06Class 11 Physics

System of Particles and Rotational Motion formulas

Centre of mass, torque and rotational formulas. Most of rotation mirrors straight-line motion, so learn the two side by side.

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Centre of mass

Discrete particles: $\displaystyle \vec r_{\text{cm}} = \frac{\sum m_i\vec r_i}{\sum m_i}$
Continuous body: $\displaystyle \vec r_{\text{cm}} = \frac{1}{M}\int \vec r\,dm$

Torque: $\displaystyle \vec\tau = \vec r\times\vec F, \quad \tau = rF\sin\theta$
Angular momentum: $\displaystyle \vec L = \vec r\times\vec p, \quad L = I\omega$
Rotational second law: $\displaystyle \tau = I\alpha = \frac{dL}{dt}$
Rotational kinetic energy: $\displaystyle K = \tfrac12 I\omega^2$
Rotational power: $\displaystyle P = \tau\omega$
Equations of rotation: $\displaystyle \omega = \omega_0 + \alpha t, \quad \theta = \omega_0 t + \tfrac12\alpha t^2, \quad \omega^2 = \omega_0^2 + 2\alpha\theta$

Definition: $\displaystyle I = \sum m_i r_i^2$
Parallel axis theorem: $\displaystyle I = I_{\text{cm}} + Md^2$
Perpendicular axis theorem: $\displaystyle I_z = I_x + I_y$
For flat bodies only.
Ring, about its central axis: $\displaystyle I = MR^2$
Disc, about its central axis: $\displaystyle I = \tfrac12 MR^2$
Solid sphere, about a diameter: $\displaystyle I = \tfrac25 MR^2$
Hollow sphere, about a diameter: $\displaystyle I = \tfrac23 MR^2$
Rod, about its centre: $\displaystyle I = \tfrac{1}{12}ML^2$
Rod, about one end: $\displaystyle I = \tfrac13 ML^2$

Condition: $\displaystyle v_{\text{cm}} = R\omega$
Total kinetic energy: $\displaystyle K = \tfrac12 mv^2\left(1 + \frac{k^2}{R^2}\right)$
k is the radius of gyration, from I = mk².
Acceleration down an incline: $\displaystyle a = \frac{g\sin\theta}{1 + k^2/R^2}$

Using the perpendicular axis theorem on a 3D body. It only applies to flat bodies, such as discs and thin plates.
Using the parallel axis theorem from an axis that isn't through the centre of mass. One of the two axes must pass through it.
Forgetting the rotational part of kinetic energy on rolling bodies.

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