Calculate the work done by tension

[physicslover14]

Homework Statement

Consider the following arrangement
Calculate the work done by tension on 2kg block during its motion on circular track from point A to point B.

Homework Equations

The Attempt at a Solution

We know that work done by a force is product of force and displacement.
We know the displacement of point of application as 4 m. How to find the work done by the tension as it is not constant it is variable.

 

[haruspex]

Homework Statement

Consider the following arrangement
Calculate the work done by tension on 2kg block during its motion on circular track from point A to point B.

Homework Equations

The Attempt at a Solution

We know that work done by a force is product of force and displacement.
We know the displacement of point of application as 4 m. How to find the work done by the tension as it is not constanit it is variable.

Conservation of work will do the integration for you. Think about the changes in potential and kinetic energy of the two blocks. (Don't assume the system will be at rest when point B is reached.)

 

[physicslover14]

how?

 

[Tanya Sharma]

You need a couple of steps to solve the problem

1 Consider the two blocks and the string together as the system.Apply work energy theorem.Work done by tension will cancel out.This will give you the kinetic energy of the 2 kg block when it reaches point B.

2 Now,consider the 2 kg block in isolation.Again apply work energy theorem.This time the work done by tension will be equal to the change in the mechanical energy of the 2 kg block .Note that the work of the normal force by the circular track on the 2 kg block will be zero.

Hope this helps

 

[haruspex]

1 Consider the two blocks and the string together as the system.Apply work energy theorem.Work done by tension will cancel out.This will give you the kinetic energy of the 2 kg block when it reaches point B.

Not directly. It will give the total kinetic energy. You then need to work out the ratio of the speeds of the two blocks when 2kg is at point B. From the diagram, I believe 2kg should moving horizontally at that point, so you can deduce this ratio from the geometry and a bit of calculus.

 

[physicslover14]

can you show how to calculate the ratios of velocities with geometry and calculus..?

 

[Tanya Sharma]

Not directly. It will give the total kinetic energy. You then need to work out the ratio of the speeds of the two blocks when 2kg is at point B.

Well,I meant the same .

 

[haruspex]

can you show how to calculate the ratios of velocities with geometry and calculus..?

Consider the movement after reaching point B. Let X be the ground-level point directly below point A. If x is the distance of the 2kg block from X, find an equation relating x to the height of the 1kg block that's valid for x>=3.

 

[Tanya Sharma]

From the diagram, I believe 2kg should moving horizontally at that point, so you can deduce this ratio from the geometry and a bit of calculus.

haruspex,i beg to differ.Why do we need calculus to get the work done by tension ? Yes,the tension is variable but we are not required to find tension ,rather work done by tension .

Work done by tension when the 2 kg block moves from A to B =\int_{0}^{\pi R/2} T dx,where T is variable.

This work done by tension will cancel when we consider the two blocks and the string together as a system and apply work energy theorem.Now the speed of the two blocks will remain same.From this we can deduce the kinetic energy of 2kg block when it reaches B.There is no need to consider any ratio.

Next,the work done by tension is simply equal to the change in mechanical energy of the 2kg block between points A and B .

Please let me know if i am missing something obvious .

 

[physicslover14]

how can u say that their velocities would be equal?

 

[ehild]

The 2 kg block moves along the circle, so its speed is Rdθ/dt. It pulls the string, the length of the string between point O and the block can we obtained with simple geometry at any position θ (ignoring the size of the pulley). The total length of the string is unchanged, so the speed of the 1kg block is dL/dt.

ehild

 

[physicslover14]

After that what to do?

 

[ehild]

After what? Have you the expression of dL/dt in terms of theta?

ehild

 

[physicslover14]

YES ,dL/dt=Rdθ/dt.BUT THAT'S SAME AS V=RW HERE W DENOTES ANGULAR VELOCITY.

 

[haruspex]

Now the speed of the two blocks will remain same.

No. At B, the 2kg block is traveling horizontally. The string is, presumably, still taut, so makes a straight line diagonally up to the pulley. The speed of the 2kg block will not be the same as the speed of the string over the pulley.

 

[haruspex]

YES ,dL/dt=Rdθ/dt.

No. See my reply just above to Tanya Sharma.

The 2 kg block moves along the circle

I wasn't sure it was supposed to be an arc of a circle. It doesn't need to be. Just has to be sufficiently smooth, starting vertical and finishing horizontal.

 

[ehild]

I wasn't sure it was supposed to be an arc of a circle. It doesn't need to be. Just has to be sufficiently smooth, starting vertical and finishing horizontal.

From the OP:

Consider the following arrangement
Calculate the work done by tension on 2kg block during its motion on circular track from point A to point B.

ehild

 

[physicslover14]

so what is the final equation to calculate the velocity?

 

[ehild]

Try to figure it out. First find the length of string between the 2kg block and point O. Then take the derivative with respect to time.

ehild

 

[physicslover14]

first the length is 1m then 4m.

 

[physicslover14]

please help...

 

[ehild]

first the length is 1m then 4m.

No. The block does not travel vertically. It goes also sideway. And what is in between?

ehild

 

[physicslover14]

i wanted to say that the change in length is 4 m.After that?

 

[ehild]

Well, the length of the right piece changes by 4 m. But what is between the initial and final position?

ehild

 

[ehild]

Find L at a given theta.

 

[haruspex]

Find L at a given theta.

Is that necessary? I think what's needed is to find the length of the string as a function of the 2kg's distance along the baseline from, say, point B, once it is moving horizontally. (Then differentiate that and evaluate at B.)

 

[physicslover14]

dl/dt = 6cos(theta) +18sin(theta)..is it correct

 

[ehild]

Is that necessary? I think what's needed is to find the length of the string as a function of the 2kg's distance along the baseline from, say, point B, once it is moving horizontally. (Then differentiate that and evaluate at B.)

You are right. The final speed is only needed, and it can be obtained using the horizontal speed at the moment when the 2 kg mass reaches the floor.


ehild

 

[ehild]

dl/dt = 6cos(theta) +18sin(theta)..is it correct

How did you get it?

ehild

 

[physicslover14]

na its wrong i found my mistake..so what were u saying above can u explain that with a figure.

 

[ehild]

How does the length L change if the block displaces horizontally by Δx?

ehild

 

[physicslover14]

ldash^2 = 25 +dx^2 + 6dx.
ldash =root(25 +dx^2 + 6dx.)we can ignore dx^2
so ldash =root(25 + 6dx.)
change in length = root(25 + 6dx.)-5

now?

 

[ehild]

What do you think L'-L is if dx is very small?

ehild

 

[physicslover14]

negligible..but there would be some velocity at that point

 

[ehild]

Try. Let dx=0.01, 0.001... Are you familiar with Calculus?ehild

 

[physicslover14]

do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right

 

[ehild]

do we get 3(velocityof 1kg block) =5(velocity of block of 2kg).!if i did the calculation right

Just the opposite dL/dt=3/5 dx/dt. dx/dt = v2, velocity of the 2 kg block, dL/dt = v1, velocity of the 1 kg block.


ehild

 

[physicslover14]

thnx.we could directly say by string constraint no need of calculus.

 

[Tanya Sharma]

ehild...This approach gives the correct result when we consider 2 kg block at point B (which is what is asked in the OP).

But what if , instead of point B ,we had to find speed of 2kg block at any other point(say,point C mid point of the quarter circle AB).Then the length of the string would not have been function of horizontal distance alone.It would have depended on the vertical distance too.

Do we have to use partial differentiation then ?

 

[ehild]

See post #26 and find length L, the hypotenuse of the blue shaded right triangle in terms of the angle theta. Take the derivative with respect to time. V1=dL/dt and v2=Rdθ/dt.

ehild

 

[Tanya Sharma]

See
post #26 and
find length L, the hypotenuse of the blue shaded right triangle in terms
of the angle theta. Take the derivative with respect to time. V1=dL/dt
and v2=Rdθ/dt.

ehild

Thanks...Since my initial response to the problem was incorrect,i have a
doubt whether the net work done on the two blocks is zero ?

 

[ehild]

Thanks...Since my initial response to the problem was incorrect,i have a
doubt whether the net work done on the two blocks is zero ?

You mean the net work of the string, don't you? I think your argument was correct, except the equality of the speeds, and the total work of the tension is zero, but you can derive the work from the forces and displacements.

ehild

 

[Tanya Sharma]

You mean the net work of the string, don't you? I think your argument was correct, except the equality of the speeds, and the total work of the tension is zero, but you can derive the work from the forces and displacements.

ehild

Well.its pretty ironical.I had a correct reasoning about the speeds,but expressed it incorrectly .What I wanted to communicate was that the component of speeds along the direction of length of the string would be equal .

Whereas,I had a wrong understanding of work done by tension ,and wrote the correct result.

Earlier I had a flawed reasoning which prompted me to think that the net work done by the tension would be zero.But now after having a better perspective ,i am not convinced that how the work done by tension on 1 kg block would be equal and opposite to the work done by tension on 2 kg block.Afterall they cover unequal distances,with varying angle between tension and displacements in case of 2 kg block.

Why would the net work done by tension be zero ?

I had read somewhere that the net work done by the internal forces is zero .Is this a general statement or applicable in special circumstances ?

 

[haruspex]

i am not convinced that how the work done by tension on 1 kg block would be equal and opposite to the work done by tension on 2 kg block.

The tensions each side of the pulley are the same at all times, the distance moved by the string in each elemental time interval is in the direction of the corresponding tension (except one is negative), and the total distance moved by the string is the same each side. So the integral must be equal and opposite.

 

[haruspex]

thnx.we could directly say by string constraint no need of calculus.

How did you get the 3/5 without using calculus - or something equivalent?

 

[Tanya Sharma]

How did you get the 3/5 without using calculus - or something equivalent?

The component of velocity of 2 kg block along the direction of length of string will be equal to the component of velocity of 1 kg block along the length of string.

Let θ be the angle between the velocity of 2 kg block (V₂) and the direction of length of string.

When 2Kg block is at B ,V₂ cosθ = V₁ ,where cosθ =3/5.

Hence,V₁/V₂ =3/5 .

This works in this case nicely as we had the angle θ known .

 

[Tanya Sharma]

The tensions each side of the pulley are the same at all times, the distance moved by the string in each elemental time interval is in the direction of the corresponding tension (except one is negative), and the total distance moved by the string is the same each side. So the integral must be equal and opposite.

The tension acting at the 2 kg block is not tangential at every instant,whereas the displacement is tangential to the arc.The angle between tension and the displacement changes continuously.The displacement is surely not in the direction of tension in case of 2 kg block

Yes, the tension is same on both the blocks.But I still can't figure out why the work done on one block will be equal and opposite to the other block.

 

[ehild]

Why would the net work done by tension be zero ?

I had read somewhere that the net work done by the internal forces is zero .Is this a general statement or applicable in special circumstances ?

The block moves along an arc of circle, its displacement is not parallel with the string. The tension acts at the end of the string, and the end moves along the arc. At small displacement Δr the work is the dot product δW=TΔr, T multiplied by the component of the displacement along the string. The component of displacement along the string is the same for both blocks, as the length of the string does not change, but the forces act in opposite directions, so the net work is zero.

The work of internal forces is not zero in general. Think of two masses connected by a spring. Let they be equal. The spring is stretched, then relaxed. Both masses move inward by Δx. The spring does positive work on both masses, the net work is 2(1/2 kΔx²).

ehild

 

[Tanya Sharma]

The block moves along an arc of circle, its displacement is not parallel with the string. The tension acts at the end of the string, and the end moves along the arc. At small displacement Δr the work is the dot product δW=TΔr, T multiplied by the component of the displacement along the string. The component of displacement along the string is the same for both blocks, as the length of the string does not change, but the forces act in opposite directions, so the net work is zero.


ehild

Yes...that makes complete sense . Brilliant!