Vertical Circular motion- A confusing question

[palaphys]

Homework Statement

A simple pendulum, consisting of a light inextensible string of length lattached to a small heavy bob of
mass m, is at rest. The bob is imparted a horizontal impulsive force which gives it a velocity of
sqrt4gl.
The speed of the bob at its highest point is

Relevant Equations

energy conservation, basics of circular motion

Now this question really startled me. We all know that from simple energy conservation, the ball can reach a height of 2l, i.e reach the top point of the vertical circle if a speed of is given at the bottom $\sqrt{4gl}$ as mentioned in the question.
Hence, I expected the answer to be A. However the key says that the answer is D.
Very confused, help is appreciated.

 

[kuruman]

. . . the ball can reach a height of 2l, i.e reach the top point of the vertical circle with the speed $\sqrt{4gl}$ as mentioned in the question.

That is not what is mentioned in the question. The speed mentioned, $\sqrt{4gl}$, is at the bottom, not the top. That said, the key is wrong. The correct answer is (A), zero speed at the top.

 

[palaphys]

That is not what is mentioned in the question. The speed mentioned, $\sqrt{4gl}$, is at the bottom, not the top. That said, the key is wrong. The correct answer is (A), zero speed at the top.

pl check edited question. this is what I meant, but conveyed it incorrectly

 

[haruspex]

That said, the key is wrong. The correct answer is (A), zero speed at the top.

It's a string, not a rigid rod.
@palaphys, will it go higher than the top of the string? What forces will act on it then?

 

[Lnewqban]

Now this question really startled me. We all know that from simple energy conservation, the ball can reach a height of 2l, i.e reach the top point of the vertical circle if a speed of is given at the bottom $\sqrt{4gl}$ as mentioned in the question.

It seems to me that you have incorrectly assumed that the kinetic energy of the small heavy bob at the bottom of the vertical circle is the very minimum to reach the top (option a).

If in fact, option d is the correct answer, shouldn't the potential energy gained by the bob at the top be equal to the difference of its kinetic energies at bottom and top?

 

[kuruman]

It's a string, not a rigid rod.
@palaphys, will it go higher than the top of the string? What forces will act on it then?

Indeed it is.

 

[Gavran]

Very confused, help is appreciated.

A circular motion of the bob ends when a tension force in the string becomes zero. At that moment another kind of motion of the bob begins. The correct answer is (d).

 

[palaphys]

A circular motion of the bob ends when a tension force in the string becomes zero. At that moment another kind of motion of the bob begins. The correct answer is (d).

how is it D? not able to get why it is D.. I know that there is going to be projectile motion of the bob. But doesn't the tension in the string become zero at the TOPMOST point? so velocity there would be zero right?

 

[Steve4Physics]

how is it D? not able to get why it is D.. I know that there is going to be projectile motion of the bob. But doesn't the tension in the string become zero at the TOPMOST point? so velocity there would be zero right?

If the bob were mounted on a (light) rigid rod, the bob would just have enough energy to reach the top position. In this case, at the top position, the rod would be in a state of compression – the compressive force would equal the bob’s weight.

In fact the rod would have become compressed even before it reached the top position, because the speed would have been too slow to maintain the rod in a state of tension.

With string, rather than a rigid rod, the string will become slack before it reaches the top position.

 

[PeroK]

how is it D? not able to get why it is D.. I know that there is going to be projectile motion of the bob. But doesn't the tension in the string become zero at the TOPMOST point? so velocity there would be zero right?

The mass is not constrained to move in a circle. In fact, it can't reach the top of the circle and have no KE at that point.

 

[Gavran]

how is it D? not able to get why it is D.. I know that there is going to be projectile motion of the bob. But doesn't the tension in the string become zero at the TOPMOST point? so velocity there would be zero right?

The tension force becomes zero when the centripetal force becomes equal to the perpendicular component of the gravitational force.

 

[haruspex]

But doesn't the tension in the string become zero at the TOPMOST point? so velocity there would be zero right?

If you still are stuck on that after all the above replies, consider the bob as it approaches the top of the circle. It would be moving nearly horizontally, ever more slowly. What has happened to gravity?

 

[palaphys]

If the bob were mounted on a (light) rigid rod, the bob would just have enough energy to reach the top position. In this case, at the top position, the rod would be in a state of compression – the compressive force would equal the bob’s weight.

In fact the rod would have become compressed even before it reached the top position, because the speed would have been too slow to maintain the rod in a state of tension.

With string, rather than a rigid rod, the string will become slack before it reaches the top position.

not understanding what you mean by compression. is it stress/ tension?

 

[palaphys]

The tension force becomes zero when the centripetal force becomes equal to the perpendicular component of the gravitational force.

I will try to use this fact to proceed with the problem

 

[Herman Trivilino]

But doesn't the tension in the string become zero at the TOPMOST point?

Have you done the math necessary to answer this question? Learning how to do that is the purpose.

 

[Steve4Physics]

not understanding what you mean by compression. is it stress/ tension?

Have you not met the term 'compression'?

Hold a pencil horizontally – one end in your left hand, the other end in your right hand.

Pull your hands outwards – to stretch the pencil. The pencil is in a state of tension.
Push your hands inwards – squashing the pencil. The pencil is in a state of compression.

A piece of string can be in a state of tension but not compression (it just goes slack).

Think about using a bob+rigid rod in your question. To supply the required force, the rod must change from being in tension at the bottom to being in compression at the top. (So at some point, the tension is zero.)

Now imagine what would happen with the bob+string, remembering that the string can’t be in a state of compression.

 

[haruspex]

I will try to use this fact to proceed with the problem

Using this fact given to you is one thing, but you need to understand why it is true.
If you do not, draw a free body diagram for the forces on the bob at some point where the string has risen above the horizontal. What is the force balance along the line of the string?

 

[palaphys]

Have you not met the term 'compression'?

Hold a pencil horizontally – one end in your left hand, the other end in your right hand.

Pull your hands outwards – to stretch the pencil. The pencil is in a state of tension.
Push your hands inwards – squashing the pencil. The pencil is in a state of compression.

A piece of string can be in a state of tension but not compression (it just goes slack).

Think about using a bob+rigid rod in your question. To supply the required force, the rod must change from being in tension at the bottom to being in compression at the top. (So at some point, the tension is zero.)

Now imagine what would happen with the bob+string, remembering that the string can’t be in a state of compression.

okay, so you are saying that the string can only reach the top slack, with the given speed in the question, that too in a hypothetical case. Whereas a rigid massless rod given the same speed would reach the top always, as tension would prevent it from collapsing at the bottom. (please correct me if I'm wrong anywhere)
But then, why do textbooks emphasize on the speed $\sqrt{4gl}$ for a bob-string system?

I have tried to proceed with the problem. After some analysis(energy+ FBD) I have found that the string will go slack at exactly $arccos(2/3)$ from the horizontal. At this point, it would have a horizontal component of speed as $\sqrt{8gl/27}$ (from energy consv. between bottom point and reqd. point). We also know that this horizontal component is not going to change, as there is no external force in the x-direction during the bob's projectile motion.

Also, at the turning point of the bob's motion, it will have only a horizontal component, hence the ans. is option D.
Will post my working soon

 

[PeroK]

okay, so you are saying that the string can only reach the top slack, with the given speed in the question, that too in a hypothetical case. Whereas a rigid massless rod given the same speed would reach the top always, as tension would prevent it from collapsing at the bottom. (please correct me if I'm wrong anywhere)

Yes. A rigid structure will continue in a circle as long as it has sufficient KE to reach the top. The mass is constrained to move in a circle - the rigid structure prevents the mass falling in parabolic projectle motion. The mass on a string is not constrained to move in a circle for the upper half. You need to be careful to check the criteria that would allow it to move in a circle. You could, for example, calculate the minimum speed at the bottom in order for the mass to complete a circle. To do this, the speed at the top must be sufficient for gravity to be the required centripetal force.

I have tried to proceed with the problem. After some analysis(energy+ FBD) I have found that the string will go slack at exactly $arccos(2/3)$ from the horizontal. At this point, it would have a horizontal component of speed as $\sqrt{8gl/27}$ (from energy consv. between bottom point and reqd. point). We also know that this horizontal component is not going to change, as there is no external force in the x-direction during the bob's projectile motion.

Also, at the turning point of the bob's motion, it will have only a horizontal component, hence the ans. is option D.
Will post my working soon

Sounds good, even without the calculations.

 

[haruspex]

why do textbooks emphasize on the speed 4gl for a bob-string system?

Most of the textbook work on such systems is for small oscillations. If they fail to mention "as long as the string remains taut” when quoting $\sqrt{4gl}$ then that is a serious omission. Of course, it will still happen that some students will have forgotten that proviso anyway ;-).

 

[PeroK]

Most of the textbook work on such systems is for small oscillations. If they fail to mention "as long as the string remains taut” when quoting $\sqrt{4gl}$ then that is a serious omission.

I'd like to see the evidence of the textbook(s) that say any such thing. Otherwise, it's just a vacuous statement.

 

[Steve4Physics]

okay, so you are saying that the string can only reach the top slack, with the given speed in the question, that too in a hypothetical case. Whereas a rigid massless rod given the same speed would reach the top always, as tension would prevent it from collapsing at the bottom. (please correct me if I'm wrong anywhere)

I think you’ve got it but I would have said it slightly differently, such as:

"The bob+string can not reach the top of the circle because the string goes slack before reaching the top. Whereas a bob+rigid massless rod would reach the top because the rod cannot go ‘slack’ – the rod will be in a compressed state."

(BTW it’s not difficult to show that if the initial speed is $\sqrt {5gl}$ or more, the bob+string would reach the top of the circle.)

But then, why do textbooks emphasize on the speed $\sqrt{4gl}$ for a bob-string system?

I’d guess that $\sqrt {4gl}$ has been chosen for this specific problem for couple of reasons:

i) The initial kinetic energy is $\frac 12 mv_0^2 = \frac 12 m(\sqrt {4gl})^2 = 2mgl$
This is exactly the increase in gravitional potential energy needed to reach the top of the circle. The kinetic energy at the top would then be zero - this makes it easier to tell that that string will have gone slack before reaching the top.

ii) It makes the maths simpler because it leads to a convenient value for the (cosine of) the angle at which the tension in the string becomes zero.

I have tried to proceed with the problem. After some analysis(energy+ FBD) I have found that the string will go slack at exactly $arccos(2/3)$ from the horizontal.

I think the string goes slack when the angle between the string and the vertical (not the the horizontal) is $\arccos (\frac 23)$. The direction of the bob's velocity is then $\arccos (\frac 23)$ to the horizontal.

At this point, it would have a horizontal component of speed as $\sqrt{8gl/27}$ (from energy consv. between bottom point and reqd. point). We also know that this horizontal component is not going to change, as there is no external force in the x-direction during the bob's projectile motion.

Also, at the turning point of the bob's motion, it will have only a horizontal component, hence the ans. is option D.

Well done!

Edit - typo'

 

[haruspex]

I'd like to see the evidence of the textbook(s) that say any such thing. Otherwise, it's just a vacuous statement.

It’s a conditional statement. $\neg p\implies(p\implies q)$. As in, if pigs could fly, perhaps.

 

[PeroK]

It’s a conditional statement. $\neg p\implies(p\implies q)$. As in, if pigs could fly, perhaps.

If you are saying the answer is a), then you're wrong!

 

[Herman Trivilino]

okay, so you are saying that the string can only reach the top slack, with the given speed in the question, that too in a hypothetical case. Whereas a rigid massless rod given the same speed would reach the top always, as tension would prevent it from collapsing at the bottom.

As long as you understand that "top" in the case of a string may not be as high as "top" in the case of the rod.

 

[haruspex]

If you are saying the answer is a), then you're wrong!

No, of course I am not saying that. In post #20 I was merely addressing the question quoted there. If a textbook states $\sqrt{4gl}$ as the generic solution to a question like the one in this thread, without the proviso "as long as the string remains taut", then it is at fault.
I was also hinting, gently, that perhaps the textbooks @palaphys is recalling did make that proviso, but the reader hasn’t registered that.

 

[palaphys]

I think you’ve got it but I would have said it slightly differently, such as:

"The bob+string can not reach the top of the circle because the string goes slack before reaching the top. Whereas a bob+rigid massless rod would reach the top because the rod cannot go ‘slack’ – the rod will be in a compressed state."

Yes, this is my understanding

I think the string goes slack when the angle between the string and the vertical (not the the horizontal) is $\arccos (\frac 23)$. The direction of the bob's velocity is then $\arccos (\frac 23)$ to the horizontal.

Here is my working:

yes, I meant to say that it is $\arccos (\frac 23)$ with the horizontal.
thanks for the help

 

[Steve4Physics]

yes, I meant to say that it is $\arccos (\frac 23)$ with the horizontal.
thanks for the help

Your correction repeats the mistake! You mean $\arccos (\frac 23)$ to the vertical - as you correctly show in your diagram.

A general point: it's helpful (to whoever is marking your work) to include few words of explanation. Then, even if you make some arithmetic/algebraic mistake(s), the marker can understand your method and can award some marks. In this question, two key points I would have explicitly stated in my answer are:

a) at some point, the tension in the string will be zero and the bob’s motion will change from circular to parabolic;

b) at the top of the parabolic path, the bob's vertical velocity component will be zero; then the bob's velocity will simply be the (constant) horizontal velocity component of the parabolic motion.

Also, a free-body diagram of the bob is generally recommended - it could help you and might get you marks.

 

[palaphys]

yes, I meant to say that it is $\arccos (\frac 23)$ with the horizontal.
thanks for the help

EDIT- yes, I meant to say that it is $\arccos (\frac 23)$ with the vertical

 

[Sarinle]

EDIT- yes, I meant to say that it is $\arccos (\frac 23)$ with the vertical

On another note, to help you understand the difference between a rigid rod and a string; imagine you picked up a string with the ball attached to it. Now imagine you also had a stick/rod which has a ball attached to it.

Now for the string+ball system, if you aim the ball towards the floor; the string applies a tension inward towards your hand equivalent to the gravitational attractive force between the ball and the earth. However, if you point the ball towards the roof; it simply falls down and the string goes slack thus it is effectively in free fall (that's what they mean by a string is not compressive).

Meanwhile for a stick/rod, pointing the ball up or down does not matter as we can always apply a counter force (compressive/tension). This is what they mean when they state the stick/rod is compressive.

Naturally, one must ask why a string is not able to provide a compressive force?

This is because a string is made of intertwined fibres that hold together and resist when pulled, but simply buckle outward and provide no support when a compressive force is applied.

I can think of an example where a system has poor tensile strength with strong compressive strength. Can you?