Finding the Velocity of the Bob on a Pendulum

  • Thread starter Thread starter LukeStarChief
  • Start date Start date
  • Tags Tags
    Pendulum Velocity
Click For Summary
SUMMARY

The problem involves calculating the speed of a bob on a 2.0 m pendulum released from rest at a 25-degree angle with the vertical. The key equations used are kinetic energy (KE = 1/2 mv²) and potential energy (PE = mgh). The correct approach to find the height is to use the formula h = 2 - 2cos(25), leading to a final velocity of 1.92 m/s at the bottom of the pendulum. The initial calculation of 0.59 m/s was incorrect due to misunderstanding the height calculation.

PREREQUISITES
  • Understanding of kinetic energy and potential energy concepts
  • Familiarity with trigonometric functions, specifically cosine
  • Basic knowledge of pendulum motion and energy conservation
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about the derivation and application of pendulum motion equations
  • Explore trigonometric identities and their applications in physics problems
  • Investigate the effects of varying angles on pendulum speed and height
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of pendulum motion calculations.

LukeStarChief
Messages
2
Reaction score
0

Homework Statement


A 2.0 m pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the string?


Homework Equations


KE=1/2mv^2
PE=mgh


The Attempt at a Solution


I thought that height would just be 2cos25. Since at the bottom of the pendulum, all PE is converted to KE, PE would equal KE. That makes sense. So my equation was gh=1/2v^2 (Mass cancels). My velocity came out to be .59 m/s and I'm unsure if this is correct or not. Some other sites for this problem talked about height being 2-2cos25. If that is true, why would that be height as opposed to just 2cos25?
 
Physics news on Phys.org
attachment.php?attachmentid=41262&stc=1&d=1322276156.jpg
 

Attachments

  • Fig1.jpg
    Fig1.jpg
    6.3 KB · Views: 699
gneill said:
attachment.php?attachmentid=41262&stc=1&d=1322276156.jpg

I've Calculated the velocity as 1.92m/s now. The diagram helped, thank you!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 31 ·
2
Replies
31
Views
4K