# Does a hinged body rotates over its centre of mass if seen from the frame of centre of mass?

I know that if an unhinged body have some constant net force and some net torque it will rotate about its centre of mass and translate at the same time.

Does a rotating hinged body also rotates about its centre of mass if seen from the frame of centre of mass?

What will be the axis of rotation if see the body from the frame of a point on the body other than centre of mass?

A hinged body is constrained to rotate about the hinge and therefore needs to generate reaction forces in order to accelerate the center of mass.

The only time a body purely rotates about the center of mass is when a pure torque is applied, or the net forces acting add up to zero.

Consider a hinged body whose center of mass is located at $$\boldsymbol{c}$$ relative to the hinge. The rotation at the same instant is $$\boldsymbol{\omega}$$.

• Kinematics the velocity and acceleration of the center of mass is \begin{aligned} \boldsymbol{v} & = \boldsymbol{\omega} \times \boldsymbol{c} \\ \boldsymbol{a} & = \boldsymbol{\dot{\omega}} \times \boldsymbol{c} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{c}) \end{aligned}
• Momentum translational and rotational momentum about the pivot are \begin{aligned} \boldsymbol{p} & = m \boldsymbol{v} = m (- \boldsymbol{c} \times \boldsymbol{\omega} ) \\ \boldsymbol{L}_H &= \mathbf{I}_C\, \boldsymbol{\omega} + \boldsymbol{c} \times \boldsymbol{p} = \mathbf{I}_C\, \boldsymbol{\omega} - \boldsymbol{c} \times ( \boldsymbol{c} \times \boldsymbol{\omega} ) = \mathbf{I}_H \boldsymbol{\omega} \end{aligned} This defines the mass moment of inertia tensor about the hinge $$\mathbf{I}_H$$.
• Dynamics forces and torques about the hinged are \begin{aligned} \boldsymbol{F} & = m \boldsymbol{a} = m (-\boldsymbol{c} \times \boldsymbol{\dot \omega} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{c})) \\ \boldsymbol{\tau}_H & = \mathbf{I}_H\,\boldsymbol{\dot \omega} + \boldsymbol{\omega} \times \mathbf{I}_H \boldsymbol{\omega} \end{aligned}

As you can see the hinge forces $$\boldsymbol{F}$$ have two components, the centrifugal $$-\boldsymbol{\omega} \times (\boldsymbol{\omega} \times m \boldsymbol{c})$$ and the Eulerian $$\boldsymbol{\dot \omega} \times m \boldsymbol{c}$$.

Consider the above case where the hinge is also accelerating with $$\boldsymbol{a}_H$$ then the dynamics have the following form

\begin{aligned} \boldsymbol{F} &= m \left( \boldsymbol{a}_H - \boldsymbol{c} \times \boldsymbol{\dot \omega} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{c} ) \right) \\ \boldsymbol{\tau}_H & = \boldsymbol{c} \times m \boldsymbol{a}_H + \mathbf{I}_H \boldsymbol{\dot \omega} + \boldsymbol{\omega} \times \mathbf{I}_H \boldsymbol{\omega} \end{aligned}

This first equation shows a direct effect between hinge acceleration and reaction forces, and in the second equation there is an equivalent torque produced $$\boldsymbol{c} \times m \boldsymbol{a}_H$$ which needs to be compensated by the hinge torque.

If you will see it from the centre of the mass frame then it will seem as if it is rotating around you and the frequency will be the same as frequency doesn't depend upon the frame of reference.In fact, this would be true for any point on the disc. I am taking into account that alpha=0