Yes, that is correct. This is known as the critical velocity for a pendulum.

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Homework Help Overview

The discussion revolves around a mechanics problem involving a planar pendulum, focusing on the concept of critical velocity necessary for the pendulum bob to reach the upward vertical position. The problem involves conservation of energy principles and the relationship between kinetic and potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy to determine the critical velocity, with references to kinetic and potential energy formulas. Questions arise regarding the setup of the problem and the interpretation of energy states at different positions of the pendulum.

Discussion Status

Some participants have provided guidance on using energy conservation principles, while others are exploring the implications of their calculations. There is an ongoing examination of how to relate kinetic energy at the bottom of the swing to potential energy at the top, with no explicit consensus reached yet.

Contextual Notes

The problem constraints include neglecting friction forces and the need to define a reference height for potential energy calculations. Participants are also considering the implications of different velocities on the pendulum's motion.

ElDavidas
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Got another question for everyone.

I've been looking over some exam past papers for mechanics and I'm stuck on a problem.

Question reads:

"Suppose that a planar pendulum has a weightless rod of length l and a pendulum bob of mass m. The only external force acting on the pendulum is gravity of magnitude mg.

There is a number c such that, if the pendulum bob passes through the downward position with a speed of magnitude > c, it will eventually pass through the upward vertical position, and if it passes through the downward vertical position with a speed of magnitude < c, it will never reach the upward vertical position. Determine the number c. Neglect friction forces in this problem."

There is a diagram that goes with this problem but it's going to be difficult to draw. Basically comprises of a horizontal y-axis (pointing right), a vertical x-axis (pointing downwards) and a vector drawn with an angle delta between the vector and the x axis.

I'm fairly certain you have to use conservation of energy but don't really know where to begin. I understand the concept of conservation of energy but don't know how to apply it to problems.

Thanks

Dave
 
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Alright, you know the formula for kinetic energy is [itex]E_K=\frac{1}{2}mv^2[/itex] and the formula for gravitational potential energy near the surface of the Earth is [itex]E_P=mgh[/itex]. m is the mass of the object, v is its velocity, h is the hieght and g is acceleration due to gravity. You know that [itex]E_K+E_P=E=constant[/itex]. So what you need to do is find the kinetic and potential energies at the initial and final stages of this process. Keep in mind that it does not matter where you choose h to be zero, since only changes in hieght matter. A convenient place would be at the bottom of the pendulum. Think about what the kinetic energy should be at the top. Remember: we want to find the minimum
energy we need to put in for the mass to reach the top.
 
Ok, I think I follow you.

Been looking over my mechanics notes and they say:
T2 - T1 = V2 - V1

where Ti, Vi represent kinetic and potential energies respectively at times t1 and t2.

So if I make the height 0 at the bottom of the pendulum, this would mean the height of the top of the pendulum is 2l (l is the length of the rod).

Using the formula for potential energy, this implies

V2 - V1 = 2mgl - mg(0) = 2mgl ?

Not quite sure about what to do with the kinetic energies. If the velocity is > c (and therefore the pendulum reaches the top) then can I say:

T2 - T1 = 1/2mc^2 - 1/2mc^2

and if the velocity is < c then

T2 - T1 = 1/2mc^2 ?

Don't think the kinetic energy is right though.
 
ElDavidas:
Remember that the minimum velocity c that achieves this, is that all kinetic energy at the bottom position becomes converted into potential energy at the top position.
 
arildno said:
Remember that the minimum velocity c that achieves this, is that all kinetic energy at the bottom position becomes converted into potential energy at the top position.

Hmmm, ok. So does this mean you let

1/2mc^2 = 2mgl

and solve for c?
 

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