Max velocity for simple pendulum oscillation

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Homework Help Overview

The discussion revolves around deriving the relationship between the maximum velocity of a simple pendulum and its maximum angular velocity, specifically examining the equation v_max = w_max * Length of string. The subject area includes concepts from oscillatory motion and angular dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between linear and angular velocity, questioning the definitions and derivations involved. Some suggest using energy conservation principles, while others propose solving the simple harmonic motion differential equation. There is also mention of the tangential velocity equation v_T = Rω and its derivation.

Discussion Status

The discussion is active, with participants offering various approaches and questioning the definitions of terms used in the problem. There is no explicit consensus yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants note the importance of careful definitions and the need to derive equations from fundamental principles. There is an acknowledgment of potential gaps in understanding the relationship between angular and linear quantities in the context of pendulum motion.

PhizKid
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Homework Statement


Show that v_max = w_max * Length of string where v_max is the velocity of the simple pendulum and w_max is the maximum angular velocity.


Homework Equations


\omega_{velocity} = -\theta_{max} \cdot \omega_{frequency} \cdot sin(\omega_{frequency} \cdot t + \phi)


The Attempt at a Solution


The closest resemblance I could find was using energy:

\frac{1}{2}kx^2 = \frac{1}{2}mv_{max}^2 where x = some displacement. The displacement in this pendulum's clase would be the Length of the string * angular displacement because s = Lθ.

Solving for v_max gives: v_{max} = \omega_{frequency} \cdot x

But the 'x' in this case represents the displacement made by the pendulum (the arclength subtended by the angular displacement).

Is there another equation I'm unaware of?
 
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Well yes - via ##v_{T}=R\omega## perhaps?

I think, to understand the problem, you need to say where you have to start from
... possibly by deriving the equation for "v" - however it has been defined.
I don't think you have been very careful with the definitions of terms.
 
You may need to solve the SHM DE of

d2x/dt2 + ω2x = 0

Are you able to solve this DE or know the solutions to this DE? (it is similar to your relevant equation)
 
Simon Bridge said:
Well yes - via ##v_{T}=R\omega## perhaps?

I think, to understand the problem, you need to say where you have to start from
... possibly by deriving the equation for "v" - however it has been defined.
I don't think you have been very careful with the definitions of terms.

Oh, so there is an equation that states ##v_{T}=R\omega##. I did not know this. How do you derive this?

rock.freak667 said:
You may need to solve the SHM DE of

d2x/dt2 + ω2x = 0

Are you able to solve this DE or know the solutions to this DE? (it is similar to your relevant equation)
My relevant equation is just the derivative to this solution. I didn't think the angular position equation was more useful than its derivative so I only included the equation for angular velocity.
 
PhizKid said:
My relevant equation is just the derivative to this solution. I didn't think the angular position equation was more useful than its derivative so I only included the equation for angular velocity.

Yes and that is why I thought by show, you would have had to show the steps in the derivation towards the DE.
 
PhizKid said:
Oh, so there is an equation that states ##v_{T}=R\omega##. I did not know this. How do you derive this?
<puzzled> the pendulum is an example of circular motion:

The distance ##s## on an arc, radius ##r##, changes with angle ##\theta## as ##ds=r d\theta## (follows from the definition of "angle"). To get the tangential velocity ##v_{T}## just divide through by ##dt##.
 

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