Conservation of Energy in circular motion

In summary, a block with mass m is initially at rest at the height of 2R on a semicircular track. It then slides without friction. The block will leave the track at a velocity of zero, and the maximum height it will reach after leaving the track can be determined by using the equation mgh = mgh' + 0.5mv'^2. The block will continue to slide until its velocity is zero, and its motion can be described using Newton's Second Law. The angle theta at the moment the block leaves the track is unknown, but the free body diagram can be drawn assuming the block is on the upper part of the track and its angle relative to an imaginary horizontal line through the center of the track
  • #1
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Homework Statement


http://www.sumoware.com/images/temp/xzmomoeqpspoqxcq.png [Broken]

A block with mass m is static at first at the height of 2R (see picture above) and then slides without friction.
a) Determine where the block leaves the track
b) Determine the maximum height which the block reaches after it gets out of the track


Homework Equations


E = E'

The Attempt at a Solution



E = E'
mgh = mgh'
mg(2R) = mgh'
h' = 2R

But, I'm still doubt on it.
I imagine it, and think that the block won't be in the same height as the initial height (since the gravity must pull it down)..
Please help me
 
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  • #2
I assume the curve at the right is a semicircle. For your conservation law you've written E = E', but what is included in these?
 
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  • #3
Your final energy should include the kinetic energy because your system is "Block-Earth" system. (In my opinion, it's better to say "Block-Gravitational field" system, but it's not common in physics textbook.)

Therefore, mgh = mgh' + 0.5mv'^2
 
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  • #4
haruspex said:
I assume the curve at the right is a semicircle. For your conservation law you've written E = E', but what is included in these?

Ep + Ek = Ep' + Ek'
mgh + 0.5mv^2 = mgh' + 0.5mv'^2

The initial velocity is zero.
And, when the block is about to leave the track, its velocity is zero.
So, the equation become
mgh = mgh'
Philethan said:
Your final energy should include the kinetic energy because your system is "Block-Earth" system. (In my opinion, it's better to say "Block-Gravitational field" system, but it's not common in physics textbook.)

Therefore, mgh = mgh' + 0.5mv'^2

mg(2R) = mgh' + 0.5mv'^2
So, what's the velocity when the block leaves the track ?
I think it must be zero at the time the block is about to leave the track..
 
  • #5
terryds said:
Ep + Ek = Ep' + Ek'
mgh + 0.5mv^2 = mgh' + 0.5mv'^2

The initial velocity is zero.
And, when the block is about to leave the track, its velocity is zero.
So, the equation become
mgh = mgh'

mg(2R) = mgh' + 0.5mv'^2
So, what's the velocity when the block leaves the track ?
I think it must be zero at the time the block is about to leave the track..
What makes you think the velocity must be zero? What are the forces on the block while it is on the arc? What equation relating to those forces expreses the condition that the block is about to leave the track?
 
  • #6
terryds said:
Ep + Ek = Ep' + Ek'
mgh + 0.5mv^2 = mgh' + 0.5mv'^2

The initial velocity is zero.
And, when the block is about to leave the track, its velocity is zero.
So, the equation become
mgh = mgh'

mg(2R) = mgh' + 0.5mv'^2
So, what's the velocity when the block leaves the track ?
I think it must be zero at the time the block is about to leave the track..
That's great! So we must check if the final velocity is zero or not.
No matter what its final velocity is, it must obey Newton's Second Law, right?
Now, please imagine its motion.

You may think this block will continue to slide on the track until its velocity is zero.
If you think that way, then draw its free body diagram right before it's at rest on the track.
If you have another picture in your mind, then please tell us and also try to draw its FBD right before it's at rest on the track.
 
  • #7
haruspex said:
What makes you think the velocity must be zero? What are the forces on the block while it is on the arc? What equation relating to those forces expreses the condition that the block is about to leave the track?

Philethan said:
That's great! So we must check if the final velocity is zero or not.
No matter what its final velocity is, it must obey Newton's Second Law, right?
Now, please imagine its motion.

You may think this block will continue to slide on the track until its velocity is zero.
If you think that way, then draw its free body diagram right before it's at rest on the track.
If you have another picture in your mind, then please tell us and also try to draw its FBD right before it's at rest on the track.

Hmm...
I think when it is about to leave the track, the velocity becomes zero.. (Because l think that after leaving the track, the velocity becomes negative or the block will go backward)

This is the free body diagram I sketched
http://www.sumoware.com/images/temp/xzibahthtpxmektk.png [Broken]

But, the problem is that the angle theta is unknown
 
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  • #8
terryds said:
Hmm...
I think when it is about to leave the track, the velocity becomes zero.. (Because l think that after leaving the track, the velocity becomes negative or the block will go backward)

This is the free body diagram I sketched
http://www.sumoware.com/images/temp/xzibahthtpxmektk.png [Broken]

But, the problem is that the angle theta is unknown
What can you say about one of those forces at the moment it leaves the track?
 
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  • #9
terryds said:
Hmm...
I think when it is about to leave the track, the velocity becomes zero.. (Because l think that after leaving the track, the velocity becomes negative or the block will go backward)

This is the free body diagram I sketched
http://www.sumoware.com/images/temp/xzibahthtpxmektk.png [Broken]

But, the problem is that the angle theta is unknown
Excellent! First of all, I suggest you remove the W vector, and only leave its components.
Otherwise, I'm afraid the W vector will make you misunderstand something.

Now, think about if it will fall when it's on lower part of the track.
I guess you'll say no! But why? I think it's good to prove it for yourself.
Therefore, I guess you'll agree with me that this block will fall unless it's on upper part of the track.
Then, please draw its free body diagram again. Assume its angle relative the the imaginary horizontal
line which pass through the center of the track.

And the most important thing is, please write down its Newton's Second Law.
But be careful! The moment when you're analyzing is right before it falls down.
So it still moves. Well, moves very very slowly. That's ok.
All you need to do is analyzing its Newton's Second Law.

[Hint: its acceleration must be centrifugal acceleration AND check if the magnitude of force is reasonable.]
 
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  • #10
Philethan said:
Well, moves in very very small speed.
No.
 
  • #11
haruspex said:
No.
I know what the answer is, but I need to show him the velocity isn't zero when the block leaves the track :P
 
  • #12
Philethan said:
I know what the answer is, but I need to show him the velocity isn't zero when the block leaves the track :P
So why assert it's very small?
 
  • #13
Philethan said:
Excellent! First of all, I suggest you remove the W vector, and only leave its components.
Otherwise, I'm afraid the W vector will make you misunderstand something.

haruspex said:
So why assert it's very small?
He said "I think it must be zero at the time the block is about to leave the track" and
I need to follow his reasoning to persuade him by his own reasoning,
so I guess he would think the velocity is very small but not equal to zero right before it falls down.
If he doesn't think that way, then I should ask him what he would think right before it falls down.
Whatever he might think, all I need to do is draw the free body diagram and let him know it's impossible that it rests on the track right before it falls down.
 
  • #14
Philethan said:
I know what the answer is, but I need to show him the velocity isn't zero when the block leaves the track :P
Why is it very very small ?
I think when it is about to leave, its velocity is zero.
Because, I think, if it leaves, its velocity must be negative..

haruspex said:
What can you say about one of those forces at the moment it leaves the track?
I can say that
What makes it leave the track is the normal force or the W sin theta
By the point of the block,
ΣFy = m ay
N - W cos Θ = m ay (with Θ is the angle from the horizontal axis of the center of semi-circle track)

ΣFx = m ax
W sin Θ = m ax

Then, how to find the position the block leaves the track ?
 
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  • #15
Philethan said:
He said "I think it must be zero at the time the block is about to leave the track" and
I need to follow his reasoning to persuade him by his own reasoning,
so I guess he would think the velocity is very small but not equal to zero right before it falls down.
If he doesn't think that way, then I should ask him what he would think right before it falls down.
Whatever he might think, all I need to do is draw the free body diagram and let him know it's impossible that it rests on the track right before it falls down.
Ok, but it was not clear that the statement was predicated on "following your reasoning", so was in danger of being misleading.
I would suggest another thought experiment. Which way was it moving just before velocity became zero?
 
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  • #16
terryds said:
Why is it very very small ?
I think when it is about to leave, its velocity is zero.
Because, I think, if it leaves, its velocity must be negative..I can say that
What makes it leave the track is the normal force or the W sin theta
By the point of the block,
ΣFy = m ay
N - W cos Θ = m ay (with Θ is the angle from the horizontal axis of the center of semi-circle track)

ΣFx = m ax
W sin Θ = m ax

Then, how to find the position the block leaves the track ?
That's right. But we need to analyze the moment right before it falls down.
So what's the block doing before it falls down?
You might think it still moves on the track, right?
Then just analyze that and you'll find the normal force does not direct inward, but outward.
That means the normal force is not repulsive force anymore, it becomes attractive force. It's impossible.
So that's what I want you to discover by yourself.
 
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  • #17
haruspex said:
Ok, but it was not clear that the statement was predicated on "following your reasoning", so was in danger of being misleading.
I would suggest another thought experiment. Which way was it moving just before velocity became zero?
thanks for suggestion. I think that's what I want to tell him.
 
  • #18
terryds said:
Why is it very very small ?
I think when it is about to leave, its velocity is zero.
Because, I think, if it leaves, its velocity must be negative..I can say that
What makes it leave the track is the normal force or the W sin theta
By the point of the block,
ΣFy = m ay
N - W cos Θ = m ay (with Θ is the angle from the horizontal axis of the center of semi-circle track)

ΣFx = m ax
W sin Θ = m ax

Then, how to find the position the block leaves the track ?
The normal force cannot make it leave the track. The normal force is merely a reaction. You can think of it as the force required to stop the block going through the track. Anyway, I didn't mean what force makes it leave the track (that is gravity, surely). I meant, what equation can you write concerning one or more of those forces to express the condition that the block is just leaving the track? Hint: what forces act on it after leaving the track? How is this different from the forced that acted on it while on the track?

You are taking the y direction as radial and the x as tangential, yes? If an object is going in a circle at constant speed, what is its acceleration?
 
  • #19
Philethan said:
That's right. But we need to analyze the moment right before it falls down.
So what's the block doing before it falls down?
You might think it still moves on the track, right?
Then just analyze that and you'll find the normal force does not direct inward, but outward.
That means the normal force is not repulsive force anymore, it becomes attractive force. It's impossible.
So that's what I want you to discover by yourself.

Philethan said:
That's right. But we need to analyze the moment right before it falls down.
So what's the block doing before it falls down?
You might think it still moves on the track, right?
Then just analyze that and you'll find the normal force does not direct inward, but outward.
That means the normal force is not repulsive force anymore, it becomes attractive force. It's impossible.
So that's what I want you to discover by yourself.
haruspex said:
The normal force cannot make it leave the track. The normal force is merely a reaction. You can think of it as the force required to stop the block going through the track. Anyway, I didn't mean what force makes it leave the track (that is gravity, surely). I meant, what equation can you write concerning one or more of those forces to express the condition that the block is just leaving the track? Hint: what forces act on it after leaving the track? How is this different from the forced that acted on it while on the track?

You are taking the y direction as radial and the x as tangential, yes? If an object is going in a circle at constant speed, what is its acceleration?
If it's going in a circle at constant speed, it will just have the gravitational acceleration, right ?

I've just re-sketched the free body diagram
http://www.sumoware.com/images/temp/xzfrssslaqaqitsk.png [Broken]

∑Fcentripetal = m acentripetal
W sin Θ = m v^2/R
m g sin Θ = m v^2/R
v = √(g R sin Θ)

So, the velocity is √(g R sin Θ) , right ?

So,
E = E'
mgh = mgh' + 0.5mv'^2
g(2R) = g(R+R sin Θ)+ 0.5 g R sin Θ
2gR = gR+gR sin Θ+0.5 g R sin Θ
gR = 1.5 g R sin Θ
sin Θ = 2/3

So, it will leave at the position R + R sin Θ = R + R (2/3) = 5/3 R
Right ?
 
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  • #20
Philethan said:
thanks for suggestion. I think that's what I want to tell him.
a) Determine where the block leaves the track
b) Determine the maximum height of the track which the block reaches after it gets out of the track


The answer to question b is the same as the answer to question a , right ?
 
  • #21
terryds said:
a) Determine where the block leaves the track
b) Determine the maximum height of the track which the block reaches after it gets out of the track


The answer to question b is the same as the answer to question a , right ?
Yes:)
Sorry I was wrong. The block's vertical component velocity is not zero after it leaves the track so that its highest point is not the same as the point at which it leaves the track.
 
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  • #22
terryds said:
a) Determine where the block leaves the track
b) Determine the maximum height which the block reaches after it gets out of the track


The answer to question b is the same as the answer to question a , right ?

No! It's vertical velocity is not 0 at the point it leaves the track, so it moves a bit higher.
 
  • #23
PeroK said:
No! It's vertical velocity is not 0 at the point it leaves the track, so it moves a bit higher.
Yup, that's right.. But, I think it doesn't matter since its limit is zero
 
  • #24
terryds said:
Yup, that's right.. But, I think it doesn't matter since its limit is zero

That doesn't make sense. The block will move up in a parabola under the ring. So, its highest point is not where it leaves the track.
 
  • #25
Philethan said:
Yes:)
Sorry I was wrong. The block's vertical component velocity is not zero after it leaves the track so that its highest point is not the same as the point at which it leaves the track.

PeroK said:
That doesn't make sense. The block will move up in a parabola under the ring. So, its highest point is not where it leaves the track.

Hmm..
I see.
The block will go up and fall down in parabolic motion.
The vertical velocity is √(g R sin Θ) = √(2/3 g R)
The highest point is reached when the vertical velocity becomes zero
V(t) ^ 2 = V(0) ^ 2 + 2 a Δy
0 = 2/3 g R - 2 g Δy
Δy = 1/3 R
y = 5/3 R + 1/3 R = 2R

Right ?
 
  • #26
terryds said:
Hmm..
I see.
The block will go up and fall down in parabolic motion.
The vertical velocity is √(g R sin Θ) = √(2/3 g R)
The highest point is reached when the vertical velocity becomes zero
V(t) ^ 2 = V(0) ^ 2 + 2 a Δy
0 = 2/3 g R - 2 g Δy
Δy = 1/3 R
y = 5/3 R + 1/3 R = 2R

Right ?

You should be able to see why it can't be 2R. Think about conservation of energy.

Your mistake was to consider all its velocity to be vertical.
 
  • #27
PeroK said:
You should be able to see why it can't be 2R. Think about conservation of energy.

Your mistake was to consider all its velocity to be vertical.

I've drawn the diagram

http://www.sumoware.com/images/temp/xzolessrnjoxfhei.png [Broken]
Its vertical velocity is Vy = V sin Θ
Vy = sin Θ √(g R sin Θ)
Vy = √(g R (sin Θ)^3 )

Then,
V(t) ^ 2 = V(0) ^ 2 + 2 a Δy
0 = g R (sin Θ)^3 - 2 g Δy
Δy = 1/2 R(sin Θ)^3
Δy = 1/2 R (2/3)^3
Δy = 8/54 R= 4/27 R
y = 5/3 R + 4/27R = 49/27 R

Right ?
Please help me
 
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  • #28
terryds said:
Its vertical velocity is Vy = V sin Θ
Not with the way you've defined Θ. Consider, as Θ approaches pi/2, is the vertical velocity increasing or tending to zero?
terryds said:
Vy = sin Θ √(g R sin Θ)
Given your previous equation, that implies V = √(g R sin Θ). Again, as Θ increases, will V increase or decrease? To get the KE at the position in the diagram, calculate how far the block is below its starting point.
 
  • #29
haruspex said:
Not with the way you've defined Θ. Consider, as Θ approaches pi/2, is the vertical velocity increasing or tending to zero?

Given your previous equation, that implies V = √(g R sin Θ). Again, as Θ increases, will V increase or decrease? To get the KE at the position in the diagram, calculate how far the block is below its starting point.

I think when It's approaching to pi/2, its vertical velocity is tending to zero. And, as Θ increases, the vertical velocity will decrease.
But, I think that the peak of parabola motion path is not the same of the peak of the semicircle. So, I think I can't say that Θ is pi/2 when the block has reached the maximum height.
And, I think I can't say that the kinetic energy is zero because the horizontal velocity is not zero.
The horizontal velocity is the x-component of its tangential velocity when it left the the track, right ? I think that the horizontal velocity is constant because there is no force acting horizontally on the block.

How far the block is below its starting point ? I don't know.
Its starting point is 2R.
And, the block will slide and the go up in the semicircle.
Then, it will leave the track and moves along a parabolic path
The peak of the parabolic path is unknown ( It will be the height when it is about to leave the track + the maximum height gained when moving in parabolic motion)

Please help me and explain the correct solution. :)
 
  • #30
terryds said:
the peak of parabola motion path is not the same of the peak of the semicircle.
Sure, but one step at a time. You cannot start analysing the parabolic arc until you've figured out at what angle the block leaves the track.
You need to obtain the speed as a function of angle (while on the track) using conservation of energy. Then you need think about forces to discover at what combination of speed and angle it will fall off the track.
 
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  • #31
haruspex said:
Sure, but one step at a time. You cannot start analysing the parabolic arc until you've figured out at what angle the block leaves the track.
You need to obtain the speed as a function of angle (while on the track) using conservation of energy. Then you need think about forces to discover at what combination of speed and angle it will fall off the track.

In my previous post in this thread, I've figured that at the angle sin-1 2/3 or at the height of R+R sin 2/3 , the block will leave the track.
Please help me.. I really have no idea to obtain the speed as function of the angle.
I'll be very glad if you give me the solution or the hint and explain it :)
 
  • #32
terryds said:
In my previous post in this thread, I've figured that at the angle sin-1 2/3 or at the height of R+R sin 2/3 , the block will leave the track.
Ok, though I still can't see where you did that - but it's the right answer.
So you can work out the KE there, and thus the speed. And you know the 'launch' angle, since it starts tangential to the arc. You've studied uniform acceleration and trajectories, yes? What equations do you know for those (particularly in relation to vertical motion)?
 
  • #33
haruspex said:
Ok, though I still can't see where you did that - but it's the right answer.
So you can work out the KE there, and thus the speed. And you know the 'launch' angle, since it starts tangential to the arc. You've studied uniform acceleration and trajectories, yes? What equations do you know for those (particularly in relation to vertical motion)?

This is how I solved the angle and the tangential velocity: https://www.physicsforums.com/threads/conservation-of-energy-in-circular-motion.795189/#post-4993826

The formulas I know are
Vy(0) = V sin Θ
Vx(0) = V cos Θ
Vy(t) = Vy(0) - g t

The Y max is reached when the velocity in y-axis is zero
Vy(t)^2 = Vy(0)^2 - 2 g Δy
0 = (V sin Θ)^2 - 2 g Δy
0 = (√(g R sin Θ) sinΘ)^2 - 2 g Δy
0 = g R (sin Θ)^3 - 2 g Δy
Δy = 1/2 R(sin Θ)^3
Δy = 1/2 R 8/27
Δy = 8/54 R
So, the max height is R + 2/3 R + 8/54 R = 49/27 R

Actually, I have figured it out in my previous post.

Am I right ? Or do I miss something ?

I'm just doubt on the launch angle value. Is it the same angle as the angle from the horizontal to the position where the block leaves the track ? My intuition says that it's same.
 
  • #34
terryds said:
This is how I solved the angle and the tangential velocity: https://www.physicsforums.com/threads/conservation-of-energy-in-circular-motion.795189/#post-4993826

The formulas I know are
Vy(0) = V sin Θ
Vx(0) = V cos Θ
Vy(t) = Vy(0) - g t

The Y max is reached when the velocity in y-axis is zero
Vy(t)^2 = Vy(0)^2 - 2 g Δy
0 = (V sin Θ)^2 - 2 g Δy
0 = (√(g R sin Θ) sinΘ)^2 - 2 g Δy
0 = g R (sin Θ)^3 - 2 g Δy
Δy = 1/2 R(sin Θ)^3
Δy = 1/2 R 8/27
Δy = 8/54 R
So, the max height is R + 2/3 R + 8/54 R = 49/27 R

Actually, I have figured it out in my previous post.

Am I right ? Or do I miss something ?

I'm just doubt on the launch angle value. Is it the same angle as the angle from the horizontal to the position where the block leaves the track ? My intuition says that it's same.
I pointed out in post #28 that you have some errors in that calculation.
 
  • #35
haruspex said:
I pointed out in post #28 that you have some errors in that calculation.

haruspex said:
Given your previous equation, that implies V = √(g R sin Θ). Again, as Θ increases, will V increase or decrease? To get the KE at the position in the diagram, calculate how far the block is below its starting point
As Θ increases, V increases since it is sinus function (in 0-90 degree quadrant).

Hmm.. I think I almost figured it out, right ? The only thing I miss is just the launch angle (Let's call this α )
http://www.sumoware.com/images/temp/xzihalbxxfsximxp.png [Broken]
Could you please tell me what is the relation between α and Θ ?
 
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