Torque and paralle-axis theorem - question

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Theorem Torque
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving torque and the parallel-axis theorem, specifically in the context of a seesaw model with point masses attached at either end. The original poster seeks to determine the angular acceleration of the rod after it is released from a horizontal position.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster considers calculating torque around the pivot and questions whether using the parallel-axis theorem to find angular acceleration from one end is valid. Some participants discuss the handling of torque and express opinions on the ease of different methods.

Discussion Status

The conversation is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the calculation of torque, but there is no explicit consensus on the best method to use.

Contextual Notes

The original poster is working under the assumption that the rod is held at rest horizontally before being released, and they are considering the implications of calculating torque from different points. There is an emphasis on understanding the material rather than simply finding a solution.

Niles
Messages
1,834
Reaction score
0

Homework Statement


The figure shows a simple model of a seesaw and/or valves used in car engines. These consist of a plank/rod of mass m_r and length 2L allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass m_2 is attached to the left end of the rod, and a small sphere of mass is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to g.

Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.)

What is the angular acceleration of the rod immediately after it is released?
________

Ok, I can find the torque around the pivot and then the only forces relevant here will be the weights of the two masses in the ends. From this I can find alpha.

But on the way home in the bus I was wondering if I can find alpha when calculating the torque around on of the ends and then use the parallel-axis theorem? Is this operation allowed and will it give the same result?

Thank you in advance.
 

Attachments

  • MRB_mi_6.jpg
    MRB_mi_6.jpg
    2.8 KB · Views: 1,077
Physics news on Phys.org
Yes, in theory this could work. Just be careful handling the torque created by the pivot now. In my opinion, it's easier the way you did it.
 
dwintz02 said:
Yes, in theory this could work. Just be careful handling the torque created by the pivot now. In my opinion, it's easier the way you did it.

Thanks. I think you are right. If we calculate the torque around the left end, is this correct (I am only doing this to prove to myself that I have understood the material):

(Positive counterclockwise):

T_z = -m_l*g*L-m_2*g*2L = I_{total}*alpha. From this I can find alpha?
 
Can you guys confirm what I wrote in #2, please?
 

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K