Masses and a Light Rod

1. Jan 3, 2009

tachu101

1. The problem statement, all variables and given/known data
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

2. Relevant equations

torque= Ia

3. The attempt at a solution
I don't know where to go with this one because gravity is ignored.

2. Jan 3, 2009

just__curious

What else is torque equal to?

3. Jan 3, 2009

tachu101

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

4. Jan 3, 2009

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?

5. Jan 4, 2009

tachu101

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

6. Jan 4, 2009

Staff: Mentor

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.

7. Jan 4, 2009

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 thats the answer???

8. Jan 4, 2009

Staff: Mentor

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