Theoretical Curve Graph vs Straight Line

AI Thread Summary
The discussion centers on understanding the difference between a theoretical curve graph and a straight line in relation to the equation τ = 2π(L/g)^(1/2). Participants explore why one representation is linear while the other is curved. It is suggested that squaring the original equation can reveal a linear relationship when plotting τ^2 against L. The conversation emphasizes the mathematical transformation that leads to this distinction. Ultimately, the relationship between the variables is clarified through the manipulation of the equation.
Richard Ros
Messages
18
Reaction score
0

Homework Statement


What could be the reason why the graph is formed the way it is?

Homework Equations



τ = 2π(L/g)^(1/2)

The Attempt at a Solution


I don't know how to explain it. Anyone know why one is a linear and another is a curve?
 

Attachments

  • 1.jpg
    1.jpg
    16.2 KB · Views: 479
  • 2.jpg
    2.jpg
    30.4 KB · Views: 445
Physics news on Phys.org
Richard Ros said:

Homework Statement


What could be the reason why the graph is formed the way it is?

Homework Equations



τ = 2π(L/g)^(1/2)

The Attempt at a Solution


I don't know how to explain it. Anyone know why one is a linear and another is a curve?

An equation of the form y=ax+b is the most general form a linear equation. If you square your original equation can you see how it must have a linear form if y=τ^2 and x=L?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top