# Falling chain fixed at one end: force at the hinge

The end B of the chain of mass per unit length (a) and length (l) is released from rest as shown in the picture given above. The force at the hinge when the end B is at $$\frac{l}{4}$$ from the ceiling is ________________

My attempt: I have tried to locate the position of the center of mass of the chain from the top after end B has fallen distance x from the ceiling. I then used the Principle of Conservation of Energy to find the velocity of the hanging part when it has fallen distance x, by equating the change in gravitational potential energy to change in kinetic energy. I cannot however figure out the relation between the force at the hinge and the velocity of hanging part.

Ideas?

Edit after JiK comment's i am writing down the equations here Suppose the free end of the chain is displaced by a distance x.thus the length of the hanging part now becomes $$\frac{l+x}{2}-x=\frac{l-x}{2}$$.now to find out the position of com

$$\frac{l-x}{2}.a.\frac {l+3 x}{4}+\frac{l+x}{2}.a.\frac{l+x}{4}=a.lx(com)$$ Now applying conservation of energy principle $$al\frac{l}{4}g=alx(com)+\frac{1}{2}.a(\frac{l-x}{4})v^2$$ Here I have considered increase in kinetic energy of only the hanging portion as it is only in motion. However writing the equation for the other portion of the chain,I am finding trouble $$a(\frac{l+x}{2})g+?=Hingeforce .$$ I couldnot find out what should be the force replacing the question mark.

• It would be nice if you showed your calculations also here so we wouldn't have to redo it or guess which way you did it. It's easier to help if the helper knows exactly the point where you're stuck.
– JiK
Sep 6 '16 at 10:00
• @JiK please check the edit.i might make you understand how i approached the problem
– Pink
Sep 6 '16 at 10:58
• Note that the connection force is what it needs to be to enforce the pin constraint. So you describe the motion in a way that does not violate the constraint and the force will come out of the equations of motion. Sep 6 '16 at 15:24

This problem has been tackled before on Physics SE, for example :

A comprehensive solution is given in Falling Chains in American Physics Teacher.

The 1st link above includes an instructive discussion of and solution to the problem from the textbook by Marion & Thornton. It is tempting to assume that the free side of the chain is in free fall, but this is incorrect. Instead, it should be assumed that energy is conserved, because this is how real chains are observed to behave.

Your approach is valid. The reaction $R$ at the hinge is related to the acceleration $\ddot y$ of the centre of mass by Newton's 2nd Law : $Mg-R=M\ddot y$. What you are unsure about is how to find $\ddot y$.

Your calculation of the position of the CM of the whole chain is correct :
$y=\frac{1}{4l}(l^2+2lx-x^2)$.
Differentiating twice gives :
$\ddot y=\frac{1}{2l}((l-x)\ddot x-\dot x^2)$

Expressions for $\ddot x$ and $\dot x^2$ can be found from the conservation of energy, as follows :

At any instant only the RHS of the chain is moving. The length of this side is $\frac12(l-x)$ and the CM has velocity $\dot x$, so its KE is $\frac14\rho (l-x)\dot x^2$. The KE gained equals the loss of PE due to the fall of the CM of the whole chain. The CM is initially at $y=\frac14 l=\frac{1}{4l}l^2$, so the loss in PE is
$l\rho g \frac{1}{4l}(l^2+2lx-x^2-l^2)=\frac14 \rho g (2lx-x^2)$.
Therefore
$\dot x^2 = g \frac{2lx-x^2}{l-x}$.
Differentiating :
$\ddot x= \frac{g(2l^2-2lx+x^2)}{2(l-x)^2}$

Substitute :
$\ddot y=\frac{g(2l^2-6lx+3x^2)}{4l(l-x)}$
$R=M(g-\ddot y)=Mg\frac{2l^2+2lx-3x^2}{4l(l-x)}$.

Substituting $x=\frac14l$ gives $R=\frac34Mg$. But note that as $x \to l$ then $R \to \infty$. This happens because of the whiplash effect.

• Thanks for your answer.I was not notified about this answer(or perhaps I overlooked it).I have one doubt.While writing down the equation of motion for the whole chain shouldnot we consider the effect of thrust force that is experienced by the left hand side of the chain when a small portion of right hand side comes instantaneously to rest.
– Pink
Jan 25 '17 at 3:05
• At any instant if we assume that the left hand side has a speed $v$ then an elementary portion of that side will have a momentum.$avdx$.where $a$ is mass per unit length of the chain.Now in time $dt$ the elementary portion will come to rest.So the change in momentum by time is simply the thrust force and it comes out to be $av^2$.But I am not sure about this as it gives me a wrong answer.The answer,as given in my module is $\frac 43 agl$
– Pink
Jan 25 '17 at 3:12
• The solution I have provided uses conservation of energy, so it is not necessary to consider what forces are involved. It is only when the acceleration of the chain as a whole is considered, ie its centre of mass, that the solution finds the reaction force at the support. ... I do not follow your calculation. I don't see how $avdx$ integrates to become $av^2$. Probably the fact that you get the wrong answer indicates that it is an incorrect calculation. Jan 26 '17 at 20:46

I have tried to locate the position of the center of mass of the chain from the top after end B has fallen distance $x$ from the ceiling. I then used the Principle of Conservation of Energy to find the velocity of the hanging part when it has fallen distance $x$, by equating the change in gravitational potential energy to change in kinetic energy.

This seems correct to me. Assuming that the horizontal length of the system is negligible, it should be quite straightforward. You should remember that the mass of the moving part changes also, as the part of the chain that points upwards gets smaller.

This gives you the velocity of the end of the chain as a function of $x$.

I cannot however figure out the relation between the force at the hinge and the velocity of hanging part.

The sum of the forces acting on any system gives the acceleration of the centre of mass of the system. The forces acting on the chain are gravity and the force at the hinge. So the sum of these gives you the acceleration of the centre of mass of the chain.

So now you know

• the velocity of the end of the rope, $\frac{d}{dt}x$, as a function of $x$
• the position of the centre of mass (let's call it $z$) as a function of $x$
• the force at the hinge in terms of $\frac{d^2}{dt^2} z$.

To get your answer, you want to find $\frac{d^2}{dt^2} z$ in terms of $x$. You know $z$ in terms of $x$, and $\frac{d}{dt} x$ in terms of $x$. I'll leave you to figure out how to continue from this.

• (I don't know what the formulas look like, but my intuition tells me the last part should be rather straightforward. I might be wrong and it might involve difficult differential equations. I don't want (nor have time) to redo your work, so I don't know.)
– JiK
Sep 6 '16 at 10:19