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

Click For Summary
SUMMARY

The discussion centers on determining the critical velocity, denoted as c, for a planar pendulum with a weightless rod of length l and a bob of mass m, under the influence of gravity. The critical velocity is defined as the minimum speed required for the pendulum bob to reach the upward vertical position after passing through the downward position. The solution involves applying the conservation of energy principle, where the kinetic energy at the lowest point (1/2 mc²) must equal the potential energy at the highest point (2mgl). Thus, the equation 1/2 mc² = 2mgl is established to solve for c.

PREREQUISITES
  • Understanding of conservation of energy principles in mechanics
  • Familiarity with kinetic energy and potential energy formulas
  • Basic knowledge of pendulum dynamics
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of energy conservation equations in mechanical systems
  • Learn about the dynamics of pendulums and their motion equations
  • Explore examples of energy transformations in physics problems
  • Investigate the effects of friction and air resistance on pendulum motion
USEFUL FOR

Students studying mechanics, physics educators, and anyone interested in understanding pendulum dynamics and energy conservation principles.

ElDavidas
Messages
78
Reaction score
0
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
 
Physics news on Phys.org
Alright, you know the formula for kinetic energy is E_K=\frac{1}{2}mv^2 and the formula for gravitational potential energy near the surface of the Earth is E_P=mgh. 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 E_K+E_P=E=constant. 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?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
23
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K