Tension in a supporting thread after a second supporting thread is cut

[Bling Fizikst]

Homework Statement

see below

Relevant Equations

see below

Since the pipe is light , net force and net torque acting on the pipe should be zero .
Initially , $T_A+T_B=3mg$ by balancing torques about each end of the rod , we get : $$T_A=\frac{4mg}{3}$$

Let the tension in $A$ just after cutting $B$ be : $T_A'$
Let the normal reactions acting on the pipe due to the balls be $N_m$ and $N_{2m}$ downwards respectively:
Hence , $$T_A'=N_{m}+N_{2m}$$ (Net force on pipe is zero)
Similarly , $$N_{m}+2N_{2m}=0$$ (writing torque about A on pipe)
Let the balls have accn $a$ downwards :
$$mg-N_m=ma$$
$$2mg-N_{2m}=2ma$$
Solving these gives normal reactions $=0$ and $a=g$ which kinda makes sense for they should be in free fall .
But this would mean there is $T_A'=0$ which is the wrong answer .
I am not sure where i went wrong

 

[Lnewqban]

Let the balls have accn $a$ downwards :

I believe that statement reflects the error in the approach.

At least during the first instants after the thread is cut, the ball of mass m will always travel at half the velocity of the ball of mass 2m.

Both can't have acceleration value equal to g.

 

[haruspex]

View attachment 354518

This is a very strange question. It doesn’t ask for the values of the tensions, but you say that they are given in the answer.

Since the pipe is light , net force and net torque acting on the pipe should be zero .

Initially , $T_A+T_B=\frac{3mg}{2}$

Try that again (but you don’t use it).

Let the balls have accn $a$ downwards :

@Lnewqban has pointed out the error there.

 

[Bling Fizikst]

I believe that statement reflects the error in the approach.

At least during the first instants after the thread is cut, the ball of mass m will always travel at half the velocity of the ball of mass 2m.

Both can't have acceleration value equal to g.

If that's the case then i would have to take two accns $a_1,a_2$ for them . But i can't think of an extra equation to solve them .

 

[haruspex]

If that's the case then i would have to take two accns $a_1,a_2$ for them . But i can't think of an extra equation to solve them .

Simple geometry if you assume the A end of the rod doesn’t rise.

 

[Bling Fizikst]

This is a very strange question. It doesn’t ask for the values of the tensions, but you say that they are given in the answer.

Try that again (but you don’t use it).

@Lnewqban has pointed out the error there'

If my initial tension is correct ($T_A=\frac{4mg}{3}$) then we can definitely find the $T_A'$?
It says it is $1/6$ times the initial tension .

 

[Bling Fizikst]

Simple geometry if you assume the A end of the rod doesn’t rise.

I am not sure if i get you , can you elaborate?

 

[haruspex]

I am not sure if i get you , can you elaborate?

Suppose the rod’s angular acceleration about end A is $\alpha$. Can you relate that to the balls' accelerations?

 

[Lnewqban]

If that's the case then i would have to take two accns $a_1,a_2$ for them . But i can't think of an extra equation to solve them .

Think of strings A and B being simultaneously cut.
What would be the rate of acceleration of each ball once both are liberated from the restriction of the string tensions A and B via the hanging horizontal pipe?

That is not our case at hand, and we can see that, due to the absence of internal friction, the only function of the pipe (which is deviating from a perfect horizontal position) is to limit the free fall of one or both of the balls.

Note that both balls are now also free to slide along the axis of the pivoting pipe.
During the first instants there is no force to deviate them from a vertical trajectory, which distance from each other and from the pivot A remains the same.

Only a force can restrain one or each of the balls from falling freely.
For that force to exist in that quasi-static initial condition, it must be transferred to the ground via the string A.

 

[haruspex]

If my initial tension is correct ($T_A=\frac{4mg}{3}$) then we can definitely find the $T_A'$?
It says it is $1/6$ times the initial tension .

Ah, the question is asking for the ratio of the tensions (“times”), not the number of different values it takes. That’s a very poor use of English.

 

[Bling Fizikst]

Suppose the rod’s angular acceleration about end A is $\alpha$. Can you relate that to the balls' accelerations?

I am sorry for my silly calculation. I have arrived at the right answer . $T_A'=\frac{2mg}{9}$

But my doubt still remains . I get through intuition that there MUST be angular accn but then again i feel like we cleverly kind of cheated math ? Since , let's say the distance between each ball and end is $x$ . Then $a_1=x\alpha , a_2=2x\alpha\implies a_2=2a_1$. But $\alpha$ itself is zero due to the net torque on a massless body is zero ?

 

[haruspex]

Doing that gives $a_2=2a_1$ and solving gives $a_1=g , a_2=2g$

That’s not what I get. Please post your working.

 

[Bling Fizikst]

If we take a pipe having mass $M$ and length $l$ instead . Will the following equations be correct?
Then the cm of the entire thing would be , (let point $A$ be the origin and the rod is along the $X$ axis)$$x_{cm}=\frac{m(\frac{l}{3})+2m(\frac{2l}{3})+M(\frac{l}{2})}{3m+M}$$

Again there will be some $\alpha$ about $A$ immediately after $B$ is cut .
So , we can write : $$a_{cm}= x_{cm}\alpha$$ (doubt here) or should it be $$a_{cm}=\frac{l}{2}\alpha$$?

 

[haruspex]

But $\alpha$ itself is zero due to the net torque on a massless body is zero ?

That doesn’t follow. If the net torque is zero and the moment of inertia is zero then $\tau=I\alpha$ says $\alpha$ can be anything at all.

we can write : $$a_{cm}= x_{cm}\alpha$$

Correct, but not terribly helpful. You would need to consider the moment of inertia of the tube.

 

[Bling Fizikst]

That doesn’t follow. If the net torque is zero and the moment of inertia is zero then $\tau=I\alpha$ says $\alpha$ can be anything at all.

So , net torque on the pipe is zero , and moment of inertia of pipe is zero . Yeah , it means $\alpha$ can be anything .
My thoughts are jumbled . I was thinking the moment of inertia would be $I=mx^2+2m(2x)^2$ about $A$ . So , when we write the torque on a general pipe (not necessarily massless) , do we ignore the contribution of balls inside it? If yes/no , why?

 

[haruspex]

So , net torque on the pipe is zero , and moment of inertia of pipe is zero . Yeah , it means $\alpha$ can be anything .
My thoughts are jumbled . I was thinking the moment of inertia would be $I=mx^2+2m(2x)^2$ about $A$ . So , when we write the torque on a general pipe (not necessarily massless) , do we ignore the contribution of balls inside it? If yes/no , why?

You can either treat the tube and balls as a unit or treat them separately. If separately, the normal forces between them must be included in the torque equation for the tube.

 

[Lnewqban]

My thoughts are jumbled.

The following alternative approach, ignoring the pipe, may help you:
Consider the 3m combined center of mass in free fall.
An extension of the system (the dashed line shown in the attached diagram) collides with the obstacle A.

That collision induces a clockwise spinning of m and 2m (linked by the pipe represented by the continuous line).

Immediately after the impact, that very rotation reduces the tension in our string A down to 1/6 times the initial tension.