Block on top of moving slab, with a rough surface - When does v_b=v_s?

[Thermofox]

Homework Statement

A wooden slab, with a mass of $M = 80 kg$, initially moves with an uniform rectilinear motion, on a horizontal surface devoid of friction. The slab has a velocity $v_{s,i}=6 m/s$. At a certain instant, a block, assumed to be a point mass with a mass of $m = 12 kg$, is placed on the right end of the slab. The block initially has a velocity of zero, relative to a frame of reference fixed to the horizontal surface. Knowing that there is a frictional force characterized by the coefficient of kinetic friction $μ_k= 0.3$ between the slab and the block, determine:
1) the minimum length of the slab such that the block does not fall from it;
2) assuming that now the length is $l \gt l_{min}$. Determine the time that elapses from when the block is left on the slab and the instant the two bodies will move together.

Relevant Equations

$\Delta E_m = 0$ $\Delta P = 0$

For the first question I thought of using an energy balance,
there is friction $\Rightarrow \Delta E_m = -W_f$. Both at the start and at the end, the block has no velocity. Therefore $E_{\text{initial}}= \frac 1 2 m_s v_{s,i}^2$ and $E_{\text{final}}= \frac 1 2 m_s v_{s,f}^2$. This means that: $$\frac 1 2 m_s v_{s,f}^2 - \frac 1 2 m_s v_{s,i}^2 = \mu_k m_b g l_{min}$$ From this I can determine $l_{min}$, but first I need to find a relationship to define $v_{s,f}$.

That's where I'm stuck at. I tried considering the system as a whole. This means that there is conservation of linear momentum, hence:
$\Delta P = 0 ;P_{\text{final}} - P_{\text{initial}} = 0$. The problem is that $P_i = m_s v_{s,i}$ and $P_f= m_s v_{s,f}$. This means that the slab maintains the same velocity, but that doesn't make sense. So there must be an error, but I can't see it.

Furthermore, for question 2, since I have to determine time, I think I need to use kinematic. Can I assume that the block moves with an uniform acceleration?

 

[kuruman]

Homework Statement: A wooden slab, with a mass of $M = 80 kg$, initially moves with an uniform rectilinear motion, on a horizontal surface devoid of friction. The slab has a velocity $v_{s,i}=6 m/s$. At a certain instant, a block, assumed to be a point mass with a mass of $m = 12 kg$, is placed on the right end of the slab. The block initially has a velocity of zero, relative to a frame of reference fixed to the horizontal surface. Knowing that there is a frictional force characterized by the coefficient of kinetic friction $μ_k= 0.3$ between the slab and the block, determine:
1) the minimum length of the rod slab? such that the block does not fall from it;
2) assuming that now the length is $l \gt l_{min}$. Determine the time that elapses from when the block is left on the rod slab? and the instant the two bodies will move together.
Relevant Equations: $\Delta E_m = 0$ $\Delta P = 0$

For the first question I thought of using an energy balance,
there is friction $\Rightarrow \Delta E_m = -W_f$. Both at the start and at the end, the block has no velocity. Therefore $E_{\text{initial}}= \frac 1 2 m_s v_{s,i}^2$ and $E_{\text{final}}= \frac 1 2 m_s v_{s,f}^2$. This means that: $$\frac 1 2 m_s v_{s,f}^2 - \frac 1 2 m_s v_{s,i}^2 = \mu_k m_b g l_{min}$$ From this I can determine $l_{min}$, but first I need to find a relationship to define $v_{s,f}$.

That's where I'm stuck at. I tried considering the system as a whole. This means that there is conservation of linear momentum, hence:
$\Delta P = 0 ;P_{\text{final}} - P_{\text{initial}} = 0$. The problem is that $P_i = m_s v_{s,i}$ and $P_f= m_s v_{s,f}$. This means that the slab maintains the same velocity, but that doesn't make sense. So there must be an error, but I can't see it.

Furthermore, for question 2, since I have to determine time, I think I need to use kinematic. Can I assume that the block moves with an uniform acceleration?

Slab or rod?

Treat this as a slow-developing perfectly inelastic collision. Consider the slab + block as the system. The difference between this and a run-of-the-mill perfectly inelastic collision, here you have a way of figuring out how much energy is lost as heat in the collision.

Also, if the block does not fall off, what is the block's final velocity?

 

[Thermofox]

Slab or rod?

It's slab, I've adjusted it. I have no idea how "rod" sneaked in the problem.

Also, if the block does not fall off, what is the block's final velocity?

Previously I thought that if it stops moving then its velocity should be zero, but now I realize that if the block stops it will have the same velocity of the slab. Then $P_f = (m_s + m_b) v_{s,f}$

Treat this as a slow-developing perfectly inelastic collision. Consider the slab + block as the system. The difference between this and a run-of-the-mill perfectly inelastic collision, here you have a way of figuring out how much energy is lost as heat in the collision.

Do you mean the collision with the block and the slab? Can't I just analyze the problem from when the block is placed on the slab?

 

[erobz]

In my estimation the block and the slab have distinct constant accelerations by Newtons Second in the horizontal direction, so I'm somewhat perplexed how the block can come to rest w.r.t. the slab? It feels like we need a relative velocity dependence in the frictional force between the block and the slab.

 

[kuruman]

Do you mean the collision with the block and the slab? Can't I just analyze the problem from when the block is placed on the slab?

Sure. The collision starts when the block makes first contact with the slab and is over when the block comes to rest relative to the slab. In the "before" picture only the slab is moving and in the "after" picture the slab and the block move together as one. Isn't that a description of a perfectly inelastic collision?

 

[Thermofox]

Isn't that a description of a perfectly inelastic collision?

Yes, it is. It just doesn't happen immediately, now I see it.

 

[Thermofox]

In my estimation the block and the slab have distinct constant accelerations by Newtons Second in the horizontal direction, so I'm somewhat perplexed how the block can come to rest w.r.t. the slab? It feels like we need a relative velocity dependence in the frictional force between the block and the slab.

Doesn't friction slow down the block?

 

[kuruman]

In my estimation the block and the slab have distinct constant accelerations by Newtons Second in the horizontal direction, so I'm somewhat perplexed how the block can come to rest w.r.t. the slab? It feels like we need a relative velocity dependence in the frictional force between the block and the slab.

Imagine yourself sitting on the rear end of the moving slab. You look around and you see, far ahead, a block flying towards you just above the top of the slab with speed $v_{s,i}$. The block lands at the front end of the slab and starts sliding towards you. Because the slab is rough, the block eventually comes to rest just before falling off the rear end of the slab.

To solve this problem, you pretend that you are the observer on the slab, find how much mechanical energy is converted into heat and subtract that from the initial mechanical energy of the cube + slab system in the frame of the frictionless surface. That's what I meant when I wrote

. . . here you have a way of figuring out how much energy is lost as heat in the collision.

 

[kuruman]

Doesn't friction slow down the block?

Yes it does.

 

[erobz]

Doesn't friction slow down the block?

Friction slows down the slab, it speeds up the block. Look at the end state, if the block and slab are moving at some constant velocity $u$, the net force acting on the small mass is? So the friction either undergoes a step change in value to zero at this instant, or it probably should be dependent on the relative velocity between the block and slab all along. To me something is fishy.

@Thermofox
What do you get when you apply free body diagrams to each mass as a free body?

Spoiler

I get a constant force acting on each block while the little block is sliding w.r.t. the slab

 

[erobz]

Imagine yourself sitting on the rear end of the moving slab. You look around and you see, far ahead, a block flying towards you just above the top of the slab with speed $v_{s,i}$. The block lands at the front end of the slab and starts sliding towards you. Because the slab is rough, the block eventually comes to rest just before falling off the rear end of the slab.

To solve this problem, you pretend that you are the observer on the slab, find how much mechanical energy is converted into heat and subtract that from the initial mechanical energy of the cube + slab system in the frame of the frictionless surface. That's what I meant when I wrote

I can imagine the math there, but to me it is fishy friction physics. I'll concede we are just supposed to ignore it, but I don't like it.

 

[Thermofox]

Friction slows down the slab, it speeds up the block. Look at the end state, if the block and slab are moving at some constant velocity $u$, the net force acting on the small mass is? So the friction either undergoes a step change in value to zero at this instant, or it probably should be dependent on the relative velocity between the block and slab all along. To me something is fishy.

@Thermofox
What do you get when you apply free body diagrams to each mass as a free body?

Spoiler

I get a constant force acting on each block while the little block is sliding w.r.t. the slab

I get that on the block I have a force, $F$ generated by the motion of the slab, that moves the block to the left and the friction force, $f$, that opposes to this motion, so it pushes the block to the right, speeding it up.

On the slab, for the third law of Newton, there is the friction force that pushes the slab to the left, slowing it down.

From my understanding, as the block moves, the slab slows down. This means that the force that pushes the block to the left reduces its intensity over time. The moment that $F=f$, the block should have a velocity of zero. When this happens the block and the slab become a whole object, but this explanation would imply that the block jumps from a velocity of 0 to the velocity of the slab at that moment, which doesn't make sense.

 

[erobz]

I get that on the block I have a force, $F$ generated by the motion of the slab, that moves the block to the left and the friction force, $f$, that opposes to this motion, so it pushes the block to the right, speeding it up.

On the slab, for the third law of Newton, there is the friction force that pushes the slab to the left, slowing it down.

From my understanding, as the block moves, the slab slows down. This means that the force that pushes the block to the left reduces its intensity over time. The moment that $F=f$, the block should have a velocity of zero. When this happens the block and the slab become a whole object, but this explanation would imply that the block jumps from a velocity of 0 to the velocity of the slab at that moment, which doesn't make sense.

Yeah, this is (kindof) what I'm talking about. Newtons Second law using $\mu N$ friction model is fishy.

They want you to take $\mu m g$ acting on the block to the right, and $\mu m g$ acting to the left on the slab. But this implies a constant distinct acceleration for each body. I don't see how them coming to rest w.r.t each other is consistent modeling.

 

[kuruman]

I can imagine the math there, but to me it is fishy friction physics.

There is nothing fishy about the physics of this. If the block slides distance $d$ on the slab before it comes to rest, is it or is it not true that the mechanical energy converted into heat is $Q=\mu_kmgd$?

Will this result be any different if the same block slides the same distance $d$ on a floor fixed firmly to the Earth having the same coefficient of kinetic friction?

 

[erobz]

There is nothing fishy about the physics of this. If the block slides distance $d$ on the slab before it comes to rest, is it or is it not true that the mechanical energy converted into heat is $Q=\mu_kmgd$?

Will this result be any different if the same block slides the same distance $d$ on a floor fixed firmly to the Earth having the same coefficient of kinetic friction?

Newtons Second says the accelerations are constant, and different from each other using the model $f = \mu N$. They can't come to rest w.r.t to each other in this way?

In the horizontal direction $\rightarrow^+$ on the small block of mass $m$

$$ \sum F = \mu m g = m a_b$$

on the slab of mass $M$

$$ \sum F = - \mu m g = M a_s $$

IMHO the simple friction model fails this problem(student). $a_s \neq a_b$

So I would say this:

"is it or is it not true that the mechanical energy converted into heat is $Q=\mu_kmgd$"

is not true or $f =\mu N$ is not true.

 

[erobz]

From my understanding, as the block moves, the slab slows down. This means that the force that pushes the block to the left reduces its intensity over time.

My point is that should decrease in intensity as the slab slows, and it must do so at some point if it's going to move along with the slab at constant velocity, but Newtons laws also seem to be quite simply written and contradictory to this. I can't tell if I'm being pedantic or a goof.

Also, is the work invariant? If I plop my butt on the slab and calculate the work of the NC's, is that the same as me standing on the ground and calculating it? In the accelerating frame we can say the distance the mass slides is $d$, but it wouldn't be $d$ in the inertial frame?

I guess I have more questions than answers (as usual). My intent was not to confuse @Thermofox , but see if that is what was confusing them too. If I'm seeing a phantom, just move on. I'll butt out.

 

[jbriggs444]

From my understanding, as the block moves, the slab slows down. This means that the force that pushes the block to the left reduces its intensity over time.

The force that pushes the block to the left retains its full intensity right up until the instant that the mass comes to relative rest.

 

[erobz]

So friction undergoes a step change:

$$ f =
\begin{cases}
\mu m g & \text{if } |v_{rel}| > 0 \\
0 & \text{if } |v_{rel}| = 0
\end{cases} $$

If you are still having issues ( which I probably had a hand in ) just write some coordinates and apply Newtons Second to each mass as I did in post 15. It's probably not the most efficient as @kuruman proposed, but it's a way.

You can use them to find the time when the velocities $\dot x_M$ and $\dot x_m$ are equal, then you can use the derived position equations to determine $\ell$.

Sorry again if I lead you down a rabbit hole, but don't give up. It will get straightened out.

 

[Thermofox]

Sorry again if I lead you down a rabbit hole, but don't give up. It will get straightened out

Don't worry. I actually think that this interaction can benefit a lot my solving skills and will surely deepen my understanding of this topic, so first of all thanks. Besides the problem asks to find when the 2 blocks acquire the same velocity, therefore I think this type of approach is necessary to solve this problem.

just write some coordinates and apply Newtons Second to each mass as I did in post 15.

If I apply Newton's second law as you did, I obtain $$\begin{cases}
\mu m g = ma_b\\
-\mu m g = Ma_s
\end{cases} \Rightarrow \begin{cases}
a_b= \frac {\mu g m} {m} = \mu g\\
a_s = -\frac {\mu g m} {M}
\end{cases}$$ The accelerations are constant because they depend only by variables that remain constant through out the sliding of the block.

I don't understand what you mean with coordinates. Are you saying to write the positions of the block and of the slab related to time [$x_M(t)$ and $x_m(t)$], since I know that their motion is uniformly accelerated?

 

[jbriggs444]

Don't worry. I actually think that this interaction can benefit a lot my solving skills and will surely deepen my understanding of this topic, so first of all thanks. Besides the problem asks to find when the 2 blocks acquire the same velocity, therefore I think this type of approach is necessary to solve this problem.


If I apply Newton's second law as you did, I obtain $$\begin{cases}
\mu m g = ma_b\\
-\mu m g = Ma_s
\end{cases} \Rightarrow \begin{cases}
a_b= \frac {\mu g m} {m} = \mu g\\
a_s = -\frac {\mu g m} {M}
\end{cases}$$ The accelerations are constant because they depend only by variables that remain constant through out the sliding of the block.

I don't understand what you mean with coordinates. Are you saying to write the positions of the block and of the slab related to time [$x_M(t)$ and $x_m(t)$], since I know that their motion is uniformly accelerated?

Following this approach, what I would do is to first write down formulas for $v_\text{M}(t)$ and $v_\text{m}(t)$ using the initial conditions for $v_\text{M}$ and $v_\text{m}$ at time zero.

Solve for the time $t$ at which the two velocities are equal.

Then you will be in a position to consider the positions of the two blocks at time $t$.

Personally, I would use the "inelastic collision" approach instead of working with accelerations. Shift to the center of mass frame, calculate the total energy prior to the interaction and consider the work drained from the system as the masses slide over each other. Figure out how much distance must be covered to make the work drained equal to the initial energy.

 

[Thermofox]

Personally, I would use the "inelastic collision" approach instead of working with accelerations. Shift to the center of mass frame, calculate the total energy prior to the interaction and consider the work drained from the system as the masses slide over each other. Figure out how much distance must be covered to make the work drained equal to the initial energy.

But with this approach can you determine the time, $t$, when the block and the slab become a whole body? I don't think there is another way to find $t$, right?

 

[jbriggs444]

But with this approach can you determine the time, $t$, when the block and the slab become a whole body? I don't think there is another way to find $t$, right?

There are a few equations that you could use for that.

You know the distance, the initial velocity and the acceleration.
You know the initial velocity, the final velocity and the acceleration.
You know the initial energy, the final energy, the average velocity and the force, hence the average power.

Yes, velocity difference divided by acceleration is the simplest way.

 

[Thermofox]

You know the initial energy, the final energy, the average velocity and the force, hence the average power.

Cool, I didn't think about using power. Thanks for the help!

 

[kuruman]

Since we are discussing alternate approaches, one might also consider solving the problem in the CM frame of slab + block (see figure below).
The velocity of the center of mass is $V_{\text{cm}}=\dfrac{Mv_0}{M+m}.$
The distance $x$ of the slab's midpoint M to the CM is easy to find in terms of $M$, $m$ and $L.$

Now, in the CM frame,
The initial velocity of the slab is $u_0=v_0-V_{\text{cm}}.$
The final velocity of the slab is $u_{\!f}=0.$
The slab is displaced by $\Delta s=2x$ under constant acceleration $a_{\text{slab}}$, therefore
$2a_{\text{slab}}\Delta s=u_{\!f}^2-u_0^2.$
This last equation can be solved for the acceleration of the slab in terms of $M$, $m$ and $L.$
Finally, one sets $Ma_{\text{slab}}=\mu_kmg$ and solves for $L$.

On edit:

Spoiler

The answer to part (2) is surprisingly simple, at least to me, but makes sense in retrospect.

 

[haruspex]

So friction undergoes a step change:

This is the standing passenger effect. At the instant the bus comes to a halt, the torque from the frictional force from the floor suddenly goes away and the passenger, who had been bracing against it, falls backwards.
The technical term is jerk, $\dddot x$.

 

[erobz]

This is the standing passenger effect. At the instant the bus comes to a halt, the torque from the frictional force from the floor suddenly goes away and the passenger, who had been bracing against it, falls backwards.
The technical term is jerk, $\dddot x$.

This is continuous though in (classical) reality?

 

[haruspex]

This is continuous though in (classical) reality?

In any real system there is some elasticity. In a sliding block, there is shear stress as the frictional force is conveyed up through the layers. When the base comes to a stop, the block will take a moment to relax internally.

 

[kuruman]

This is the standing passenger effect. At the instant the bus comes to a halt, the torque from the frictional force from the floor suddenly goes away and the passenger, who had been bracing against it, falls backwards.
The technical term is jerk, $\dddot x$.

The effect is also observed in cars with seated passengers when their car comes to a halt at a stop light. I think is has to do with the sudden change of tactile sensations when the passengers, transit (rather quickly) from being at rest in a non-inertial frame to being at at rest in an inertial frame.

The picture below left shows a schematic bus moving to the right with speed $v$ while braking. The acceleration is to the left. A plumb bob hanging from the ceiling shows the direction of the effective gravity $g_{\text{eff.}}$ which, by definition, is down. A helium balloon taped to the floor points in a direction opposite to down, i.e. up.

The vector diagram in the middle shows how $g_{\text{eff.}}$ is obtained from the acceleration of gravity $g_0= 9.8~\text{m/s}^2$ and the acceleration of the bus.

The diagram on the right shows the physical situation in the non-inertial frame. Standing passengers brace themselves as if they were standing on a floor sloping towards the front of the bus. When the slope suddenly disappears, they find themselves tilted to the left with respect to the floor. Seated passengers (facing front) also feel the change and rotate about their butts with the backrest providing additional torque to stop the rotation.