Circular Motion: Understanding Centripetal Force

[Peter G.]

Hi,

So I was reading that no work is done by the centripetal force during uniform circular motion because the force is at 90 degrees to the displacement.

I am having a hard time getting my mind around it because isn't velocity that is a tangent to the circle, and therefore, at 90 degrees to the force? I keep picturing the displacement as the circle shape itself!

Any help?

Thanks

 

[cupid.callin]

You know about Work-Energy Principle

Change in kinetic energy = net work done on the body

as kinetic energy is not changing so net work done (i.e. by tension) is 0

 

[Peter G.]

But it asks us to prove why Kinetic Energy is not changing, so we have to say about how the force and the displacement are at 90 degrees, but from my understanding, the velocity and the force are at 90 degrees. Is there any definition of work that has to do with force and velocity?

 

[gneill]

Work done per unit time is power, P. But P = F x V, where "x" is the cross product. Use your given directions for the force and velocity vectors to deduce something about the work done.

 

[Peter G.]

Ok thanks gneill, that was what I wanted. I will try it now.

 

[Peter G.]

Sorry, I never studied multiplication of vectors and I've been studying them now but I am a bit confused . Can you help me please?

Or, if I can assume that the velocity is the direction of motion, than, I am fine agreeing that the direction of motion is at 90 degrees to the force.

 

[gneill]

Sorry, I never studied multiplication of vectors and I've been studying them now but I am a bit confused . Can you help me please?

Or, if I can assume that the velocity is the direction of motion, than, I am fine agreeing that the direction of motion is at 90 degrees to the force.

Your math text should have an explication of cross product. But actually I misspoke. The power is the DOT PRODUCT of the force and velocity, not the cross product. This is what comes from typing faster than thinking!

This makes sense because power is a scalar quantity, not a vector, and the result of a dot product is a scalar value.

To make a long story short, the dot product of two vectors, A and B, is given by |A||B|cos(θ),
where the | | represents the magnitude of the vector, and θ is the angle between the vectors.

For two perpendicular vectors θ = 90°, and cos(θ) = 0...

Also, yes, you can assume that the velocity vector is in the (instantaneous) direction of motion. After all, it's more or less the definition of velocity!

 

[Peter G.]

Ok, cool!

Feels so much better when I understand!

Thanks gneill