Energy analysis of this system (curved ramp up and a spring)

[Thermofox]

Homework Statement

I need to determine
1)the velocity that the block needs to have to compress the spring by 40cm
2)The normal reaction on the circular path (BC) when the block has described an arc of 45°

Relevant Equations

See below

For point one it's clear that I have to use energy
=> $ΔE_{AB} = W_{friction}$ ; $\frac 1 2 mv_0^2 - \frac 1 2 mv_1^2 = mgμ_d d$
After that there is the path BC, but I don't know how to analyze it from an energetic standpoint.
Then after BC the block will now have a different velocity, I assume, I'm calling $v_2$.
=> If I set the zero of the gravitational potential on $Δ_{y,max}$ I have that:
$ΔE_{CSpring}= 0$ ; $\frac 1 2 mv_2^2 + mgh = \frac 1 2 kΔ_{y,max}^2$.

As for point 2 I don't know how I should handle it. The fact that the block covers an arc of $45°$ means that it will cover an eight of a circumference, but I don't understand how I should determine the normal reaction.

 

[jbriggs444]

After that there is the path BC, but I don't know how to analyze it from an energetic standpoint.
Then after BC the block will now have a different velocity, I assume, I'm calling $v_2$.
=> If I set the zero of the gravitational potential on $Δ_{y,max}$ I have that:

By $\Delta_{y,max}$, do you mean the position of the block when the spring is compressed by the requisite 40 centimeters?

What energy losses can you identify for the block on its path from $B$ to $C$ to the top of $h$ to the top of $\Delta y$?

What energy losses can you identify for the block on its path from $B$ to the 45 degree mark on its way to $C$?

 

[Thermofox]

By $\Delta_{y,max}$, do you mean the position of the block when the spring is compressed by the requisite 40 centimeters?

What energy losses can you identify for the block on its path from $B$ to $C$ to the top of $h$ to the top of $\Delta y$?

What energy losses can you identify for the block on its path from $B$ to the 45 degree mark on its way to $C$?

With $\Delta_{y,max}$ I mean the maximum compression of the spring which is 40cm.
From what I see the only thing that happens on BC is that the velocity of the block changes in its direction but not on its magnitude

 

[jbriggs444]

From what I see the only thing that happens on BC is that the velocity of the block changes in its direction but not on its magnitude

What about gravity?

 

[Thermofox]

What about gravity?

Right, I didn't see that. Now the weight of the block isn't countered by the normal reaction of the plane. Therefore now I have a force that pushes the block down. I suppose that this force can be decomposed into a radial and a tangential component, but I still can't quite see how it influences energy.

 

[jbriggs444]

Right, I didn't see that. Now the weight of the block isn't countered by the normal reaction of the plane. Therefore now I have a force that pushes the block down. I suppose that this force can be decomposed into a radial and a tangential component, but I still can't quite see how it influences energy.

The beauty of potential energy is that you can ignore the details of the force and only examine the starting and ending positions.

 

[Thermofox]

The beauty of potential energy is that you can ignore the details of the force and only examine the starting and ending positions.

Then I can completely disregard the arc of circumference and analyze the energy in B and when the spring reaches maximum compression?

 

[Thermofox]

Also I was thinking that when the block is halfway through the path BC, thus at $45°$, the normal reaction of the ramp will be equal and opposite to the radial component of the weight force.
=> $N= F_{weight} cos(45°)$ , where $N$ is the normal reaction.

 

[kuruman]

The beauty of potential energy is that you can ignore the details of the force and only examine the starting and ending positions.

Except if there is friction. In that case the energy dissipated by friction needs some attention because the normal force is variable as the block goes around the track. We do not have the complete statement of the problem but OP thinks there is

For point one it's clear that I have to use energy
=> $ΔE_{AB} = W_{friction}$ ; $\frac 1 2 mv_0^2 - \frac 1 2 mv_1^2 = mgμ_d d$

and what do I know without the complete statement of the problem?

To @Thermofox: For future reference please provide the statement of the problem as was given to you. Telling us what you need to determine is your own interpretation and that could be faulty or you could have overlooked something. Let us be the judge and agree or (disagree) with you about what you need to determine.

 

[haruspex]

Except if there is friction.

Based on the diagram, the arc is smooth.

$\frac 1 2 mv_2^2 + mgh = \frac 1 2 kΔ_{y,max}^2$.

A couple of problems there.
As @jbriggs444 hinted in post #2, gravity doesn’t switch off when the spring is reached.
More significantly, gravity will slow the rise, not speed it up.

 

[jbriggs444]

Except if there is friction. In that case the energy dissipated by friction needs some attention because the normal force is variable as the block goes around the track.

Indeed. I am reading into the diagram that the horizontal section of the track is rough but that the curved portion is smooth.

 

[jbriggs444]

Also I was thinking that when the block is halfway through the path BC, thus at $45°$, the normal reaction of the ramp will be equal and opposite to the radial component of the weight force.
=> $N= F_{weight} cos(45°)$ , where $N$ is the normal reaction.

Consider that force balance again. You are missing something.

You are apparently attempting a force balance in the normal direction. You have identified two forces with non-zero components in the normal direction. What is the resulting acceleration in that direction?

 

[kuruman]

Based on the diagram, the arc is smooth.

Indeed. I am reading into the diagram that the horizontal section of the track is rough but that the curved portion is smooth.

Thanks, I stand corrected. Still, I dislike having to deduce the problem description from a diagram.

 

[Thermofox]

Except if there is friction. In that case the energy dissipated by friction needs some attention because the normal force is variable as the block goes around the track. We do not have the complete statement of the problem but OP thinks there is

and what do I know without the complete statement of the problem?

To @Thermofox: For future reference please provide the statement of the problem as was given to you. Telling us what you need to determine is your own interpretation and that could be faulty or you could have overlooked something. Let us be the judge and agree or (disagree) with you about

I'm sorry, it wasn't my intention to cause uncertainty. I, unfortunately, have this rut of omitting the context of what I'm talking about. People never fully understand what I'm trying to say and it is frustrating both for me and the person I'm talking to. But I'm aware of that. Next time I'll try to be as concise as possible and will try my best not to take anything for granted. Thanks for saying it explicitly.

 

[Thermofox]

Consider that force balance again. You are missing something.

You are apparently attempting a force balance in the normal direction. You have identified two forces with non-zero components in the normal direction. What is the resulting acceleration in that direction?

There is also a centripetal acceleration $a_c=\frac {v_1^2} r$
=>$N=F_W + ma_c\frac {v_1^2} r$?

 

[haruspex]

There is also a centripetal acceleration $a_c=\frac {v_1^2} r$

Yes, there is centripetal acceleration, but also tangential acceleration.
If you are trying to write the general equation along the ramp, you need to allow an arbitrary speed since it will decrease.

Are you trying to solve the problem via forces out of interest? It is far easier just using energy.

=>$N=F_W + ma_c\frac {v_1^2} r$?

I hope you meant $N=F_W + ma_c=F_W+m\frac {v_1^2} r$, but that's not right either.
First, Newton's equation is $F_{net}=ma$. To be safe against sign errors, always write it in that form, not mixing forces and ma on the same side.
Assuming Fw is the weight, the overall acceleration is given by $m\vec a=\vec N+\vec F_W$.
If you want to break that into components, you can either do it by horizontal and vertical or by tangential and normal. In each case, you have to choose which directions are positive.
For tangential/normal, up positive:
$m a_c= N- F_W\cos(\theta)$
$m a_t= - F_W\sin(\theta)$
where theta is the angle to the horizontal.

 

[Thermofox]

Yes, there is centripetal acceleration, but also tangential acceleration.
If you are trying to write the general equation along the ramp, you need to allow an arbitrary speed since it will decrease.

Are you trying to solve the problem via forces out of interest? It is far easier just using energy.

I hope you meant $N=F_W + ma_c=F_W+m\frac {v_1^2} r$, but that's not right either.
First, Newton's equation is $F_{net}=ma$. To be safe against sign errors, always write it in that form, not mixing forces and ma on the same side.
Assuming Fw is the weight, the overall acceleration is given by $m\vec a=\vec N+\vec F_W$.
If you want to break that into components, you can either do it by horizontal and vertical or by tangential and normal. In each case, you have to choose which directions are positive.
For tangential/normal, up positive:
$m a_c= N- F_W\cos(\theta)$
$m a_t= - F_W\sin(\theta)$
where theta is the angle to the horizontal.

How can I find $N$ with energy and avoid such a hustle? I can't see how I would do it.

 

[haruspex]

How can I find $N$ with energy and avoid such a hustle? I can't see how I would do it.

What forms of mechanical energy are involved here? What is the change in each?

 

[Thermofox]

Kinetic and gravitational potential.
$ΔE_K= E_{K,f}-E_{K,i}= \frac 1 2 mv_2^2 - \frac 1 2 mv_1^2$
$ΔE_U= E_{U,f}-E_{U,i}= mg\frac R 2 - 0$

 

[haruspex]

Kinetic and gravitational potential.
$ΔE_K= E_{K,f}-E_{K,i}= \frac 1 2 mv_2^2 - \frac 1 2 mv_1^2$
$ΔE_U= E_{U,f}-E_{U,i}= mg\frac R 2 - 0$

Why R/2?

 

[Thermofox]

Why R/2?

Because if goes till $45°$, so halfway through the arc, doesn't this mean that the block is $\frac R 2$ high from the ground?

 

[haruspex]

Because if goes till $45°$, so halfway through the arc, doesn't this mean that the block is $\frac R 2$ high from the ground?

First, I thought we were still on part 1. Is that done with? If so, which post completed that?
Secondly, at 45° it is not R/2 from the ground. That would be after 60°.

 

[Thermofox]

First, I thought we were still on part 1. Is that done with? If so, which post completed that?
Secondly, at 45° it is not R/2 from the ground. That would be after 60°.

Then I can completely disregard the arc of circumference and analyze the energy in B and when the spring reaches maximum compression?

$\begin{cases} \frac 1 2 mv_1^2 + mg(h+R) = \frac 1 2 kΔ_{y,max}^2~ \text{where y=0 is positioned at} \Delta_{y,max}\\ \frac 1 2 mv_1^2 - \frac 1 2 mv_0^2 = -mgμ_d d \end{cases}$
I don't have the velocities but I can derive from this system since I have 2 equations and 2 unknowns.
The first equation is $\Delta E(\text{from B to}\space \Delta_{y,max})$
the second equation is $\Delta E(\text{from A to B})$
This should be point 1.

As for point 2 you're right, now I see it. Then $h_{45°} = R\space cos(45°)$ Right?

 

[haruspex]

$\frac 1 2 mv_1^2 + mg(h+R) = \frac 1 2 kΔ_{y,max}^2~ \text{where y=0 is positioned at} \Delta_{y,max}$

You are still getting wrong two things I pointed out in post #10.

$h_{45°} = R\space cos(45°)$ Right?

Depends how you are defining $h_{45°}$

 

[Thermofox]

You are still getting wrong two things I pointed out in post #10.

$\frac 1 2 mv_1^2 - mg(h+R) = \frac 1 2 kΔ_{y,max}^2$
At the end there is no gravitational potential because I'm saying that at that height, $h=0$. Since I'm using this condition, the height in B is negative hence $h_B= - (h + R)$.

Depends how you are defining $h_{45°}$

It is the height that the block has from the ground when the block describes an arc of 45°

 

[haruspex]

$\frac 1 2 mv_1^2 - mg(h+R) = \frac 1 2 kΔ_{y,max}^2$
At the end there is no gravitational potential because I'm saying that at that height, $h=0$. Since I'm using this condition, the height in B is negative hence $h_B= - (h + R)$.

What is the total height through which it ascends from the bottom of the ramp to when the spring reaches its maximum compression?

$h_{45°} = R \cos(45°)$

It is the height that the block has from the ground when the block describes an arc of 45°

Draw the triangle with R as the hypotenuse and $\theta$ as the angle the block has described. Which length in the diagram is $R\cos(\theta)$?

LaTeX tip: for standard trig functions, like cos, put a backslash in front. It looks better, making the script roman instead of italic, and you don't then need \space.

 

[Thermofox]

What is the total height through which it ascends from the bottom of the ramp to when the spring reaches its maximum compression?

I should've said $h_B= -(h+R+\Delta_{y,max})$.

Draw the triangle with R as the hypotenuse and as the angle $\theta$ the block has described. Which length in the diagram is $R\cos\theta$ ?

$R\cos\theta$ tells me the horizontal displacement from $B$. Therefore $R\sin\theta$ should give me the vertical displacement, but if i draw the triangle it doesn't look like it. $R\sin\theta$ doesn't give me all $h_{\theta}$, it just gives me a portion of it.

 

[SammyS]

( I wrote this up and had it ready some time beforw @haruspex post. I simply failed to Post it.)

Depends how you are defining $h_{45°}$

It is the height that the block has from the ground when the block describes an arc of 45°

Then $\displaystyle h_{45°} = R\ \cos(45^\circ)$ is not correct. What that is is the vertical distance (downward) of the block from the center of the arc, when the block is at the $45^\circ$ position.

 

[Thermofox]

( I wrote this up and had it ready some time beforw @haruspex post. I simply failed to Post it.)

Then $\displaystyle h_{45°} = R\ \cos(45^\circ)$ is not correct. What that is is the vertical distance (downward) of the block from the center of the arc, when the block is at the $45^\circ$ position.

So is $h_{45°}= R-R\cos(45°)$ right? When the blocks reaches $C$, $\theta = 90°$.
=> $h_{90°}= R - 0$ which is true. So even if it is wrong I must be close.

 

[haruspex]

So is $h_{45°}= R-R\cos(45°)$ right?

Yes