed. You were actually correct from the beginning. This was my error.
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@James Wilson – Forces directed toward its center will only make it slide, but not rotate. It is the friction force, like you said that would make it roll. My apologies! I was completely wrong.
@Arjen Vreugdenhil – Thank you for sharing your solution. I'm glad I could communicate with some very smart people on this site like yourself :)
In general, the horizontal forces, vertical forces, and torques must cancel. Since there is no friction at the bottom, it is obvious that the torques must cancel, since all lines of force pass through the center of the cylinder. The vertical normal force can take up the slack to balance the vertical forces. As it turns out, we can solve this problem by focusing our attention on the horizontal forces. The magnitudes of the net horizontal forces on the left and right sides must be equal. Derive a general expression for the horizontal force on one side, using the angular convention shown in the diagram (angle is with respect to the vertical). Use σ as the general density variable. The parameter θ increases from θ 0 at the fluid surface to π at the bottom of the cylinder.
Pressure expression:
P = σ g R ( cos θ 0 − cos θ )
Differential area (L is the cylinder length):
d A = R L d θ
Differential force magnitude:
d F = P d A = R 2 L σ g ( cos θ 0 − cos θ ) d θ
Differential horizontal force:
d F x = d F sin θ = R 2 L σ g ( cos θ 0 sin θ − sin θ cos θ ) d θ
Integrate the horizontal force (I won't go into details since it is fairly trivial):
∫ θ 0 π d F x = R 2 L σ g ( 2 1 + cos θ 0 + 2 cos 2 θ 0 )
Substitutions on the left ( θ 0 l is unknown at this point) :
θ 0 = θ 0 l σ l = ρ l
Substitutions on the right :
θ 0 = θ 0 r = 2 π σ r = ρ r
Equating the horizontal force expressions on left and right gives:
R 2 L g ρ l ( 2 cos 2 θ 0 l + cos θ 0 l + 2 1 ) = 2 1 R 2 L g ρ r which leads to cos θ 0 l h = ρ l ρ r − ρ l = ( 1 + cos θ 0 l ) R = ρ l ρ r R
Cant we do it without calculus?
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I don't know. It seems like a calculus problem to me.
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Every problem seemsto be calculus to u. Sir please can u give me ur phone no. Or mobile no. I want to contact u through whatsapp as its difficult for me to send prblms using slack
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@Md Zuhair – I am reachable on Slack or Brilliant. I think query problems would be good to post as notes.
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@Steven Chase – There are methods of solving it easily without using so much of calculus.
Just calculate the forces on these rectangles instead on the curved surface of the cylinder and equate them.
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@Rohit Gupta – Sir, Can you post an solution with this method?
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@Md Zuhair – Ok, sure. I will post it in some time.
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@Rohit Gupta – Sir please, can you post? I need the solution, if you dont mind
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@Md Zuhair – Arjen has already posted what I had in mind.
Anyways, let me do it again for you.
Imagine a rectangular sheet dipped in water, the force on one side of the sheet will be,
∫ 0 h ( P 0 + y ρ g ) l d y = P 0 l h + 2 1 ρ g l h 2
Now, imagine the rectangles on both the sides of the cylinders as shown in the above comments and instead of analyzing the equilibrium of the cylinder alone, consider the equilibrium of cylinder + plus between these rectangles.
These rectangles will experience horizontal force on both the sides according to the formula above. Equate these two forces to obtain your result.
@Rohit Gupta – Ah, yes. That's a clever way.
Can't we use the concept of projected area that p(l) pi (h^2)/4 = p(r)+pi*(r^2)/4? Here p stands for density.
Water pressure is perpendicular to the cylinder, so the line of force passes perpendicularly through its central axis: there is no torque. Vertically, the cylinder is kept in equilibrium by normal force. Therefore we only need to consider the horizontal forces.
Let z be the length of the cylinder. Let y = 0 at the top of a surface and increase in downward direction to depth h . There is no need for trigonometry, because the horizontal force component depends only on the horizontal projection of the surface: d F x = d F cos θ = p d A cos θ = p d y d z . With p = ρ g y , we find F x = ∫ d z ∫ 0 h d y ρ g y = g z ρ ∫ 0 h y d y = 2 1 g z ρ h 2 . In order for the horizontal forces to be balanced, we need F x , 1 = F x , 2 ∴ 2 1 g z ρ ℓ h 2 = 2 1 g z ρ r R 2 ∴ ρ ℓ h 2 = ρ r R 2 ∴ h = ρ ℓ ρ r R .