- #1
Circular Movement Homework: Part A Done, Part B Troubles
Click For Summary
Homework Help Overview
The discussion revolves around a problem related to circular motion, specifically focusing on the tension in a string as a function of angle in a scenario involving a ball moving along a circular path. The original poster has completed part A of the assignment but is encountering difficulties with part B.
Discussion Character
- Exploratory, Assumption checking, Conceptual clarification
Approaches and Questions Raised
- Participants discuss the need to express velocity as a function of angle and question the constancy of velocity throughout the motion. There are hints about using energy conservation principles and considerations regarding the role of tension in the energy equation.
Discussion Status
The conversation is ongoing, with participants exploring various approaches to the problem. Some guidance has been provided regarding the conservation of energy and the relationship between potential and kinetic energy. Questions about the nature of tension as a non-conservative force and its implications for energy calculations have also been raised.
Contextual Notes
The original poster mentions a lack of familiarity with the conservation of energy concept, which may impact their understanding of the problem. There is also an acknowledgment that tension does not perform work on the ball due to its perpendicular relationship to the motion.
- #2
Doc Al
Mentor
- 45,584
- 2,459
You need to find the tension as a function of angle. (Forget the tension at the top.)
You have the correct equation. Hint: Express v^2 as a function of angle.
You have the correct equation. Hint: Express v^2 as a function of angle.
- #3
- #4
Doc Al
Mentor
- 45,584
- 2,459
The equation I had in mind was: mv^2/r = T + mg cos(theta).
Another hint: What's conserved as the ball continues on its path?
Another hint: What's conserved as the ball continues on its path?
- #5
asi123
- 254
- 0
Doc Al said:
Ok, I already wrote that equation at the beginning in the solution part I uploaded and got stuck there.
I got stuck because I don't know the velocity at this point, I mean, the velocity is not constant, right? there is a mgsin(theta) that keep changing it...I think.
The velocity at the top is not the velocity at the bottom, right?
10x.
- #6
Doc Al
Mentor
- 45,584
- 2,459
The velocity is definitely not constant. Reread my hint in post #4.asi123 said:
Given the velocity at the top, you should be able to find the velocity at any point as a function of angle.
- #7
asi123
- 254
- 0
Doc Al said:The velocity is definitely not constant. Reread my hint in post #4.
Given the velocity at the top, you should be able to find the velocity at any point as a function of angle.
I'm totally stuck, should I use trigo, or there is some equation for the velocity that I don't know about...Oh, maybe u mean to use energy calculation?
- #8
Doc Al
Mentor
- 45,584
- 2,459
Getting warmer!asi123 said:
- #9
asi123
- 254
- 0
Doc Al said:Getting warmer!
The potential energy turns into kinetic energy, no?
- #10
Doc Al
Mentor
- 45,584
- 2,459
You got it. Keep going.
- #11
asi123
- 254
- 0
- #12
Doc Al
Mentor
- 45,584
- 2,459
Looks good to me.asi123 said:
Excellent question! Ask yourself: Does the tension force do any work on the ball?
- #13
asi123
- 254
- 0
Doc Al said:
Oh, right, it's vertical to his movement.
10x a lot.
- #14
Doc Al
Mentor
- 45,584
- 2,459
The ball's movement is always perpendicular to the tension, thus the tension does no work on the ball.
Similar threads
- · Replies 12 ·
- · Replies 10 ·
- · Replies 1 ·
- · Replies 5 ·
- · Replies 3 ·
- · Replies 1 ·
- · Replies 3 ·