Centripetal Force and Work: Understanding the Relationship

[EthanVandals]

Homework Statement

A 5kg ball is swung in a circle of radius .5 meters at a constant speed of 2 meters per second, how much work does the centripetal force do as the ball does once in a circle?If the ball were pushed the same distance as the circumference of the circle with a force equivalent to the centripetal force but in a straight line on a friction-less surface, what is the work this force would do? Explain why the answers are different. (Hint: They are, review the definition of work)

Homework Equations

Unknown

The Attempt at a Solution

I looked through all of the slides and notes that the professor posted, as well as my own notes, but I couldn't find anything relating to this style of problem. There were many equations talking about Omega and all sorts of different ways to solve similar problems, but I could not figure out how to translate it. Could anybody give me a hand with this problem? Thanks in advance!

PS: Sorry about the way the problem is worded. I just typed it in, word for word as my professor has it on our homework.

 

[CWatters]

Answer the straight line case first. You should be able to do that and it will help with the circular case.

 

[CWatters]

You should also review the definition of work. Its not just "force times distance", there is a bit more to it and that's important for this problem.

 

[EthanVandals]

You should also review the definition of work. Its not just "force times distance", there is a bit more to it and that's important for this problem.

Ah, I see. I found the equation for Centripetal Force (mv^2)/r, and I found that the Centripetal Force (and therefore the force that would be pushing the ball in a straight line in) is equivalent to about 40 Newtons. Applying that to the frictionless surface, using the same formula would give me the same answer. The work done by both would be 40pi Joules. Looking at the definition of work, it appears that Displacement would need to occur for there to be ANY work done at all. Therefore, since the ball on the string returned to its starting point, it could be said that no work was done at all. Is this correct? Thanks!

 

[CWatters]

It's not because it returns to the starting position. The work would still be zero if it only moved say quarter of a revolution. There is another reason.

 

[EthanVandals]

It's not because it returns to the starting position. The work would still be zero if it only moved say quarter of a revolution. There is another reason.

Oh, is it because the force itself isn't causing the displacement? Since the force is the tension pulling towards the center, but the movement is in another direction? Other than that, I can't really think of what it could be...My apologies, I'm not very good at physics.

 

[haruspex]

is it because the force itself isn't causing the displacement?

Yes. Vectorially, ##W=\int \vec F.\vec{ds}##, where the . denotes dot product. So it is only the displacement in the direction of the force that matters. (Or, equivalently, use the whole displacement but only the component of the force in that direction.)

 

[CWatters]

Since the force is the tension pulling towards the center, but the movement is in another direction?

Specifically the force and displacement are orthogonal (at right angles to each other). So the component of force in the direction of the displacement is zero.

 

[EthanVandals]

Specifically the force and displacement are orthogonal (at right angles to each other). So the component of force in the direction of the displacement is zero.

Awesome, I was hoping that was what it was. Thank you so much for the help!