How to find the center of rotation? (2D)

There is a easy semi-geometrical way of finding the center of rotation due to a force.

1. Find the moment arm $c$ of the force through A. $$ c = r \cos \theta$$
2. Find the radius of gyration about the center of mass C $$\rho = \sqrt{ \frac{I_C}{m} }$$
3. Measure the distance $\ell$ away from the center of mass and mark point R $$\ell = \frac{\rho^2}{c}$$

Point R is the instance center of rotation.

See also https://physics.stackexchange.com/a/212939/392 for more details.

Example

A plank of length $a$ and and width $b$ is struck on one end with a force $F$. Place a coordinate system at the center of mass and measure the force moment arm as $c=\frac{a}{2}$. The mass moment of inertia of a plank is $$I_C = \frac{m}{12} \left( a^2+b^2\right)$$ and hence the radius of gyration about the center is $$\rho = \sqrt{ \frac{a^2+b^2}{12} }$$

With the method above the center of rotation is at a distance $$\boxed{\ell = \frac{\rho^2}{c} = \frac{ \frac{a^2+b^2}{12} } {\frac{a}{2} } = \frac{a^2+b^2}{6 a} }$$

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Lets find the same answer using the equations of motion.

• The sum of the forces of the body are $$\sum \boldsymbol{F} = (0,F,0)$$
• The sum of moment about the center are $$\sum \boldsymbol{M} = (0,0,\frac{a}{2} F)$$
• The linear acceleration of the center is $$\boldsymbol{a} = \frac{\sum \boldsymbol{F}}{m} = (0,\frac{F}{m},0)$$
• The angular acceleration of the body is $$\boldsymbol{\alpha} = \frac{\sum \boldsymbol{M}}{I_C} = (0,0,\frac{6 F a}{m (a^2+b^2)})$$
• Find the center of rotation R such that $\boldsymbol{a}_y=\ell \boldsymbol{\alpha}_z$ $$ \boxed{\ell = \frac{ \frac{F}{m} }{ \frac{6 F a}{m (a^2+b^2)} } = \frac{a^2+b^2}{6 a}}$$

Same answer!! No need to do equations of motion to get the center of rotation. The point is purely a result of the inertial properties and geometry.