Tension in a Pendulum String - Oh my

Click For Summary
SUMMARY

The discussion focuses on calculating the tension in a pendulum string when a 0.030 kg mass is released from a 25° angle. The correct formula used is T = mg sin(65°), leading to a tension value of 0.27 N. The conversation clarifies that while centripetal force is zero at the extreme position, this does not imply that tension is zero. The net force in the radial direction must account for gravitational forces.

PREREQUISITES
  • Understanding of basic physics concepts such as tension and centripetal force.
  • Knowledge of trigonometric functions, specifically sine, in relation to angles.
  • Familiarity with Newton's laws of motion.
  • Ability to perform calculations involving mass, gravity, and angles.
NEXT STEPS
  • Study the derivation of tension in pendulum systems using different angles.
  • Learn about the effects of mass and length on pendulum dynamics.
  • Explore the relationship between centripetal force and tension in circular motion.
  • Investigate the role of gravitational force in pendulum motion at various angles.
USEFUL FOR

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

Engineering101
Messages
6
Reaction score
0
So I have one attempt left and want to make sure my answer is correct, so double checking me would be super awesome!

You are given a pendulum composed of a 0.030 kg mass on the end of a 0.42 m long massless string. The pendulum is moved 25° from the vertical and released. Find the tension in the string when the mass is at the release point,
θ = 25°.

Tsin(65)-mg=0
Tsin(65)=mg
T=.32 N

This is one solution I came up with.

The other option I am unsure of is if the T force would be 0 because "there is no net force directed along the axis that is perpendicular to the motion. Since the motion of the object is momentarily paused, there is no need for a centripetal force." -http://www.physicsclassroom.com/class/waves/u10l0c.cfm

Thanks for the guidance.
 
Physics news on Phys.org
The centripetal force is indeed zero in the extreme position. However this does not imply T=0.
The net force along the radial direction should be zero. And this net force includes gravity's contribution as well.

The first "option" is not good because the body's acceleration along the vertical direction is not zero.
 
  • Like
Likes harmyder
So,
Fradial= -T + mgsin(65)
T=mgsin(65)
T=.27 N
 
So I have one attempt left and want to make sure my answer is correct, so double checking me would be super awesome!

You are given a pendulum composed of a 0.030 kg mass on the end of a 0.42 m long massless string. The pendulum is moved 25° from the vertical and released. Find the tension in the string when the mass is at the release point,
θ = 25°.


"The centripetal force is indeed zero in the extreme position. However this does not imply T=0.
The net force along the radial direction should be zero. And this net force includes gravity's contribution as well."--nasu




So,
Fradial= -T + mgsin(65)
T=mgsin(65)
T=.27 N
 
Engineering101 said:
So I have one attempt left and want to make sure my answer is correct, so double checking me would be super awesome!

You are given a pendulum composed of a 0.030 kg mass on the end of a 0.42 m long massless string. The pendulum is moved 25° from the vertical and released. Find the tension in the string when the mass is at the release point,
θ = 25°.


"The centripetal force is indeed zero in the extreme position. However this does not imply T=0.
The net force along the radial direction should be zero. And this net force includes gravity's contribution as well."--nasu




So,
Fradial= -T + mgsin(65)
T=mgsin(65)
T=.27 N

It is correct.

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
21
Views
889
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 21 ·
Replies
21
Views
3K