Centripetal Force: Poll, Rope, Cart & 80kg Person

[UrbanXrisis]

Lets say there was a poll in the middle of the ground and a 2m rope attached to it. Attached to the rope was a cart that sat a 80kg person. The person had a constant force being exerted on them so they the person went in a circle. There is a spring scale attached to the rope and poll to show the spring scale force. What would be the relationship between this spring scale force and the centripetal force?

 

[Skomatth]

Is the cart massless? Is there an angle between the ground and the rope?

 

[UrbanXrisis]

massless cart, parallel between ground and rope

 

[UrbanXrisis]

actually, I think I am supposed to find the relationship between the centripetal force and v^2/r

 

[nolachrymose]

Well, if that's the case, then you know $$v^2/r$$ is equal to the centripetal acceleration. By Newton's Second Law, you know that $$F=ma$$. That should be enough for you to finish the problem on your own.

 

[UrbanXrisis]

I need to plot a graph of (v^2/r) vs Centripetal force. There is a linear relationship, why is that?

 

[UrbanXrisis]

maybe it's just me, but I understand all these concepts already. However, it might just be the way the question is worded. All that's different between v^2/r and the centripetal force is that the centripetal force has a mass multiplied into it. The problem asks for is the "functional relationship." Cf=v^2/r times the mass. I thought it was asking for something more than that

 

[Skomatth]

Its the same as asking you to plot F vs a, a is just determined differently in this case.

 

[Diane_]

I need to plot a graph of (v^2/r) vs Centripetal force. There is a linear relationship, why is that?

You know that C = m v^2/r. Let C = y and v^2/r = x, then the law becomes

y = mx

This is a line, of course, with slope m. The centripetal force is linear in (v^2/r).