Hot answers tagged rotational-dynamics
To balance the system with respect to the lowest point of the rotating thin rod, ensure the balance of moments with respect to the $y$ axis is zero. That simply yields: $m_1r_1 + m_3r_3=m_2r_2$. To ensure no net centripetal force acts on the rod: $m_1 r_1 \omega^2y_1+m_3 r_3 \omega^2y_3=m_2 r_2 \omega^2y_2$. $\omega^2$ drops out, so we get: $m_1 ...
Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ...
The internal forces don't cancel each other they are just equal and in opposite directions. This results in an internal momentum change of zero. The same is true for angular momentum. As one object in the isolated system begins to rotate another object must rotate with exactly opposite angular momentum so there is a net change of zero.
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