Falling and sliding stick (David Morin)

[Father_Ing]

Homework Statement

See attachment

Relevant Equations

F= ma

In the solution for question $(a)$, it is written that the equation of translational motion for the center of mass is $N-mg=ma_y$
Why $N$ is also included inside of the equation? In my opinion, the rail does not exerting force (N) to slow down the mass' acceleration. Instead, the purpose of $N$ is only to keep the pivot so that it is still on the rail.

Can anyone please explain to me why $N$ is also included in the calculation?

 

[PeroK]

There must be an upward force at the pivot. Otherwise, the stick would simply be in freefall and the CoM would fall vertically under the force of gravity alone.

 

[Father_Ing]

Alright...
But, after thinking it over again, why not $N - mg$ = 0?
The stick does not falling freely,but it only rotates. Which means that there is no translational motion.Hence $a_y = 0$

 

[PeroK]

Alright...
But, after thinking it over again, why not $N - mg$ = 0?
The stick does not falling freely,but it only rotates. Which means that there is no translational motion.Hence $a_y = 0$

If $a_y = 0$ then the stick will move horizontally! The CoM of the stick must have vertical motion.

 

[haruspex]

the purpose of N is only to keep the pivot so that it is still on the rail.

It is not still. The rail is frictionless, so can only exert a vertical force. What does that tell you about the trajectory of the mass centre of the stick?
And it must exert a vertical force to cause the stick to rotate.

 

[Father_Ing]

If $a_y = 0$ then the stick will move horizontally! The CoM of the stick must have vertical motion.

Isn't $a_y$ in post#1 means translational acceleration?
If yes, then having $a_y= 0$ does not means it doesn't have vertical motion since the total acceleration is the sum of translational acc. + rotational acc..

 

[Father_Ing]

What does that tell you about the trajectory of the mass centre of the stick?

I think it is the same as a case where the pivot is restrained (because in the post#1 problem, there is only $mg$ and $N$ forces, and no horzontal forces), which is 1/4 of a circle

 

[PeroK]

Isn't $a_y$ in post#1 means translational acceleration?
If yes, then having $a_y= 0$ does not means it doesn't have vertical motion since the total acceleration is the sum of translational acc. + rotational acc..

The stick falls, therefore $a_y \ne 0$. Assuming $a_y$ refers to the vertical acceleration of the CoM. The vertical acceleration of the bottom of the stick is zero.

If you are talking about rotation about the CoM, then that cannot cancel the translation of the CoM. In this case, the CoM translates (falls); and, the stick rotates about its CoM.

 

[Father_Ing]

why is the rotation not about the pivot, like in the case where the pivot is restrained?

 

[haruspex]

why is the rotation not about the pivot, like in the case where the pivot is restrained?

It doesn't matter whether you want to think of it as a rotation about one end, or as a rotation about the mass centre plus a vertical motion of that. You cannot get away from the fact that the mass centre accelerates according to the net force. $\Sigma F=ma$. The mass centre does accelerate downwards, so there is a net vertical force.

 

[PeroK]

why is the rotation not about the pivot, like in the case where the pivot is restrained?

You can model the system as motion of the pivot and rotation about the pivot. That should give you the same physical solution. In general, you can choose any point on the stick and decompose the motion into translation of that point plus rotation about that point.

In any case, the external forces on the stick are the same: they are $mg$ and $N$, both vertical. And, that determines the translational (linear) motion of the CoM.

 

[haruspex]

I think it is the same as a case where the pivot is restrained

It is not. In the fixed pivot case, some of the energy goes into horizontal motion, slowing the descent.

 

[Father_Ing]

I see..
So, If I want to assume that it rotates about the pivot, then the eq of the torque will be:
$mg sin\theta \frac L 2 = \frac {mL^2} 3 \alpha$, and we still use the same equation the book uses for translational motion.
Is this true?

 

[PeroK]

I see..
So, If I want to assume that it rotates about the pivot, then the eq of the torque will be:
$mg sin\theta \frac L 2 = \frac {mL^2} 3 \alpha$, and we still use the same equation the book uses for translational motion.
Is this true?

You need to be careful there. The pivot itself is accelerating horizontally. In that reference frame, you would need to add a fictitious horizontal force through the CoM.

 

[Father_Ing]

Where does the force in horizontal axis come from?
From the body diagram that i made, the only forces that works on the system are $N$ from the floor and $mg$(both are in vertical axis). Did I miss something?

 

[PeroK]

Where does the force in horizontal axis come from?
From the body diagram that i made, the only forces that works on the system are $N$ from the floor and $mg$(both are in vertical axis). Did I miss something?

You wanted to use the pivot, which is accelerating to the left. That's a non-inertial frame. Newton's laws do not directly apply in a non-inertial frame. You need to include relevant fictitious forces.

Consider this: imagine the stick is falling horizontal (no pivot, just simple freefall). Now, do an analysis of forces about one end of the stick. There is a torque $mgL/2$ about the end. But, there is no rotation. That illustrates the problem of using an accelerating point on an object (that is not the CoM) and trying to apply Newton's laws.

 

[Father_Ing]

You wanted to use the pivot, which is accelerating to the left.

Oh! It accelerates to the left so that the position of CoM in X direction doesn't move, right?
Also, when calculating the torque, is the fictious force always located at the CoM?

 

[PeroK]

Oh! It accelerates to the left so that the position of CoM in X direction doesn't move, right?
Also, when calculating the torque, is the fictious force always located at the CoM?

The same fictitious force must be applied to each point in the system, so the effect is a single force through the CoM. And, therefore, the addition of the fictitious force does not affect the torque about the CoM - you only need the real forces.

Note: if the pivot is fixed, then it's also an inertial point and we may apply Newton directly.

 

[Father_Ing]

Consider this: imagine the stick is falling horizontal (no pivot, just simple freefall). Now, do an analysis of forces about one end of the stick. There is a torque $mgL/2$ about the end. But, there is no rotation. That illustrates the problem of using an accelerating point on an object (that is not the CoM) and trying to apply Newton's laws.

When analyzing the torque about the end, do we take $\frac 1 3$ as the moment of inertia?

 

[PeroK]

When analyzing the torque about the end, do we take $\frac 1 3$ as the moment of inertia?

The only sensible way to solve this problem is to analyse the motion of the CoM and rotation relative to the CoM.

Once you have the correct answer, there is no harm in trying to solve the problem using the accelerating pivot, but given the conceptual problems you have been having that's not a recommended approach.

 

[Father_Ing]

Thanks for the advice!
The question that I asked in post #19 is when we are solving the freefall stick with assumption that it rotates about the end. We take 1/3 as the moment inertia, right?

 

[PeroK]

We take 1/3 as the moment inertia, right?

That information is readily available online:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html

 

[Father_Ing]

Alright! Thanks for your help

 

[ergospherical]

When the rod is horizontal at $\theta = \pi/2$, the vertical acceleration of the mass centre is $-l\dot{\omega}/2$. The vertical force is $N - mg$, so\begin{align*}
N - mg &= -\dfrac{ml\dot{\omega}}{2}
\end{align*}The moment about the mass centre is $G = N l/2$, so \begin{align*}
\dfrac{N l}{2} = \dfrac{1}{12} ml^2 \dot{\omega} \implies \dot{\omega} = \dfrac{6 N}{ml}
\end{align*}which allows you to eliminate $\dot{\omega}$. For $(\mathrm{b})$, you should consider energy.

 

[Father_Ing]

Now that I think about it...
Since there is no external forces, the position of CoM in X direction is maintained. To realize this, the end of the rod needs to accelerate to the left. But, where does the force that accelerates the rod's end come from?

 

[haruspex]

Now that I think about it...
Since there is no external forces, the position of CoM in X direction is maintained. To realize this, the end of the rod needs to accelerate to the left. But, where does the force that accelerates the rod's end come from?

The rod as a whole is not going left, so you do not need a force going left. The motion is vertical for the rod as a whole, plus a rotation. The rotation comes from the torque that arises because the normal force is left of the gravitational force.

If it still bothers you when considering the rod as made of small elements, you could say the leftward force on the leftmost element comes from the reaction from the rightward acceleration of the rightmost element, and v.v.

 

[PeroK]

... for example, if the rod is spinning with no external forces, then the centripetal force that accelerates each particle must come from internal forces between the particles.

 

[PeroK]

@Father_Ing I think this is quite a tough question!

 

[Father_Ing]

I am a bit confused..
If we see the the rod as a rigid body that is formed by many particles, N is exerted only to the particle at the end of the rod.In my opinion, N only affects the acceleration of the particle this point (the pivot, or we can also say, the end of the rod). However,in the equation $N -mg = ma$ , I can conclude that N affects the acceleration of the whole particles.

How can N affects the whole particles' acceleration?

 

[haruspex]

I am a bit confused..
If we see the the rod as a rigid body that is formed by many particles, N is exerted only to the particle at the end of the rod.In my opinion, N only affects the acceleration of the particle this point (the pivot, or we can also say, the end of the rod). However,in the equation $N -mg = ma$ , I can conclude that N affects the acceleration of the whole particles.

How can N affects the whole particles' acceleration?

Because the rod is rigid, so each particle transfers force and torque to the next.

 

[PeroK]

I am a bit confused..
If we see the the rod as a rigid body that is formed by many particles, N is exerted only to the particle at the end of the rod.In my opinion, N only affects the acceleration of the particle this point (the pivot, or we can also say, the end of the rod). However,in the equation $N -mg = ma$ , I can conclude that N affects the acceleration of the whole particles.

How can N affects the whole particles' acceleration?

If you lift a suitcase up by the handle, then the handle doesn't necessarily break off from the rest of the suitcase. And, if you hold a tennis racket by the handle, then the racket head (the bit with the strings) moves as well when you swing the handle.

Have you made any progress at solving this problem?

 

[Father_Ing]

Have you made any progress at solving this problem?

Yep! I have made it through the 2nd question. I still have some problems regarding the concepts, though.

Because the rod is rigid, so each particle transfers force and torque to the next.

So, each particle has a force $dN$ exerted on it, and the value of $dN$ differs from each particle?

 

[PeroK]

So, each particle has a force $dN$ exerted on it, and the value of $dN$ differs from each particle?

Yes, each particle on the rod must be subject to the force required by the constraint that the rod remains rigid. $N$ is an external force applied at the pivot. $dN$ doesn't make sense in this context.

There will be a distribution of internal forces along the rod. If you have calculated the motion of the rod, you may calculate the motion of every point on the rod and hence the force on that point.

The point is that the internal forces cancel (Newton's third law) and the external forces determine the motion. Unless the rod breaks, the internal forces are irrelevant.

 

[Father_Ing]

Alright! Thanks.