Rotational Dynamics & Non-Inertial Systems

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Centre of Mass (CoM)

The Centre of Mass of a system of particles is the point where the entire mass of the body can be assumed to be concentrated for analyzing motion.

For a system of particles:$$\vec{R} = \frac{1}{M}\sum_i m_i \vec{r_i}$$R=M1​i∑​mi​ri​​

where $M = \sum m_i$M=∑mi​ is total mass.

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Motion of Centre of Mass

• The motion of CoM is independent of the internal forces.
• External forces determine the motion:

$$M\vec{a}_{CM} = \sum \vec{F}_{ext}$$MaCM​=∑Fext​

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Centre of Mass and Laboratory Frames

• Laboratory Frame: The usual reference frame where we observe motion.
• CoM Frame: Frame moving with the centre of mass.

Relationship between momenta:$$\vec{P}_{lab} = M\vec{V}_{CM}$$Plab​=MVCM​

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Angular Momentum

For a particle:$$\vec{L} = \vec{r} \times \vec{p}$$L=r×p​

For a system of particles:$$\vec{L}_{total} = \sum \vec{L_i}$$Ltotal​=∑Li​​

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Principle of Conservation of Angular Momentum

If net external torque is zero:$$\frac{d\vec{L}}{dt} = 0 \Rightarrow \vec{L} = \text{constant}$$dtdL​=0⇒L=constant

Example: Figure skater spins faster by folding arms inward.

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Rotation About a Fixed Axis

Moment of Inertia (MI)

MI is the rotational equivalent of mass:$$I = \sum m_i r_i^2$$I=∑mi​ri2​

Dependence:

• Shape and size of the body
• Axis of rotation
• Distribution of mass

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Theorems Used for MI Calculation

Theorem	Statement
Parallel Axis Theorem	$I = I_{CM} + Md^2$I=ICM​+Md2
Perpendicular Axis Theorem	$I_z = I_x + I_y$Iz​=Ix​+Iy​ (for planar bodies)

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Routh Rule

A useful rule to find MI of symmetrical lamina:

Shape	MI
Rectangular plate	$\frac{1}{3}ML^2$31​ML2 (about an edge)

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Moment of Inertia for Common Bodies

Body	Axis	MI
Solid cylinder	about its axis	$\frac{1}{2}MR^2$21​MR2
Solid sphere	about diameter	$\frac{2}{5}MR^2$52​MR2
Hollow sphere	about diameter	$\frac{2}{3}MR^2$32​MR2

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Kinetic Energy of Rotation

$$K = \frac{1}{2} I \omega^2$$K=21​Iω2

If translation + rotation:$$K = \frac{1}{2} I \omega^2 + \frac{1}{2}Mv^2$$K=21​Iω2+21​Mv2

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Euler’s Equations for Rigid Body

$$\begin{aligned} I_1 \dot{\omega_1} – (I_2 – I_3)\omega_2\omega_3 &= \tau_1 \\ I_2 \dot{\omega_2} – (I_3 – I_1)\omega_3\omega_1 &= \tau_2 \\ I_3 \dot{\omega_3} – (I_1 – I_2)\omega_1\omega_2 &= \tau_3 \end{aligned}$$I1​ω1​˙​−(I2​−I3​)ω2​ω3​I2​ω2​˙​−(I3​−I1​)ω3​ω1​I3​ω3​˙​−(I1​−I2​)ω1​ω2​​=τ1​=τ2​=τ3​​

Used for bodies rotating without symmetry (e.g., satellites).

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Moment of Inertia of a Flywheel

A flywheel is a heavy rotating wheel used to store rotational energy.

Energy stored:$$E = \frac{1}{2} I \omega^2$$E=21​Iω2

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2️⃣ Non-Inertial Systems

A non-inertial frame is one that is accelerating relative to an inertial frame. In such frames, fictitious forces appear.

Examples: Accelerating car, rotating Earth.

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Uniformly Rotating Frame

Consider a body in a frame rotating with angular velocity $\vec{\omega}$ω. Two significant forces appear:

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Centrifugal Force

A force acting outward from the center in a rotating frame:$$\vec{F}_{centrifugal} = m(\vec{\omega} \times (\vec{\omega} \times \vec{r}))$$Fcentrifugal​=m(ω×(ω×r))

Example: Mud thrown off a rotating tyre.

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Coriolis Force

Acts on a body when it moves within a rotating frame:$$\vec{F}_{coriolis} = -2m(\vec{\omega} \times \vec{v})$$Fcoriolis​=−2m(ω×v)

Example:

• Deflection of winds on Earth (cyclones rotate anti-clockwise in Northern Hemisphere).
• Path of a falling object shifts sideways.

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Laws of Physics in Rotating Systems

To apply Newton’s laws in rotating frames, add fictitious forces:$$\vec{F}_{effective} = \vec{F}_{real} + \vec{F}_{centrifugal} + \vec{F}_{coriolis}$$Feffective​=Freal​+Fcentrifugal​+Fcoriolis​

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Conclusion

Rotational dynamics connects linear and rotational motion using moment of inertia, torque, and angular momentum. Non-inertial reference frames help explain many natural phenomena like atmospheric motion and satellite movement.