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Conservation of Energy in circular motion

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  • #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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Answers and Replies

  • #2
haruspex
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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?
 
  • #3
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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
 
  • #4
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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'


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
haruspex
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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
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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
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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?
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
haruspex
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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
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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
haruspex
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  • #11
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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
haruspex
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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
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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.
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
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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..

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 ?
 
  • #15
haruspex
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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?
 
  • #16
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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.
 
  • #17
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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
haruspex
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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
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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.
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.
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
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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
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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
PeroK
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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
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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
PeroK
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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
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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.
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 ?
 

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