Question on AP Physics C FRQ 2013. #3

In summary, the conversation discusses a physics problem involving a disk in equilibrium and the use of torque and force equations to solve it. In part A, the left rope's tension is equal to Fa because the disk is at rest and the tension balances out the torque from Fa on the right side. In part B, the disk is no longer in equilibrium and a separate value for tension (T) is needed to account for the rotational equilibrium. This is because the torque from Fa on the right side is countered by the tension on the left side. The conversation concludes with a clarification on the reasoning behind this approach.
  • #1
RoboNerd
410
11

Homework Statement


http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_physics_c_mechanics.pdf

The problem is above. It is number 3.

The solutions are below:
http://media.collegeboard.com/digit...entral/ap13_physics_cm_scoring_guidelines.pdf

Homework Equations


Torque= I * angular acceleration, F=ma

The Attempt at a Solution

My questions are the following:

Why do they in part a of the solution do the expression 2Fa-mg=0?
I understand that they have Fa acting on the object and mg, but why do they have the left rope's tension equal Fa?And then in part B, why do they have the expression Fa + T - Mg = Ma, meaning why do they have Fa != T. And then they substitute the T from the torque equation. Why is doing the problem in this manner necessary, with a separate value for T, which does not equal Fa?
 
Physics news on Phys.org
  • #2
RoboNerd said:
Why do they in part a of the solution do the expression 2Fa-mg=0?
I understand that they have Fa acting on the object and mg, but why do they have the left rope's tension equal Fa?
You are told that the disk is at rest (in equilibrium).

In part B the disk is no longer in equilibrium.
 
  • #3
Yeah, but when it is at equilibrium, then couldn't I just write it as Fa-mg = 0.

Maybe it is because the Fa is causing a torque on the right side and we need an equal force to counteract it from the left side to have sum of torques cancel?
 
  • #4
RoboNerd said:
Yeah, but when it is at equilibrium, then couldn't I just write it as Fa-mg = 0.
That assumes the tension on the left segment is zero. And it wouldn't satisfy rotational equilibrium.

RoboNerd said:
Maybe it is because the Fa is causing a torque on the right side and we need an equal force to counteract it from the left side to have sum of torques cancel?
Right!
 
  • #5
Yeah that makes sense. Thanks a lot!
 

Related to Question on AP Physics C FRQ 2013. #3

1. What was the context of the 2013 AP Physics C FRQ #3?

The context of the 2013 AP Physics C FRQ #3 was a free response question that focused on rotational dynamics and conservation of energy in a system of a rotating rod and a hanging mass.

2. What was the main concept being tested in this FRQ?

The main concept being tested in this FRQ was the application of rotational kinematics and dynamics, as well as the principle of conservation of energy, to solve a complex problem involving a rotating system.

3. How many points was this question worth?

This question was worth a total of 10 points out of a possible 45 points on the AP Physics C exam.

4. What was the most challenging aspect of this FRQ for students?

The most challenging aspect of this FRQ for students was understanding and applying the concept of rotational motion and energy conservation in a multi-step problem. Additionally, some students may have struggled with identifying the correct equations and understanding how to use them in the context of the problem.

5. What are some tips for approaching this type of FRQ?

Some tips for approaching this type of FRQ include carefully reading and understanding the problem, drawing clear and accurate diagrams, identifying the relevant equations, and breaking the problem down into smaller, more manageable parts. It is also important to check your units and make sure they are consistent throughout the solution.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
4K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
11K
Back
Top