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Motion of System of Particles and Rigid Body - Angular Momentum

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Angular momentum (LL) of a particle is defined as the moment of its linear momentum. It is a vector quantity given by the cross product of the position vector and linear momentum vector: L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p}.

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For a rigid body rotating about a fixed axis, the angular momentum is the product of its moment of inertia (II) and its angular velocity (ω\omega), expressed as L=IωL = I\omega.

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The SI unit of angular momentum is kg m2s−1\text{kg m}^2 \text{s}^{-1} or J s\text{J s}, and its dimensional formula is [ML2T−1][ML^2T^{-1}].

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The Principle of Conservation of Angular Momentum states that if the total external torque acting on a system is zero (τ⃗ext=0\vec{\tau}_{ext} = 0), the total angular momentum of the system remains constant in magnitude and direction.

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The relation between torque and angular momentum is analogous to Newton's Second Law for linear motion: τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}.

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In the absence of external torque, if the moment of inertia (II) of a body changes, its angular velocity (ω\omega) changes such that the product IωI\omega remains constant (I1ω1=I2ω2I_1\omega_1 = I_2\omega_2).

📐Formulae

L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p}

L=mvrsin⁡θL = mvr \sin \theta

L=IωL = I\omega

τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}

I1ω1=I2ω2 (when τ=0)I_1 \omega_1 = I_2 \omega_2 \text{ (when } \tau = 0)

K=L22I (Rotational Kinetic Energy in terms of L)K = \frac{L^2}{2I} \text{ (Rotational Kinetic Energy in terms of L)}

💡Examples

Problem 1:

A thin uniform circular disc of mass MM and radius RR is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω\omega. If another disc of same dimensions but mass M/4M/4 is placed gently on the first disc coaxially, find the new angular velocity of the system.

Solution:

Initial moment of inertia I1=12MR2I_1 = \frac{1}{2}MR^2. Initial angular velocity is ω1=ω\omega_1 = \omega. Final moment of inertia I2=12MR2+12(M4)R2=12MR2(1+14)=58MR2I_2 = \frac{1}{2}MR^2 + \frac{1}{2}(\frac{M}{4})R^2 = \frac{1}{2}MR^2(1 + \frac{1}{4}) = \frac{5}{8}MR^2. By conservation of angular momentum: I1ω1=I2ω2  ⟹  (12MR2)ω=(58MR2)ω2  ⟹  ω2=45ωI_1\omega_1 = I_2\omega_2 \implies (\frac{1}{2}MR^2)\omega = (\frac{5}{8}MR^2)\omega_2 \implies \omega_2 = \frac{4}{5}\omega.

Explanation:

Since no external torque is applied to the system (the second disc is placed gently), the total angular momentum is conserved. The increase in the moment of inertia results in a proportional decrease in angular velocity.

Problem 2:

A particle of mass 0.5 kg0.5 \text{ kg} is moving with a velocity v⃗=4i^ m/s\vec{v} = 4\hat{i} \text{ m/s} along the line y=2 my = 2 \text{ m}. Calculate its angular momentum about the origin.

Solution:

Position vector r⃗=xi^+2j^\vec{r} = x\hat{i} + 2\hat{j}. Linear momentum p⃗=mv⃗=0.5×4i^=2i^ kg m/s\vec{p} = m\vec{v} = 0.5 \times 4\hat{i} = 2\hat{i} \text{ kg m/s}. Angular momentum L⃗=r⃗×p⃗=(xi^+2j^)×(2i^)\vec{L} = \vec{r} \times \vec{p} = (x\hat{i} + 2\hat{j}) \times (2\hat{i}). Using cross product rules: i^×i^=0\hat{i} \times \hat{i} = 0 and j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}. Therefore, L⃗=2(2)(−k^)=−4k^ kg m2/s\vec{L} = 2(2)(-\hat{k}) = -4\hat{k} \text{ kg m}^2\text{/s}. Magnitude L=4 kg m2/sL = 4 \text{ kg m}^2\text{/s}.

Explanation:

Angular momentum is calculated using the cross product of the position vector and the momentum vector. Since the velocity is strictly in the xx-direction, only the yy-component of the position contributes to the torque/momentum about the origin.

Angular Momentum - Revision Notes & Key Formulas | ICSE Class 11 Physics