Question

Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol I, 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?

Take the counterclockwise direction to be positive. Express ? in terms of some or all of the variables mr, m1, m2, x, I, and g.

Answer 1

Isn't there supposed to be more to this problem?

The torque about the pivot point due to a force is where

(F)$\tiny T=rpivotFsin\Theta=rpivotFperpindicular=(moment arm)F$

rpivot...is the vector from the pivot point to the point where the force is applied. The other symbols have their usual meanings. If you are using any of the latter two expressions, you must remember that if the force tends to cause a clockwise rotation, you need to include a negative sign in your expression since the torque due to such a force is taken to be negative.

So

T1= m1gx

T2= -m2gx

Tpivot= Ia

SO,

a=   (m₁-m₂)gx        substitue I and you get a=         (m₁-m₂)gx            cancel the X's so a=   (m₁-m₂)g   

               I                                                      (m_r/3 + m₁ + m₂) x²                           (m_r/3 + m₁ + m₂)x

Answer 2

[(m_1-m_2)g]/[(1/3m_r+m_1+m_2)x]