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How can a particle in circular motion about a fixed point accelerate, if the point doesn't too?
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When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But this is not possible according to the string constraint. Where am I wrong?
newtonian-mechanics acceleration
edited 4 mins ago
knzhou
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Harsh Somani
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up vote
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When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But this is not possible according to the string constraint. Where am I wrong?
newtonian-mechanics acceleration
edited 4 mins ago
knzhou
37.3k9104180
asked 5 hours ago
Harsh Somani
211
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Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
What do you mean by string constraint?
– ayc
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But this is not possible according to the string constraint. Where am I wrong?
newtonian-mechanics acceleration
edited 4 mins ago
knzhou
37.3k9104180
asked 5 hours ago
Harsh Somani
211
New contributor
Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But this is not possible according to the string constraint. Where am I wrong?
newtonian-mechanics acceleration
newtonian-mechanics acceleration
edited 4 mins ago
knzhou
37.3k9104180
asked 5 hours ago
Harsh Somani
211
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Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 4 mins ago
knzhou
37.3k9104180
asked 5 hours ago
Harsh Somani
211
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edited 4 mins ago
knzhou
37.3k9104180
edited 4 mins ago
knzhou
37.3k9104180
edited 4 mins ago
knzhou
37.3k9104180
37.3k9104180
asked 5 hours ago
Harsh Somani
211
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Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 5 hours ago
Harsh Somani
211
asked 5 hours ago
Harsh Somani
211
211
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Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Harsh Somani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
What do you mean by string constraint?
– ayc
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago
add a comment |Â
2
What do you mean by string constraint?
– ayc
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago
2
2
What do you mean by string constraint?
– ayc
4 hours ago
What do you mean by string constraint?
– ayc
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago
add a comment |Â
5 Answers
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I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?
So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?
So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.
If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.
Hope that answered your question!
answered 4 hours ago
CuriousHegemon
314
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actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
 |Â
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accelerations of both the ends along the string is same if the string is not slacked
Now, I understand your problem.
If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.
In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.
Now,since string isn't slackened, is the acceleration of both ends same?
I want to explain what's happening in terms of forces rather than acceleration:
Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.
The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.
The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.
answered 4 hours ago
ayc
38418
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I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
A lot of good answers already, but I will start by discussing what context you heard this in.
This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.
Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $omega=v/r$,
$$a=fracv^2r=omega^2r$$
So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.
The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($0$), so the acceleration must be $0$ as well.
So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.
answered 3 hours ago
Aaron Stevens
5,9692830
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I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$beginalign
mathbfv_0(t) =& 0
\ mathbfv_1(t) =& rcdotomegacdotbeginpmatrix-sin(omegacdot t)\cos(omegacdot t)endpmatrix
endalign$$
The radial component of $mathbfv_1$ is indeed always zero
$$
mathbfv_1(t)cdotmathbfe_mathrmr
= rcdotomegacdotBigl(-cos(omegacdot t)cdotsin(omegacdot t) + sin(omegacdot t)cdotcos(omegacdot t)Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
fracmathrmdmathbfv_1mathrmdt = rcdotomega^2cdotbeginpmatrix-cos(omegacdot t)\-sin(omegacdot t)endpmatrix
= -rcdotomega^2cdot mathbfe_mathrmr.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
answered 40 mins ago
leftaroundabout
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To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving in one direction with no rotation, all points in the system move with the same acceleration.
In any other case (curved motion, rotation of the system, etc.) this is no longer true.
answered 12 mins ago
Ben51
2,614523
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?
So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?
So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.
If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.
Hope that answered your question!
answered 4 hours ago
CuriousHegemon
314
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actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
 |Â
show 2 more comments
up vote
3
down vote
I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?
So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?
So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.
If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.
Hope that answered your question!
answered 4 hours ago
CuriousHegemon
314
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CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
 |Â
show 2 more comments
up vote
3
down vote
up vote
3
down vote
I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?
So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?
So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.
If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.
Hope that answered your question!
answered 4 hours ago
CuriousHegemon
314
New contributor
CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm guessing you mean the string constraint that Tension must be equal in both directions at all points in the string except the endpoints, where the tension at the endpoints must be equal and opposite?
So for an object moving in circular motion around a fixed point attached to a string, you're right that the object is moving in a circle because of the tension from the rope giving centripetal force. I think your confusion is coming from, shouldn't the center point also feel a tension and thus accelerate?
So the answer comes from the definition of a "fixed point"! In real life this means nailing something to the ground, or gluing it down, or placing it between a rock and a hard place, etc. This means that the center will indeed feel tension, but it will also feel some resistive force (usually normal or frictional forces) that will keep it from accelerating.
If the center point was not "fixed", then the circular motion would immediately stop, the string would go immediately slack, and the problem would become much more complex.
Hope that answered your question!
answered 4 hours ago
CuriousHegemon
314
New contributor
CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 4 hours ago
CuriousHegemon
314
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CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
CuriousHegemon
314
answered 4 hours ago
CuriousHegemon
314
314
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CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CuriousHegemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
 |Â
show 2 more comments
actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
actually I meant the accelerations of both the points along the string should be same by spring constraint
– Harsh Somani
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
I think I'm a little confused by your comment. "Strings" in the conventional sense don't obey Hookean physics, i.e. the tension in a rope is not given by $F = -k(x - x_0)$. Indeed, this makes sense because Tension is not a conservative force, and hookean forces are the product of a derivative of a potential, which Tension is not.
– CuriousHegemon
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
The string constraint is the condition that the length of an ideal string is constant. For solving problems, the string constraint translates to Sum of acceleration for all ends of a system of strings is zero.
– Harsh Somani
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
@HarshSomani But, if I spin a ball in a string, then clearly the outer end of the string accelerates a lot more than the inner end that I am holding fixed. The string constraint you describe is obviously not the case here. Maybe only in certain situations such as during translation, but not during rotation.
– Steeven
4 hours ago
2
2
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
@HarshSomani I think I see what you mean. So the spring constraint just means that the forces on the ends are the same. But that doesn't mean that the net force at the center is equal to the tension! There is always some normal or frictional force canceling out the pull of the tension at the center when we are looking at central motion.
– CuriousHegemon
4 hours ago
 |Â
show 2 more comments
up vote
3
down vote
accelerations of both the ends along the string is same if the string is not slacked
Now, I understand your problem.
If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.
In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.
Now,since string isn't slackened, is the acceleration of both ends same?
I want to explain what's happening in terms of forces rather than acceleration:
Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.
The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.
The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.
answered 4 hours ago
ayc
38418
add a comment |Â
up vote
3
down vote
accelerations of both the ends along the string is same if the string is not slacked
Now, I understand your problem.
If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.
In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.
Now,since string isn't slackened, is the acceleration of both ends same?
I want to explain what's happening in terms of forces rather than acceleration:
Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.
The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.
The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.
answered 4 hours ago
ayc
38418
add a comment |Â
up vote
3
down vote
up vote
3
down vote
accelerations of both the ends along the string is same if the string is not slacked
Now, I understand your problem.
If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.
In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.
Now,since string isn't slackened, is the acceleration of both ends same?
I want to explain what's happening in terms of forces rather than acceleration:
Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.
The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.
The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.
answered 4 hours ago
ayc
38418
accelerations of both the ends along the string is same if the string is not slacked
Now, I understand your problem.
If you have a string placed in the shape of s in vacuum and if you start pulling it from one end it finally becomes l i.e straight.Here you can say that string isn't slackened because acceleration of both ends in same.
In case of circular motion i.e particle rotating about a fixed centre the string provides the necessary force to keep the particle moving around the centre and this force is called as tension.
Now,since string isn't slackened, is the acceleration of both ends same?
I want to explain what's happening in terms of forces rather than acceleration:
Newton's third law of action and reaction states that if the string exerts an inward centripetal force on the particle, the particle will exert an equal but outward reaction upon the string,the reactive centrifugal force.
The string transmits the reactive centrifugal force from the particle to the centre, pulling upon the centre. Again according to Newton's third law, the centre exerts a reaction upon the string, pulling upon the string. The two forces upon the string are equal and opposite, exerting no net force upon the string (assuming that the string is massless), but placing the string under tension i.e no slackening of the string.
The reason the centre appears to be "immovable"(not accelerating) is because it is fixed. If the rotating ball was tethered to the mast of a boat, for example, the boat mast and ball would both experience rotation about a central point.
answered 4 hours ago
ayc
38418
answered 4 hours ago
ayc
38418
answered 4 hours ago
ayc
38418
answered 4 hours ago
ayc
38418
38418
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I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
A lot of good answers already, but I will start by discussing what context you heard this in.
This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.
Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $omega=v/r$,
$$a=fracv^2r=omega^2r$$
So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.
The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($0$), so the acceleration must be $0$ as well.
So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.
answered 3 hours ago
Aaron Stevens
5,9692830
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1
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I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
A lot of good answers already, but I will start by discussing what context you heard this in.
This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.
Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $omega=v/r$,
$$a=fracv^2r=omega^2r$$
So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.
The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($0$), so the acceleration must be $0$ as well.
So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.
answered 3 hours ago
Aaron Stevens
5,9692830
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up vote
1
down vote
up vote
1
down vote
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
A lot of good answers already, but I will start by discussing what context you heard this in.
This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.
Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $omega=v/r$,
$$a=fracv^2r=omega^2r$$
So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.
The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($0$), so the acceleration must be $0$ as well.
So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.
answered 3 hours ago
Aaron Stevens
5,9692830
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
A lot of good answers already, but I will start by discussing what context you heard this in.
This was probably told to you in the scenario of two objects connected by a string, and then you pull horizontally on one object to pull the entire system. In this case you are right. If the string is not slacked, then the accelerations of each end of the string must be the same. At this point the string acts like a massless rigid body. Since the whole system will have a single acceleration, it must be that all points (not just the ends) of the string have the same acceleration. If this were not the case, then the string would either be stretching at some points, or folding in on itself at some points.
Now, in the case of the rotating object attached to a string, you actually have a bigger "issue" than you realize. Technically all points along the string have a different acceleration! This is because, assuming a constant angular velocity $omega=v/r$,
$$a=fracv^2r=omega^2r$$
So if the heart of your question is asking why the "string constraint" doesn't apply here, you should be looking at all points along the string. Not just the end in the center.
The reason this occurs is because as you move closer to the center of the circle, the linear velocity becomes smaller and smaller. Since acceleration is change in velocity, this means that the acceleration will also become smaller. At the center of the circle, the velocity is constant ($0$), so the acceleration must be $0$ as well.
So the resolution to your question really is just that the "string constraint" is not a true constraint in this system (unless you find a way to reword it to be more general). What you learned in school was probably just said in the context of a particular problem, and was not meant to be a generalization to how strings behave in all contexts.
answered 3 hours ago
Aaron Stevens
5,9692830
answered 3 hours ago
Aaron Stevens
5,9692830
answered 3 hours ago
Aaron Stevens
5,9692830
answered 3 hours ago
Aaron Stevens
5,9692830
5,9692830
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I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$beginalign
mathbfv_0(t) =& 0
\ mathbfv_1(t) =& rcdotomegacdotbeginpmatrix-sin(omegacdot t)\cos(omegacdot t)endpmatrix
endalign$$
The radial component of $mathbfv_1$ is indeed always zero
$$
mathbfv_1(t)cdotmathbfe_mathrmr
= rcdotomegacdotBigl(-cos(omegacdot t)cdotsin(omegacdot t) + sin(omegacdot t)cdotcos(omegacdot t)Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
fracmathrmdmathbfv_1mathrmdt = rcdotomega^2cdotbeginpmatrix-cos(omegacdot t)\-sin(omegacdot t)endpmatrix
= -rcdotomega^2cdot mathbfe_mathrmr.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
answered 40 mins ago
leftaroundabout
9,78732243
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0
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I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$beginalign
mathbfv_0(t) =& 0
\ mathbfv_1(t) =& rcdotomegacdotbeginpmatrix-sin(omegacdot t)\cos(omegacdot t)endpmatrix
endalign$$
The radial component of $mathbfv_1$ is indeed always zero
$$
mathbfv_1(t)cdotmathbfe_mathrmr
= rcdotomegacdotBigl(-cos(omegacdot t)cdotsin(omegacdot t) + sin(omegacdot t)cdotcos(omegacdot t)Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
fracmathrmdmathbfv_1mathrmdt = rcdotomega^2cdotbeginpmatrix-cos(omegacdot t)\-sin(omegacdot t)endpmatrix
= -rcdotomega^2cdot mathbfe_mathrmr.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
answered 40 mins ago
leftaroundabout
9,78732243
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$beginalign
mathbfv_0(t) =& 0
\ mathbfv_1(t) =& rcdotomegacdotbeginpmatrix-sin(omegacdot t)\cos(omegacdot t)endpmatrix
endalign$$
The radial component of $mathbfv_1$ is indeed always zero
$$
mathbfv_1(t)cdotmathbfe_mathrmr
= rcdotomegacdotBigl(-cos(omegacdot t)cdotsin(omegacdot t) + sin(omegacdot t)cdotcos(omegacdot t)Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
fracmathrmdmathbfv_1mathrmdt = rcdotomega^2cdotbeginpmatrix-cos(omegacdot t)\-sin(omegacdot t)endpmatrix
= -rcdotomega^2cdot mathbfe_mathrmr.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
answered 40 mins ago
leftaroundabout
9,78732243
I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$beginalign
mathbfv_0(t) =& 0
\ mathbfv_1(t) =& rcdotomegacdotbeginpmatrix-sin(omegacdot t)\cos(omegacdot t)endpmatrix
endalign$$
The radial component of $mathbfv_1$ is indeed always zero
$$
mathbfv_1(t)cdotmathbfe_mathrmr
= rcdotomegacdotBigl(-cos(omegacdot t)cdotsin(omegacdot t) + sin(omegacdot t)cdotcos(omegacdot t)Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
fracmathrmdmathbfv_1mathrmdt = rcdotomega^2cdotbeginpmatrix-cos(omegacdot t)\-sin(omegacdot t)endpmatrix
= -rcdotomega^2cdot mathbfe_mathrmr.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
answered 40 mins ago
leftaroundabout
9,78732243
answered 40 mins ago
leftaroundabout
9,78732243
answered 40 mins ago
leftaroundabout
9,78732243
answered 40 mins ago
leftaroundabout
9,78732243
9,78732243
add a comment |Â
add a comment |Â
up vote
0
down vote
To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving in one direction with no rotation, all points in the system move with the same acceleration.
In any other case (curved motion, rotation of the system, etc.) this is no longer true.
answered 12 mins ago
Ben51
2,614523
add a comment |Â
up vote
0
down vote
To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving in one direction with no rotation, all points in the system move with the same acceleration.
In any other case (curved motion, rotation of the system, etc.) this is no longer true.
answered 12 mins ago
Ben51
2,614523
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving in one direction with no rotation, all points in the system move with the same acceleration.
In any other case (curved motion, rotation of the system, etc.) this is no longer true.
answered 12 mins ago
Ben51
2,614523
To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving in one direction with no rotation, all points in the system move with the same acceleration.
In any other case (curved motion, rotation of the system, etc.) this is no longer true.
answered 12 mins ago
Ben51
2,614523
answered 12 mins ago
Ben51
2,614523
answered 12 mins ago
Ben51
2,614523
answered 12 mins ago
Ben51
2,614523
2,614523
add a comment |Â
add a comment |Â
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2
What do you mean by string constraint?
– ayc
4 hours ago
I was taught in school that accelerations of both the ends along the string is same if the string is not slacked a.k.a the string constraint.
– Harsh Somani
4 hours ago