Masses and a Light Rod: Solving for Linear Acceleration Without Gravity

  • Thread starter Thread starter tachu101
  • Start date Start date
  • Tags Tags
    Light Rod
Click For Summary

Homework Help Overview

The problem involves two equal masses connected by a light rod, pivoted at its center, with a force applied to the rod while ignoring gravity. The objective is to determine the linear acceleration of the masses.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between torque and angular acceleration, questioning how to apply the torque equation in the absence of gravity. There are attempts to clarify the definitions of torque and rotational inertia, as well as how to relate angular acceleration to linear acceleration.

Discussion Status

Some participants have offered guidance on using torque to find angular acceleration and relating it to linear acceleration. Multiple interpretations of the problem are being explored, with some expressing confusion about the next steps.

Contextual Notes

There is a noted lack of clarity regarding the rotational inertia of the system and how to proceed without considering gravitational effects.

tachu101
Messages
74
Reaction score
0

Homework Statement


Two equal masses (m) are connected by a light rod of length L that is pivoted about its center. A downward Force F is applied to the rod at a distance L/4 from the pivot. If you ignore gravity, the linear acceleration of the mass is...

O--------o--------O

Homework Equations



torque= Ia

The Attempt at a Solution


I don't know where to go with this one because gravity is ignored.
 
Physics news on Phys.org
What else is torque equal to?
 
torque = r*F... torque= (L/4)(F) right?
 
ya but torque doesn't equal to linear acceleration but angular acceleration. That might be what you mean. And can you find I for the object?
 
I'm really lost at what to do in this question.
 
Use the torque to find the angular acceleration of the rod. (What's the rotational inertia of this system?) Then relate angular acceleration to linear acceleration of the masses.
 
a=torque/I == (F*(L/4))/ (2(m(L/2)^2)) = a= F/2mL ? a(linear)= a(rot)*radius == F/2ml= a*(L/2) solve for a and that's the answer?
 
Yes, that's all there is to it. Use different symbols for angular acceleration (α) versus linear acceleration (a) to avoid confusion.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 62 ·
3
Replies
62
Views
14K
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K