Toppling an Object over a Table

In summary, you solved the problem by finding the minimum velocity such that the block arrives at G before G reaches A.
  • #1
SilverSoldier
26
3
Homework Statement
A uniform rectangular rod of length ##L##, having mass ##M## is placed at the edge of a frictionless table so that some portion of the rod rests on the table, and the other portion extends out of the table, and its center of gravity ##G## is a distance ##x## away from the edge of the table.
A block of mass ##m## is initially placed at the far end of the part of rod resting on the table, and given a velocity ##v##. If the coefficient of kinetic friction between the rod and the object is ##mu##, what must be the minimum value of ##v##, so that the rod can be made to topple over the table?
Relevant Equations
Equations of motion
Kinetic friction = ##\mu_k\cdot N##
Momentum = ##mv##
Torque = ##F\times d##
Here is a diagram of the situation.
1586167895185.png
I first marked the forces acting on the system.
1586168786042.png
I understand that for this rod to topple over, there must be a net moment about point ##A##. But how exactly should I approach this problem?

Isn't it so, that irrespective of the velocity given to the block, at some point both the rod and block will acquire equal velocities as a result of their acceleration and deceleration, and then just continue moving forward at that velocity?

If by any chance that were not the case, I tried solving it as follows.

First I calculate the distance to the center of gravity, say ##G'##, of the system from ##A##. We have
$$
GG'=\dfrac{m}{m+M}\cdot\dfrac{L}{2},
$$
and so we have
$$
\begin{align*}
AG'&=x+GG'\\
\\
&=\dfrac{2Mx+2mx+mL}{2(m+M)}.
\end{align*}
$$
Then, I observe that momentum in the system is conserved horizontally as there is no external horizontal force. Therefore,
$$
mv=(m+M)w\\
\text{ }\\
w=\dfrac{mv}{m+M},
$$
where ##w## is the velocity at which the center of gravity of the system will move after the block is given a velocity ##v##.

Next, I calculate the acceleration ##a## of the little block and get ##a=\dfrac{\mu mg}{m}=\mu g##.

If ##t## is the time that the block will take to come to rest, I have ##0=v-\mu gt##, so ##t=\dfrac{v}{\mu g}##.

If the distance that the center of gravity of the system moves in time ##t## is equal to ##AG'##, then the system will be at the verge of toppling over. Therefore,
$$
\begin{align*}
w\cdot t&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\
\\
\dfrac{mv}{m+M}\cdot \dfrac{v}{\mu g}&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\
\\
\dfrac{mv^2}{\mu g}&=\dfrac{2Mx+2mx+mL}{2}\\
\\
v^2&=\dfrac{\mu g}{2m}\cdot\left(2Mx+2mx+mL\right)\\
\\
v&=\sqrt{\mu g\left(x+\dfrac{L}{2}+\dfrac{Mx}{m}\right)}.
\end{align*}
$$
Is this correct?

Thank you very much for your assistance.
 
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  • #2
the table is frictionless the center of mass of the system moves towards the edge of the table... what could stop it?
 
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  • #3
SilverSoldier said:
If ##t## is the time that the block will take to come to rest,
As you noted, when it has the same velocity as the rod the acceleration will cease. It will never come to rest in the lab frame.
This bizarre question is familiar, but I cannot find it in the archives.
Where does it come from?

It is as though something has been left out, like toppling before the mass reaches the table edge, or is between there and the end of the rod, or after it has left the rod. But nothing seems to make sense.
 
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  • #4
haruspex said:
Where does it come from?

It was on an A/L past paper book :smile:

haruspex said:
It is as though something has been left out, like toppling before the mass reaches the table edge, or is between there and the end of the rod, or after it has left the rod. But nothing seems to make sense.

I believe the question has an error in it or something then. Anyway, the velocity ##\sqrt{\mu g\left(x+\dfrac{L}{2}+\dfrac{Mx}{m}\right)}## I have found is the velocity that must be given to the block so that the system will be on the verge of toppling over, when its center of gravity will have reached the edge of the table?
 
  • #5
SilverSoldier said:
the velocity ##\sqrt{\mu g\left(x+\dfrac{L}{2}+\dfrac{Mx}{m}\right)}## I have found is the velocity that must be given to the block so that the system will be on the verge of toppling over, when its center of gravity will have reached the edge of the table?
Not sure what you mean. It is always true that the system will be on the verge of toppling when its mass centre reaches the edge.

What you have found, I think, is the minimum velocity such that the block stops sliding before the system topples.
Another reasonable interpretation is to find the minimum velocity such that the block arrives at G before G reaches A. I.e, it topples before it would have done without the block.

Btw, your diagram shows a ball or cylinder, not a block.
 
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  • #6
haruspex said:
What you have found, I think, is the minimum velocity such that the block stops sliding before the system topples.

Oh yes. I did mean to say that. I just noticed that I had left the "block stops sliding" part in my previous reply 🤦‍♂️😜.

haruspex said:
Another reasonable interpretation is to find the minimum velocity such that the block arrives at G before G reaches A. I.e, it topples before it would have done without the block.

If put another way, this means to find the velocity so that when the block stops moving, point ##G## and the ball would both be exactly above point ##A##?

haruspex said:
Btw, your diagram shows a ball or cylinder, not a block.

Well, it was just easier to draw a circle than a square 😇. It hasn't got to do anything with rolling.
 
  • #7
SilverSoldier said:
If put another way, this means to find the velocity so that when the block stops moving, point G and the ball would both be exactly above point A?
Yes.
 
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Related to Toppling an Object over a Table

1. How does the height of the object affect its likelihood of toppling over a table?

The height of an object does not necessarily affect its likelihood of toppling over a table. Factors such as the weight distribution, shape, and stability of the object play a bigger role in determining its stability.

2. Is there a specific angle at which an object is more likely to topple over a table?

Yes, there is a specific angle at which an object is more likely to topple over a table. This angle, known as the center of mass, is when the object's weight is evenly distributed on all sides, making it more stable and less likely to topple over.

3. Can the material of the table affect the stability of an object placed on it?

Yes, the material of the table can affect the stability of an object placed on it. A smooth and flat surface, such as a wooden table, provides more stability compared to a rough and uneven surface, such as a carpeted floor.

4. How does the shape of an object affect its stability on a table?

The shape of an object can greatly affect its stability on a table. Objects with a wider base, such as a pyramid, have a lower center of mass and are more stable compared to objects with a narrow base, such as a pencil.

5. What are some ways to prevent an object from toppling over a table?

There are a few ways to prevent an object from toppling over a table. One way is to ensure that the object's weight is evenly distributed on all sides, making its center of mass lower. Another way is to place the object closer to the table's edge, providing more support and stability. Additionally, using a non-slip mat or adding weight to the base of the object can also help prevent toppling.

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