Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Here the big circle denotes the circular path of a stone (small circle on path) tied to a string from the centre of the circular path.

This is COMPLETELY HORIZONTAL This is COMPLETELY HORIZONTAL

enter image description here

At an instant the velocity in horizontal direction is $0$, and tension acts horizontally. Then it is said tension will do no work.

Now, at the very next instant of time, enter image description here

Supposedly, all this is horizontally happening, this is a horizontal circle. Then, tension increased velocity in horizontal direction, no one decreased velocity in vertical direction. Then as a result, net velocity should increase. No other force, other than tension is acting. Then it means as net velocity didn't increase, this is a violation of newton's second law (?) Now in the vertical direction velocity didn't change and in the horizontal direction velocity increased since $a=\frac{T}{m}$ where $T$ is Tension in string. So as it acted for a very small amount of time, it did increase velocity in horizontal direction. and velocity in vertical direction remained same, thus,

$|v| = \sqrt{(v_y)^2+(v_x)^2} > v_{earlier}$, thus speed must have increased in the next instant a/c to newton's laws. But it didn't according to the fact that tension did no work--why?

When I say x $/$ horizontal direction, it means parallel to the tension shown in the image. And vertical $/$ y means perpendicular to tension shown in image .

share|improve this question

closed as unclear what you're asking by Manishearth Jan 9 at 17:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

9  
The tension does no work because it acts perpendicularly to the motion. That doesn't mean that it causes no acceleration in the horizontal direction. Acceleration $\neq$ work. The only thing that changes the kinetic energy here is gravity. Energy doesn't have a direction - it is a scalar. There is no such thing as "kinetic energy in its direction" - kinetic energy is just a number related to the magnitude of the velocity. Acceleration and velocity are both vectors and can change direction without a change in energy. –  Michael Brown May 3 '13 at 9:02
2  
There is no such thing as "Kinetic Energy in horizontal direction" and no, it does not mean "Tension changed Kinetic Energy". This is what I tried to explain in my comment. What happens to the verticle component of velocity? Work that out then compute the kinetic energy. Then compute the work done by the gravitation field. You'll see that they are the same. No work is being done by the tension. –  Michael Brown May 3 '13 at 9:20
    
No, the kinetic actually decreases when the object moves up. Tension and gravity aren't balancing each other in any way here. They act seperately in different directions. Some kinetic energy becomes gravitational potential energy due to gravity and, independently, the tension changes the direction of motion. Imagine the string broke at the moment where the object is horizontal. There would be no more tension and it would go straight up instead of turning around, but it would slow down under gravity at exactly the same rate as before. –  Michael Brown May 3 '13 at 9:53
    
It doesn't. Tension does. That doesn't mean tension has done work. You're clearly confused about the definition of work. Forces orthogonal to the motion do no work. The only role of tension here is to change the direction of the velocity vector, but energy is a scalar - it only depends on the magnitude of the velocity and not on the direction. A force orthogonal to the velocity does not change the energy: $\dot{E} = \vec{F}\cdot\vec{v} = 0$. –  Michael Brown May 3 '13 at 10:41
6  
Nonagon, you seem to be sticking quite persistently to a mental model of what is going on here that is simply wrong. You persistence in this matter is probably the cause of the downvotes. Please, please for your own understanding of this stuff stop, back up and start over. –  dmckee May 4 '13 at 6:30

4 Answers 4

up vote 13 down vote accepted
+50

This does not contradict newton, because the error is in the calculation:

You are calculating in discrete time steps. In that case the calculated speed will increase, as if energy is not preserved. However, Newton's laws apply to continuous time. In the mathematical world in which newton laws are described, speed and acceleration are NOT defined by comparing two nearby moments in time. Instead they are instant properties of objects, mathematically defined as derivative functions.

Considering two nearby moments will get you only an approximation of speed and acceleration. In this specific example, this approximation is enough to produce a seemingly paradox.

Another way to look at it:

  • Imagine you would go around the circle in 10-degree increments (a bit like in your drawings). You will get a considerable increase in speed.

  • Now do the same in 1-degree increments (360 steps around the circle). The error will be significantly less.

  • Now do the same in 0,00000001 degree increments. There will be almost no increase of speed.

  • If you had the time and calculation resources to divide your circle in INFINITE steps, then newton law would be spot on.

PS: Computer programmers (eg. for physics in games), have the same challenge as you present in your question. If you evaluate newtons laws in discrete time-steps, you are likely to get violations of preservation of energy. Luckily there are different techniques to avoid this. But that goes beyond the scope of my answer.

share|improve this answer
1  
@007 How do you say that ?, tension is being applied horizontally , that means acceleration in vertical direction has to be 0. I think Kris' answer was spot on . –  nonagon May 4 '13 at 8:49
2  
@nonagon $\vec T\perp \vec v$. So, no net work is done. –  Mr.ØØ7 May 4 '13 at 8:50

Kris Van Bael's answer has the right idea, and that is a problem for computer simulations. If you want to simulate this on a computer, you have to discretize time, and that causes problems with conservation laws.

Another way to look at it is to look at an infinitesimally small amount of time (this is where Kris Van Bael's answer is going). The instant that the mass moves the tiniest bit up, the string starts to pull down with a tiny amount of force. That tiny force slows the upward motion of the mass just slightly enough to keep it moving in the circle.

share|improve this answer

Kris's answer is correct.

Here's another way to look at it. An object is moving at speed v. You exert a sharp force from behind. It's velocity increases. An object is moving at speed v. You exert a sharp force from the front (i.e. against the motion). It's velocity decreases.

An object is moving at speed v. You exert a sharp force EXACTLY AT RIGHT ANGLES to v. It's speed neither increases, nor decreases. The object changes direction. (By "sharp" I was meaning instantaneously brief time...what Kris was describing mathematically.)

Uniform circular motion results when you continuously exert the (instantaneous) force at right angles to the velocity. Changes of direction only...no change in speed.

share|improve this answer

Many ways to prove why work done is zero:

1)I am pretty sure that your initial premise, namely "magnitude of the vector sum of the 'vertical' and 'horizontal' velocities is increasing" is flawed. You have written "no one increased velocity in vertical direction" is what is causing confusion (acc to me). I am sure that the vertical velocity is decreasing by the same amount by which horizontal velocity increases, in order to keep the vector sum the same.

2)MOREOVER, you must agree that this is Uniform Circular Motion?? In that case, speed is same? Therefore the magnitude of the vector sum (ie the speed) must be the same? SO, the resultant velocity IS NOT changing...hence, even acc to the work energy theorem, work done is zero.

3)work done is the product of the magnitude of force and the displacement in the direction of the force. There is NO displacement in direction of force, it is, at ALL moments of time (however discrete Or infinitesimal), perpendicular!

(If you are not satisfied, and feel like downvoting, then maybe I havent understood ur question, so please state the crux of it in the comments)

share|improve this answer
1  
Yes , you haven't understood . And Kris' answer is very satisfactory . Basically , what you are doing is making an assumption ahead of time . You are assuming that the stone will take the circular path whereas all you know is newton's laws . And if using that you try to explain taking time to be discrete , there is a paradox which can't be resolved .Don't worry , I'd never downvote an answer to my own question . –  nonagon May 8 '13 at 12:49
1  
The stone HAS to take the circular path...because, there is a centripetal force acting –  Saurabh Raje May 8 '13 at 13:16
2  
You are using circular reasoning here , centripetal acts because of circular motion and not the other way around . –  nonagon May 8 '13 at 13:22
1  
@nonagon, what I meant is that you see this happening in the world, so you know that circular motion HAS to take place. Given that assumption(based on observation), we can prove it very well using the previously proven work energy laws. I agree that it appears like circular reasoning, but give it some thought, and I hope you will see. –  Saurabh Raje Jul 7 '13 at 16:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.