# Block moving on a circular track-Work energy Circular Motion problem

Tanya Sharma

## Homework Statement

A block of mass 1 Kg initially at rest at point B is being pulled by a constant force F= 100N along a smooth circular track of radius 10 m . Find the velocity of the block when it reaches point P making an angle 60 with the vertical.

## The Attempt at a Solution

KE1+PE1+W[F]=KE2+PE2

Now KE1=PE1=0 assuming reference level to be at B.
KE2=1/2(mv[2]
PE2=mgl(1-cos60°)
W[F]=?

How do we calculate work done by force F? My understanding says that angle between constant force and displacement changes continuously as the block moves along the track.Please help me.Thanks...

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• block_circular.jpg
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azizlwl
F is a conservative force.

Tanya Sharma
No... F is force applied by some external agent.

Tanya Sharma
The block is pulled via a light inextensible string by external force F.

Mentor
How do we calculate work done by force F? My understanding says that angle between constant force and displacement changes continuously as the block moves along the track.
Yes, the angle continually changes but luckily the force is constant. What's the displacement in the direction of that constant force?

Tanya Sharma
Thanks Doc Al ...I have thought about displacement in the direction of constant force but not able to comprehend ...Sorry...please explain

Tanya Sharma
I feel the direction of force is continuously changing as the block moves.

Mentor
I feel the direction of force is continuously changing as the block moves.
So it's not a constant force? Did you misstate something in the problem statement?

Homework Helper

## Homework Statement

A block of mass 1 Kg initially at rest at point B is being pulled by a constant force F= 100N along a smooth circular track of radius 10 m .

Tanya, is the magnitude of the force constant and acting along the track or is it really a constant horizontal force? The drawing suggest a constant horizontal force.

ehild

Tanya Sharma
The block is being pulled by a light inextensible string which passes over a small frictionless massless pulley . The string is pulled by a constant horizontal force of magnitude 100 N .

Tanya Sharma
What i meant by change of direction of force is that the magnitude of tension in the string pulling the block will remain constant but the direction of this tension force changes.Whereas the displacement of the block is always tangential to the circular track.

Mentor
The string is pulled by a constant horizontal force of magnitude 100 N .
Good. So the direction of that force F is kept horizontal. (Makes things easier.)

Mentor
What i meant by change of direction of force is that the magnitude of tension in the string pulling the block will remain constant but the direction of this tension force changes.Whereas the displacement of the block is always tangential to the circular track.
Does the tension do any work on the block?

As far as the work done by that constant force F, go back to my earlier question: What's the displacement in the direction of that constant force?

Tanya Sharma
Please correct me if i am wrong...the constant force is being applied on the string which in turn pulls the block.So , now external force on the block is the tension .Now since the string is massless the force applied on the string = Tension in the string. But the angle between this tension force and the displacement of the block changes continuously isn't it ??

Doc Al ...We have to relate displacement of the block with the tension not the constant horizontal force... isn't it ?

Mentor
Please correct me if i am wrong...the constant force is being applied on the string which in turn pulls the block.So , now external force on the block is the tension .Now since the string is massless the force applied on the string = Tension in the string. But the angle between this tension force and the displacement of the block changes continuously isn't it ??
Yes, you are correct. I guess I wasn't paying close enough attention to your diagram. (I was thinking of something else. ) Nonetheless, there's a much easier way to solve this and my suggestion stands as is.
Doc Al ...We have to relate displacement of the block with the tension not the constant horizontal force... isn't it ?
Ask yourself: How much work is done on the system by that constant force? That's all you need to know.

Tanya Sharma
Displacement in the direction of constant force should be 10sin60° .Work done accordingly is 100 × 10(√3/2) ? is this the work done ??

Mentor
Displacement in the direction of constant force should be 10sin60° .Work done accordingly is 100 × 10(√3/2) ? is this the work done ??
Exactly. Whatever is exerting that constant force is doing work on the system.

Edit: My mistake! Please use the displacement of the applied force--how much the string moves--not the displacement of the block.

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Homework Helper
If the work is equal to the product of the force acting on the rope and the displacement of the body in the direction of that force, what is the work done on the block in the picture by the force F when it raises with height h? Is it zero because the force and displacement enclose 90°angle?

ehild

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Tanya Sharma
echild... sir... please help me with the problem...What you have asked ...according to Doc Al... that should be zero...But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F... Please help with my problem...

Homework Helper
The tension acts along a chord which makes angle β with the tangent of the circle. The work during displacement ds along the arc is Tds cos(β). β=(\pi/3-φ)/2. The total work of the rope is $$W=\int_0^{\pi/3}{TR\cos(\pi/6-0.5 \phi)d\phi}$$

There is an easier approach: The rope is pulled by the horizontal constant force F. The end of the rope moves horizontally along a distance L, the work done by the pulling force is FL. The rope is inextensible, so L is equal to the length of the chord between the initial and final position of the block.

ehild

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• track.JPG
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Homework Helper
echild... sir... please help me with the problem...What you have asked ...according to Doc Al... that should be zero...But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F... Please help with my problem...

You are exactly right. See my next post.

ehild

Tanya Sharma
brilliant...echild...this was what i was looking for ...how have you calculated β ?

Tanya Sharma
sir...how have you calculated β in terms of ∅ ?

Homework Helper
beta is the tangent chord angle, half the central angle alpha.
http://www.regentsprep.org/Regents/math/geometry/GP15/CircleAngles.htm
Do you understand the other method, too?

ehild

Mentor
Not at all. The force applied times the displacement of the point of application of that force is certainly not zero.
But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F... Please help with my problem...
It's an excellent exercise to think in terms of the tension and compute the work done by that force directly. But there's an easier way.

Tanya Sharma
echild...i perfectly understand the first method...regarding second method...work done by pulling force is FL which is work done on the system...but system comprises of block and string wherein string is massless so its mechanical energy is zero .Total energy of system is equal to total energy of block ...?am i right ...this means if the string were having mass then we would not be able to apply the second method?because here

Tanya Sharma
because here tension on the block would not be equal to the force applied on the string...please reply

Mentor
Yikes! Silly me. Earlier I meant to use the displacement of the applied force and not that of the block. My apologies. (Can I blame it on lack of sufficiently strong coffee? )

Homework Helper
because here tension on the block would not be equal to the force applied on the string...please reply

You are right again. If the string has mass, the force have to accelerate it. There would act different forces at the ends of a the string.

So what is L? how is it related to the radius and the 60°angle?
Can you do the integral method? You will see that both methods give the same result.

ehild

Homework Helper
Yikes! Silly me. Earlier I meant to use the displacement of the applied force and not that of the block. My apologies. (Can I blame it on lack of sufficiently strong coffee? )

Well, have you got that coffee since then? You totally confused me, so as I deleted my post and had to rewrite it when the example of the simple pulley occurred to me. And I was surprised by the result of the integral method. (I blame it to the hot damp weather here. It is difficult to think.)

ehild

Homework Helper
echild...i perfectly understand the first method...regarding second method...work done by pulling force is FL which is work done on the system...but system comprises of block and string wherein string is massless so its mechanical energy is zero .Total energy of system is equal to total energy of block ...?am i right ...this means if the string were having mass then we would not be able to apply the second method?because here

When you calculate the speed of the block, do not forget the work of gravity. The change of KE is equal to the work of all forces...

ehild

Tanya Sharma
echild ...thanks a ton... what an explanation...u have made physics enjoyable...i m very very grateful

thanks to Doc Al ...u were correct but the language confused me...

Homework Helper
Thanks Tania. You are really very kind. I am pleased that you feel Physics enjoyable. It is.
Remember Doc Al sentence: The work done on a system by a force is
The force applied times the displacement of the point of application of that force
. It is very important.

ehild

Mentor
Well, have you got that coffee since then? You totally confused me, so as I deleted my post and had to rewrite it when the example of the simple pulley occurred to me. And I was surprised by the result of the integral method. (I blame it to the hot damp weather here. It is difficult to think.)

ehild
Sorry about that, ehild! And yes, I've been drinking a lot of coffee since then. :tongue:

Homework Helper
Sorry about that, ehild! And yes, I've been drinking a lot of coffee since then. :tongue:

Take care, too much coffee is bad to hearth. (I also had a coffee since then. )

ehild