A slanted cylinder full of liquid about to fall

In summary: What are the lengths of the base and height of the red triangle? In particular, what is the height of the center of mass in terms of...The base is ##L=3.5m## and the height of the center of mass is ##H=2.7m##.
  • #1
lorenz0
148
28
Homework Statement
A slanted cylinder of negligible mass has radius ##R=0.3m## and an internal angle of ##\alpha=70°## lies on a flat surface and is filled up with a liquid up to a height ##H##. Find the maximum value of ##H## such that the cylinder remains in equilibrium
Relevant Equations
##\vec{\tau}=\vec{r}\times\vec{F}, \sum\vec{\tau}=\vec{0}, \sum{\vec{F}}=\vec{0}##
The cylinder will cease to be in equilibrium when the sum of the torques on the cylinder calculated with respect to the rightmost point of contact of the cylinder with the plane will be unbalanced. Now, the liquid is homogeneous and the cylinder has negiglible mass so the forces (normal force of the surface and gravitational force) will act on the center of mass which is in the geometric center of the figure so the lever arm of this torque with respect to said point should be ##2R-(R+\frac{L}{2}\sin(30°))=2R-(R+\frac{H}{2\sqrt{3}})=R-\frac{H}{2\sqrt{3}}##.

Now, the thing that confuses me is that these two torques have the same ##F## and the same ##r## so their algebraic sum is always ##0##.

There must be something about setting up this problem that I don't understand so I would appreciate an hint about how to reason about it.
 

Attachments

  • cylinder.png
    cylinder.png
    17.4 KB · Views: 83
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
Where does the 30° come from?
lorenz0 said:
normal force of the surface and gravitational force) will act on the center of mass which is in the geometric center of the figure
The normal force from the ground acts on the flat surface, but at what point of it?
 
  • #3
lorenz0 said:
There must be something about setting up this problem that I don't understand so I would appreciate an hint about how to reason about it.
If the centre of gravity is over the base, the cylinder is stable (why?).

If the centre of gravity is to the right of the base, the cylinder will topple over (why?)

So where would the centre of gravity be when the cylinder is on the verge of toppling over?
 
  • Like
Likes rsk
  • #4
lorenz0 said:
Homework Statement:: A slanted cylinder of negligible mass has radius ##R=0.3m## and an internal angle of ##\alpha=70°## lies on a flat surface and is filled up with a liquid up to a height ##H##. Find the maximum value of ##H## such that the cylinder remains in equilibrium
Relevant Equations:: ##\vec{\tau}=\vec{r}\times\vec{F}, \sum\vec{\tau}=\vec{0}, \sum{\vec{F}}=\vec{0}##

The cylinder will cease to be in equilibrium when the sum of the torques on the cylinder calculated with respect to the rightmost point of contact of the cylinder with the plane will be unbalanced.
Yes. So draw a picture of that situation (in your head or on paper) and notice the symmetry about a vertical line from the pivot which must then be true. The rest is just a little bit of geometry. Do you see it?
 
  • Like
Likes lorenz0
  • #5
hutchphd said:
Yes. So draw a picture of that situation (in your head or on paper) and notice the symmetry about a vertical line from the pivot which must then be true. The rest is just a little bit of geometry. Do you see it?
from the drawing I would say that the maximum ##H## is the one such that ##mg## and the normal force ##N## acting on the pivot are collinear, otherwise the torque due to ##mg## wrt the pivot point will cause the cylinder to gain a clockwise angular acceleration that will make it fall. EDIT: it seems to me now that it would be sufficient to impose that the x-coordinate of the center of mass be at most ##R## to the right of the center of the cylinder so it should be sufficient to impose that ##H\tan(20°)=R## i.e. ##H=\frac{R}{\tan(20°)}##. Is this correct?
 

Attachments

  • cylinder_falling.png
    cylinder_falling.png
    33.4 KB · Views: 87
Last edited:
  • Like
Likes bob012345
  • #6
I would write it as $$\frac H R=\tan \alpha$$ but that is the same thing. Good.
 
  • Like
Likes bob012345 and lorenz0
  • #7
lorenz0 said:
EDIT: it seems to me now that it would be sufficient to impose that the x-coordinate of the center of mass be at most ##R## to the right of the center of the cylinder so it should be sufficient to impose that ##H\tan(20°)=R##
Does ##H## here represent the height of the top surface of the water, or the height of the center of mass of the water?
 
  • Like
Likes haruspex
  • #8
TSny said:
Does ##H## here represent the height of the top surface of the water, or the height of the center of mass of the water?
The height of the top surface of the water
 
  • #9
lorenz0 said:
The height of the top surface of the water
Then "double" check your answer.
 
  • Like
Likes lorenz0
  • #10
Or realize that I am correct about "half" the time. For reasons that escape me I assumed that R was the diameter of the container. Physics correct, geometry wrong. Thanks for check.
 
  • #11
haruspex said:
Then "double" check your answer.
Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose ##H=\frac{T}{\tan(20°)}##.
 
  • #12
lorenz0 said:
Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose ##H=\frac{T}{\tan(20°)}##.
1631296264147.png


What are the lengths of the base and height of the red triangle? In particular, what is the height of the center of mass in terms of ##H##?
 
  • #13
lorenz0 said:
Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose ##H=\frac{T}{\tan(20°)}##.
I assume you mean ##H=\frac{R}{\tan(20°)}##, which is what you posted before.
H is the full height, not the height to the mass centre. R is a radius, not a diameter.
 

Related to A slanted cylinder full of liquid about to fall

1. What is the cause of a slanted cylinder full of liquid about to fall?

The cause of a slanted cylinder full of liquid about to fall is typically due to an imbalance in forces acting on the cylinder. This could be caused by external factors such as an uneven surface or internal factors such as an uneven distribution of liquid inside the cylinder.

2. How does the angle of the cylinder affect its stability?

The angle of the cylinder plays a crucial role in its stability. As the angle increases, the center of mass shifts towards the lower end of the cylinder, making it more likely to fall. At a certain angle, known as the critical angle, the cylinder will become unstable and fall.

3. Can the liquid inside the cylinder affect its stability?

Yes, the liquid inside the cylinder can greatly affect its stability. If the liquid is evenly distributed, it can help to lower the center of mass and increase stability. However, if the liquid is unevenly distributed, it can cause an imbalance of forces and lead to the cylinder falling.

4. How can we prevent a slanted cylinder full of liquid from falling?

To prevent a slanted cylinder full of liquid from falling, we can try to evenly distribute the liquid inside the cylinder or adjust the angle of the cylinder to lower the center of mass. We can also place the cylinder on a flat and stable surface to minimize external factors that may cause it to fall.

5. What factors can affect the critical angle of a slanted cylinder full of liquid?

The critical angle of a slanted cylinder full of liquid can be affected by several factors, including the viscosity of the liquid, the shape and size of the cylinder, and the surface on which the cylinder is placed. The critical angle can also be influenced by external factors such as air resistance or vibrations.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
376
Replies
13
Views
971
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
283
  • Introductory Physics Homework Help
Replies
4
Views
821
  • Introductory Physics Homework Help
Replies
7
Views
757
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
Back
Top