View Full Version : centripetal forces
UrbanXrisis
Nov14-04, 11:13 AM
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
Nov14-04, 12:28 PM
Is the cart massless? Is there an angle between the ground and the rope?
UrbanXrisis
Nov14-04, 12:34 PM
massless cart, parallel between ground and rope
UrbanXrisis
Nov14-04, 12:35 PM
actually, I think I am supposed to find the relationship between the centripetal force and v^2/r
nolachrymose
Nov14-04, 12:46 PM
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
Nov14-04, 12:50 PM
I need to plot a graph of (v^2/r) vs Centripetal force. There is a linear relationship, why is that?
UrbanXrisis
Nov14-04, 01:03 PM
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
Nov14-04, 03:45 PM
Its the same as asking you to plot F vs a, a is just determined differently in this case.
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).
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