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