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

[tachu101]

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.

 

[just__curious]

What else is torque equal to?

 

[tachu101]

torque = r*F... torque= (L/4)(F) right?

 

[just__curious]

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?

 

[tachu101]

I'm really lost at what to do in this question.

 

[Doc Al]

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.

 

[tachu101]

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?

 

[Doc Al]

Yes, that's all there is to it. Use different symbols for angular acceleration (α) versus linear acceleration (a) to avoid confusion.