What is the Period of a Physical Pendulum with a Horizontal Knife-Edge Pivot?

Click For Summary
SUMMARY

The discussion centers on calculating the period of a physical pendulum, specifically a 0.50-kg ring with a radius of 0.60 m suspended from a horizontal knife-edge pivot. The relevant formula for the period is T = 2π / ω, where ω is the angular frequency. A key insight shared is the application of the parallel-axis theorem to determine the moment of inertia, which is crucial for solving the problem. The HyperPhysics resource is recommended for further understanding of the physical pendulum concept.

PREREQUISITES
  • Understanding of physical pendulums and their dynamics
  • Familiarity with the formula T = 2π / ω
  • Knowledge of the parallel-axis theorem in physics
  • Basic grasp of angular frequency and moment of inertia
NEXT STEPS
  • Study the application of the parallel-axis theorem in physical pendulum problems
  • Learn about calculating moment of inertia for various shapes
  • Explore the concept of angular frequency and its relation to oscillatory motion
  • Review resources on physical pendulums, such as HyperPhysics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to explain the dynamics of physical pendulums.

Joshua A
Messages
10
Reaction score
0

Homework Statement



A thin 0.50-kg ring of radius R = 0.60 m hangs vertically from a horizontal knife-edge pivot about which the ring can oscillate freely.

If the amplitude of the motion is kept small, what is the period?

Homework Equations



T = 2pi / ω

Not sure what others...

The Attempt at a Solution



I have no idea what a "horizontal knife-edge pivot" is (I tried to Google it with no success) so I actually don't know where to begin with this problem because I cannot understand what is happening. I'd appreciate an explanation as to what the question means.
 
Physics news on Phys.org
Joshua A said:

Homework Statement



A thin 0.50-kg ring of radius R = 0.60 m hangs vertically from a horizontal knife-edge pivot about which the ring can oscillate freely.

If the amplitude of the motion is kept small, what is the period?

Homework Equations



T = 2pi / ω

Not sure what others...

The Attempt at a Solution



I have no idea what a "horizontal knife-edge pivot" is (I tried to Google it with no success) so I actually don't know where to begin with this problem because I cannot understand what is happening. I'd appreciate an explanation as to what the question means.

This is the arrangement. The ring hangs from a knife, the edge of the knife is the pivot. It is a "physical pendulum". http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

upload_2017-11-25_9-15-36.png
 

Attachments

  • upload_2017-11-25_9-15-36.png
    upload_2017-11-25_9-15-36.png
    2 KB · Views: 1,575
  • Like
Likes   Reactions: Joshua A and scottdave
ehild said:
This is the arrangement. The ring hangs from a knife, the edge of the knife is the pivot. It is a "physical pendulum". http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

View attachment 215582

Thank you! I figured out the question now.

Hint for anyone who may stumble upon this thread in the future: when using the equation on the HyperPhysics page, note that it says Isupport. Use the parallel-axis theorem to help you find this.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
1
Views
2K
Replies
6
Views
3K
Replies
12
Views
11K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
6K
  • · Replies 8 ·
Replies
8
Views
3K