Work In Circular Motion With Tension

[Lori]

Homework Statement

The speed of the pendulum bob remains constant
as it travels around the circle

(a) Over one complete circle, how
much work does the tension force F do on the bob? (i) A positive amount; (ii) a negative
amount; (iii) zero.

(b) Over one complete circle, how much work does the weight do on
the bob? (i) A positive amount; (ii) a negative amount; (iii) zer

Homework Equations

W = Fs
W = 1/2kmv^2[/B]

The Attempt at a Solution

I was thinking that work of tension would be negative because it would balance out the work of weight which would be positive? Also, I just took a quiz and it was multiple choice, and the only choice that made sense was this!

The answer that i thought made most sense was that both work would be zero because work total is 0.

Can someone please explain?[/B]

 

[BvU]

Can someone please explain?

Sure. Is the circle you mention the bob traveling horizontal or vertical ?
With $W = Fs$ you mention an important formula. If the symbols stand for something. What is $W$, what is $F$ and what is $s$ (PS, I know, but I would like eto read what you think they stand for ...)

 

[Lori]

Sure. Is the circle you mention the bob traveling horizontal or vertical ?
With $W = Fs$ you mention an important formula. If the symbols stand for something. What is $W$, what is $F$ and what is $s$ (PS, I know, but I would like eto read what you think they stand for ...)

Sorry about the late response. Just finished class.

Anyway, the bob is swinging vertically. With W=Fs, W stands for Work , F stands for the force applied to an object, and s stands for the displacement. In this motion, i believe the force on the pendulumn thing would be centripetal force towards the center right which is tension i assume? Plus, there is displacement so work would equal to a positive value.For work done by force of weight, which is negative, would make work negative as well. So positive work from tension and negative from weight

 

[I like Serena]

Hey Lori! ;)

The symbol $s$ doesn't just stand for displacement, it stands for the displacement in the direction of the force.
It's also common to write it as $W=Fs\cos\theta$, where $\theta$ is the angle between the displacement and the force.
And if the displacement is opposite to the force, the work done is negative.

In other words, for work to be done, we need a displacement with a component in the direction of the force.
Do we have that?

 

[BvU]

Point is that work is a scalar, a number. Force and displacement are vectors: they have a magntude and a direction.
When both are in the same direction, work is indeed |Force| * |displacement|
But if there is an angle between them, the cosine of that angle comes in. Are you familiar with that ?

 

[Lori]

Hey Lori! ;)

The symbol $s$ doesn't just stand for displacement, it stands for the displacement in the direction of the force.
It's also common to write it as $W=Fs\cos\theta$, where $\theta$ is the angle between the displacement and the force.
And if the displacement is opposite to the force, the work done is negative.

In other words, for work to be done, we need a displacement with a component in the direction of the force.
Do we have that?

Ohhhhh. That makes it more easier to understand work. So, since tension is a force towards the center, there will be no displacement towards the center. Therefore, no work.

So, since the weight force is always downward, the ball has no displacement downward? But does the ball when it swings down count as displacement?

 

[Lori]

Point is that work is a scalar, a number. Force and displacement are vectors: they have a magntude and a direction.
When both are in the same direction, work is indeed |Force| * |displacement|
But if there is an angle between them, the cosine of that angle comes in. Are you familiar with that ?

That makes more sense that when work has to be scalar, the direction of force and displacement should be the same. Thanks! I guess now I understand why we need to take the component of a force that's in the direction of displacement!

 

[I like Serena]

Ohhhhh. That makes it more easier to understand work. So, since tension is a force towards the center, there will be no displacement towards the center. Therefore, no work.

So, since the weight force is always downward, the ball has no displacement downward? But does the ball when it swings down count as displacement?

Yep.
If the ball swings downward, gravity would indeed do some work, which would result in more kinetic energy - that is, more speed.
But apparently the speed is constant.
I'm not quite sure yet what your setup is, but if the speed is constant that implies that no work was done.

 

[Lori]

Yep.
If the ball swings downward, gravity would indeed do some work, which would result in more kinetic energy - that is, more speed.
But apparently the speed is constant.
I'm not quite sure yet what your setup is, but if the speed is constant that implies that no work was done.

Kinda confused cause if the ball is at constant speed, don't we still say that gravity did work? Constant speed implies that the total work is zero since the net force is 0. But, if we were just looking at work done by gravity, as separate measurements, wouldn't it still be negative work?

 

[I like Serena]

Kinda confused cause if the ball is at constant speed, don't we still say that gravity did work? Constant speed implies that the total work is zero since the net force is 0. But, if we were just looking at work done by gravity, as separate measurements, wouldn't it still be negative work?

If the speed doesn't change, no work was done, so the total work is 0.
The net force can still have effect and change the direction of the velocity, but if the speed doesn't change, that still means that the work was 0.
This would be consistent with a ball that makes a circle at constant height, meaning that gravity does not do any 'work'.

 

[BvU]

If the speed doesn't change, no work was done, so the total work is 0.

This leads to misunderstanding: while going around, the bob has a constant kinetic energy, but NOT a constant potential energy from gravity. So mechanical work IS being done on the bob. But not by the tension -- we have already concluded that the tension force along the pendulum is perpendicular to the motion at all times. The work is being done by the force (torque) required to keep the angular velocity and thereby the speed constant. This can not be achieved with a wire, only with a stiff pendulum. What is gained in potential energy while going up is lost when going down, and over a full cycle the net result is zero.
The potential field from gravity is conservative: returning to the same point means returning to the same potential energy. So no work done over a full cycle.

 

[Lori]

This leads to misunderstanding: while going around, the bob has a constant kinetic energy, but NOT a constant potential energy from gravity. So mechanical work IS being done on the bob. But not by the tension -- we have already concluded that the tension force along the pendulum is perpendicular to the motion at all times. The work is being done by the force (torque) required to keep the angular velocity and thereby the speed constant. This can not be achieved with a wire, only with a stiff pendulum. What is gained in potential energy while going up is lost when going down, and over a full cycle the net result is zero.
The potential field from gravity is conservative: returning to the same point means returning to the same potential energy. So no work done over a full cycle.

Ahh, so the full cycle is what explains why total work from weight is zero. The negative work as ball swings down balances work as ball swings up , simply?

My lecture today had a bunch of clicker questions and people got confused with work and force with different directions.

 

[I like Serena]

This leads to misunderstanding: while going around, the bob has a constant kinetic energy, but NOT a constant potential energy from gravity.

Not sure what you're saying here.
If the only factors are gravitational energy and kinetic energy, than a lack of change in speed is definitely a lack of work being done by gravity.
The force of tension isn't particularly relevant, since it's an internal force. That is, any form of tension is compensated by action=-reaction as Newton's 3rd law prescribes.

 

[haruspex]

Not sure what you're saying here.
If the only factors are gravitational energy and kinetic energy, than a lack of change in speed is definitely a lack of work being done by gravity.
The force of tension isn't particularly relevant, since it's an internal force. That is, any form of tension is compensated by action=-reaction as Newton's 3rd law prescribes.

I believe BvU is saying that if the speed is constant then some other mechanism is exerting a torque on the pendulum to achieve this. There is a work exchange between this mechanism and gravity.

Edit: but such a mechanism appears to violate the dictionary definition of a pendulum.

 

[conscience]

I agree with haruspex .

 

[BvU]

Metoo (Although nowadays I don't know if that wording can still be used :) ) but the problem statement tries to make things easier by stating a constant speed. Thus making life more diffficult for those who think a bit more than others ...

 

[I like Serena]

Ah, I get it now. I misunderstood the problem statement, which is badly phrased.

First off, a pendulum does not make a circle, but a section of a circle. Perhaps a cycle was meant?
And the speed is not constant, but we have the same speed at the beginning and end of a cycle.

 

[I like Serena]

Ahh, so the full cycle is what explains why total work from weight is zero. The negative work as ball swings down balances work as ball swings up , simply?

My lecture today had a bunch of clicker questions and people got confused with work and force with different directions.

Gravity does positive work as the weight swings down. It displaces in the same direction as the force of gravity after all.
Gravity does an equal amount of negative work when the weight swings up.
And yes, the total work is 0.