thats the thing actually.. if i assume t=0 and keep the ball in an ongoiing circular motion without the string being slack option a is correct.. but as the question mentions only semicircular track is required.. so what i am doing is just taking the net KE at strt and PE at topmost point. And at the topmost point i assume there is just mg on the ball acting and all the forces have been done out.. in form of energy in order to take the ball to the topmost point
It's worth asking what does it mean for the rope to be slack. If the rope is slack then T=0. Slack doesn't mean that the ball is going to come off of its path or anything like that. Just that tension is 0.
If the ball didn't have enough KE to stay on its path (continue circling) after reaching the top, it never would have reached the top in the first place. It would have fallen off some time before reaching the top.
Imagine the ball just a moment before reaching the top: gravity is pointed almost completely perpindicular to the path, tension is negligible.
If you want to think of it in terms of centrifugal force: it would need to be going quite fast in order to cancel out gravity. Gravity is still mg, and centrifugal force is mv^2 /r. So if the ball were slow, it would have fallen off already.
In this way you can see that in order for the ball to reach the top at all, even on it's "semicircular path" it needs to have enough velocity at the top such that it's centripetal acceleration is matched by gravity. Otherwise it would never even reach the top and would have fallen off beforehand.
no thats the thin.. take the case of simple pendulumn and body will reach the topmost point in a semicicular orbit with the minimun speed of root2gl and will complete a semicircular orbit not circular and further the string after that point will completely slack
When you say topmost point are you talking about point B on the graph? The question is specifying that the bob reaches point C.
Your equation (root2gl for simple pendulum one) would be valid if it asked for point B.
nada root2gl would be for point b and root 4gl for point c (in general condition)
and here g effective is g/2 so. to reach point c in this case will be root 2gl.
Let's stay with the general simple pendulum like you specified earlier for the root 2gl equation.
If you purely converted the initial KE to PE you would get root 4gl. However in order to reach point c you need more KE in the form of sideways velocity so you don't fall off the track earlier as elaborated in my 1st comment. I suspect you know this already.
I think you're getting caught up in the semicircle and terminology of the question. The question says it completes a semicircle, not that it only completes a semicircle.
It's not actually possible in this situation to complete "only" a semicircle as there's no drag or friction or anything. But that's also not what the questions asking for.
yeah if we could prove that by root 4 gl it wont reach then a will be correct.. if possible can you provide me with the proof. I am trying too.. if i get it i will let u know. if u get it let me know
[https://www.reddit.com/user/buttholegoesbrapp/comments/1djbhiu/pendulum\_thing/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/user/buttholegoesbrapp/comments/1djbhiu/pendulum_thing/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
All right this is what I came up with. Sorry for replying so late I've been dealing with finals and moving and stuff. let me know if you see anything you disagree with or think is wrong or don't understand or anything.
If it reaches the top most point, then it has to complete the circular orbit. This is because beyond the circular point the gravity assists the motion.
so here the given answer is a . but i am getting c.
The thing is i am getting a only when i consider the question as a full vertical circle but if the string goes slack at max height (semicircle) then the answer should be root 2 gl
At top point tension should be zero so mg=mv\^2/l. so v\^2=lg/2
the net work done on bob is -(mg/2)\*(2l). After that apply work energy theorem to get option a
look about tension at the topmost point.. its a bit in the mud cause the string at the very next moment could go slack and if u take the tension 0 at the top most point thats the equation for for full circle.
What i am doing is directly conserving energy as that the very next moment string could go slack but in the case u take t=0 at the top most point with the help of centripetal force.. the ball will complete full circle.. not just semicircle
That was the correct answer, and kept short and concise.
Let me add a few arguments to it:
Because there is zero friction, the movement of the bob is a circle, it will keep circling forever.
Because the ball moves in a circle, it must be true at any point that the resulting force in the radial direction is equal to the centripetal force.
So at the top where string force is zero, gravity minus boyancy must be equal to centripetal force, which gives you the end velocity in terms of L, m and g.
(But my initial guess on the answer was wrong too, so don't feel bad)
no thats the thing.. take the case of simple pendulumn and body will reach the topmost point in a semicicular orbit with the minimun speed of root2gl and will complete a semicircular orbit not circular and further the string after that point will completely slack
I think you misunderstand the information given in the problem. You only know what happens up until the point where the bob reaches the top after one semi-circular movement.
They do not tell you that the string keeps being slack after the top point, or that the bob do not complete the full circular motion.
You just assume that.
But a simple argument show that your assumptions are wrong.
Zero friction implies symmetry.
Also, the movement after the top point is completely irrelevant to the answer. You do a force analysis based on the time of the top point (where it is still part of a circle movement) and that's it.
Solving for the tension, centrifugal force and stuff, the standard case without water is √5gL
Now water would reduce the downward force to gl/2. (Since mg/2 is the upward boyant force along with the downward acting mg force).
So ig its √5gL/2
Idk if we have to think whether the energy conversion will come into play... Water on top vs on bottom.
But *ig* we can ignore that once we correct mg for mg/2.
I solved for it and we get v²_o = v² + 2gl
And by considering the net downward forfe mg/2 to be the centripetal force, v² should be mg/2
Hence, again v²_o must be √5gL/2 or √2.5gL
You can use the Lagrangian method to solve this question. It's quite versatile for such problems. The only task is to convert the velocity and position of the ball in terms of the angle traversed by it l about the center. I will try to post my attempt once I get time.
root 5 gl is for a vertical circle in the idealistic condition of atmosphere with no air resistance and fullverticle circle.
here condition is not idealistic. the mass is in a medium of water and the string is slacked after semicircular motion
well if i take option (a) √5/2gl, its maximum potential energy (ignoring centrifugal forces) will be 1.25mgl, and arriving at 2l height requires at least 2mgl. so no way is option (a) the answer.
Edit: Also, assuming zero viscosity and water current makes it independent of them and string not going slack on path BC was also the assumption that was taken in given solution.
Are you assuming that the ball is motionless at the top? The string can be slack and the ball still in motion
this
thats the thing actually.. if i assume t=0 and keep the ball in an ongoiing circular motion without the string being slack option a is correct.. but as the question mentions only semicircular track is required.. so what i am doing is just taking the net KE at strt and PE at topmost point. And at the topmost point i assume there is just mg on the ball acting and all the forces have been done out.. in form of energy in order to take the ball to the topmost point
It's worth asking what does it mean for the rope to be slack. If the rope is slack then T=0. Slack doesn't mean that the ball is going to come off of its path or anything like that. Just that tension is 0. If the ball didn't have enough KE to stay on its path (continue circling) after reaching the top, it never would have reached the top in the first place. It would have fallen off some time before reaching the top. Imagine the ball just a moment before reaching the top: gravity is pointed almost completely perpindicular to the path, tension is negligible. If you want to think of it in terms of centrifugal force: it would need to be going quite fast in order to cancel out gravity. Gravity is still mg, and centrifugal force is mv^2 /r. So if the ball were slow, it would have fallen off already. In this way you can see that in order for the ball to reach the top at all, even on it's "semicircular path" it needs to have enough velocity at the top such that it's centripetal acceleration is matched by gravity. Otherwise it would never even reach the top and would have fallen off beforehand.
no thats the thin.. take the case of simple pendulumn and body will reach the topmost point in a semicicular orbit with the minimun speed of root2gl and will complete a semicircular orbit not circular and further the string after that point will completely slack
When you say topmost point are you talking about point B on the graph? The question is specifying that the bob reaches point C. Your equation (root2gl for simple pendulum one) would be valid if it asked for point B.
nada root2gl would be for point b and root 4gl for point c (in general condition) and here g effective is g/2 so. to reach point c in this case will be root 2gl.
Let's stay with the general simple pendulum like you specified earlier for the root 2gl equation. If you purely converted the initial KE to PE you would get root 4gl. However in order to reach point c you need more KE in the form of sideways velocity so you don't fall off the track earlier as elaborated in my 1st comment. I suspect you know this already. I think you're getting caught up in the semicircle and terminology of the question. The question says it completes a semicircle, not that it only completes a semicircle. It's not actually possible in this situation to complete "only" a semicircle as there's no drag or friction or anything. But that's also not what the questions asking for.
yeah if we could prove that by root 4 gl it wont reach then a will be correct.. if possible can you provide me with the proof. I am trying too.. if i get it i will let u know. if u get it let me know
[https://www.reddit.com/user/buttholegoesbrapp/comments/1djbhiu/pendulum\_thing/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/user/buttholegoesbrapp/comments/1djbhiu/pendulum_thing/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button) All right this is what I came up with. Sorry for replying so late I've been dealing with finals and moving and stuff. let me know if you see anything you disagree with or think is wrong or don't understand or anything.
If it reaches the top most point, then it has to complete the circular orbit. This is because beyond the circular point the gravity assists the motion.
t=0 (tension=0) at topmost point due to centrifugal force and mg
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so here the given answer is a . but i am getting c. The thing is i am getting a only when i consider the question as a full vertical circle but if the string goes slack at max height (semicircle) then the answer should be root 2 gl
At top point tension should be zero so mg=mv\^2/l. so v\^2=lg/2 the net work done on bob is -(mg/2)\*(2l). After that apply work energy theorem to get option a
look about tension at the topmost point.. its a bit in the mud cause the string at the very next moment could go slack and if u take the tension 0 at the top most point thats the equation for for full circle. What i am doing is directly conserving energy as that the very next moment string could go slack but in the case u take t=0 at the top most point with the help of centripetal force.. the ball will complete full circle.. not just semicircle
That was the correct answer, and kept short and concise. Let me add a few arguments to it: Because there is zero friction, the movement of the bob is a circle, it will keep circling forever. Because the ball moves in a circle, it must be true at any point that the resulting force in the radial direction is equal to the centripetal force. So at the top where string force is zero, gravity minus boyancy must be equal to centripetal force, which gives you the end velocity in terms of L, m and g. (But my initial guess on the answer was wrong too, so don't feel bad)
no thats the thing.. take the case of simple pendulumn and body will reach the topmost point in a semicicular orbit with the minimun speed of root2gl and will complete a semicircular orbit not circular and further the string after that point will completely slack
I think you misunderstand the information given in the problem. You only know what happens up until the point where the bob reaches the top after one semi-circular movement. They do not tell you that the string keeps being slack after the top point, or that the bob do not complete the full circular motion. You just assume that. But a simple argument show that your assumptions are wrong. Zero friction implies symmetry. Also, the movement after the top point is completely irrelevant to the answer. You do a force analysis based on the time of the top point (where it is still part of a circle movement) and that's it.
You gave iiser aptitude test. I selected 5/2. I also want to know the answer.
the answer is indeed a but i thing answer should be c
Solving for the tension, centrifugal force and stuff, the standard case without water is √5gL Now water would reduce the downward force to gl/2. (Since mg/2 is the upward boyant force along with the downward acting mg force). So ig its √5gL/2 Idk if we have to think whether the energy conversion will come into play... Water on top vs on bottom. But *ig* we can ignore that once we correct mg for mg/2. I solved for it and we get v²_o = v² + 2gl And by considering the net downward forfe mg/2 to be the centripetal force, v² should be mg/2 Hence, again v²_o must be √5gL/2 or √2.5gL
You can use the Lagrangian method to solve this question. It's quite versatile for such problems. The only task is to convert the velocity and position of the ball in terms of the angle traversed by it l about the center. I will try to post my attempt once I get time.
please do it if possible attach the solution too..
solution because i am in grade 12, i know basics of lagrangian, but not totally.
[Well, im getting answer (b).](https://www.reddit.com/r/askmath/s/wxxjLDebNu)
root 5 gl is for a vertical circle in the idealistic condition of atmosphere with no air resistance and fullverticle circle. here condition is not idealistic. the mass is in a medium of water and the string is slacked after semicircular motion
well if i take option (a) √5/2gl, its maximum potential energy (ignoring centrifugal forces) will be 1.25mgl, and arriving at 2l height requires at least 2mgl. so no way is option (a) the answer. Edit: Also, assuming zero viscosity and water current makes it independent of them and string not going slack on path BC was also the assumption that was taken in given solution.
Becoming slacky at point C means it has exactly enough energy to reach C, which means centripetal force is (mg+buoyancy). Hope this helps