How to graph Fc=mv^2/r so that the gradient = velocity

Click For Summary

Homework Help Overview

The discussion revolves around the equation Fc = mv^2/r, which is being analyzed in the context of graphing data related to a pendulum's motion. Participants are exploring how to represent this relationship graphically to derive the velocity of the object.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to graph Fc against various variables, including r and v^2, and are discussing the implications of these graphs on understanding velocity. There are questions about the clarity of the measurements taken and the appropriateness of the equation in relation to pendulum motion.

Discussion Status

Some participants have provided insights into the confusion surrounding the question, with suggestions to clarify the relationship between the variables involved. Others express uncertainty about the original poster's measurements and the relevance of the equation to the context of a pendulum.

Contextual Notes

There is mention of an exam context and potential discrepancies in the question posed, with participants questioning the accuracy of the relationship between the equation and pendulum motion. The original poster indicates a lack of clarity regarding the experimental setup and measurements taken.

lachtal9
Messages
3
Reaction score
0

Homework Statement


Fc = mv^2/r represents the motion of a simple pendulum. Describe how this data could be graphed so that the gradient of a straight line could be used to determine the velocity of the object.

Homework Equations


Fc = mv^2/r

The Attempt at a Solution


I'm kinda stumped. I tried graphing Fc vs r and got nowhere. Kinda attempted @ m =0.5, r =1, Fc = v^2/2 and thus when you derive Fc with respect to v, the gradient function = v, and thus the gradient of any tangent through the curve would give the value of the velocity at that point. Didn't really make sense to me
UPDATE: tried graphing Fc vs v^2 and got a straight line. realized the slope gives m/r. not sure if this is any helpful info but pls need help on the ASAP :) thankssss
 
Last edited:
Physics news on Phys.org
Hi Lachtal, :welcome:

Unclear to me what you measured. If you can graph Fc vs v2 it seems to me that you already have the velocity of the object as observations ?
So: what did you vary, what did you measure ?

And "Fc = mv^2/r represents the motion of a simple pendulum" ? More like a circular motion, I would say.
 
Hi, this is an exam question. It was at the end of a practical exam in which we evaluated an experimental value of gravity on Earth using simple pendulum motion. So we found the period of the pendulum and plotted length of string vs period squared; the gradient gave us the experimental value of g. Not sure if this last question was a bit of a mix up by the examiner, which is probable as my teacher isn't the best, but either way that is the question we were given and for the life of me I've no clue of the answer. Also, I am sorry i didnt answer your qn bc quite frankly I am not sure
 
Sounds a lot clearer to me !
For a mathematical pendulum there is an expression for the period as a function of the length. Looks a lot like what you posted, But it seems to me a plot of T2 versus l is more sensible ...
 
yes I have plotted the T^2 vs l, it's just that the this question makes specific reference to the velocity and not gravity... I think it's an incorrect question. Thanks!
 
lachtal9 said:
Fc = mv^2/r represents the motion of a simple pendulum.
It does? That's a bit of a stretch. The motion of the pendulum is governed by the differential equation for its angular displacement.
Your title "How to graph Fc=mv^2/r so that the gradient = velocity"
Isn't quite accurate. It only asks for a graph the gradient of which can be used to find the velocity. If you could plot force against radius for the same mass and speed then you could do that, but this would mean the speed is constant, which it certainly is not for pendulum.

In short, the question is a complete bungle from start to finish.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
15
Views
2K
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 19 ·
Replies
19
Views
3K