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System of Particles and Rotational Motion - Angular Momentum in Case of Rotation about a Fixed Axis

Grade 11CBSEPhysics

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

🔑Concepts

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The angular momentum of a particle about an origin is defined as the cross product of its position vector r⃗\vec{r} and its linear momentum p⃗\vec{p}: l⃗=r⃗×p⃗\vec{l} = \vec{r} \times \vec{p}.

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For a rigid body rotating about a fixed axis (e.g., the zz-axis), the component of the total angular momentum along the axis of rotation is given by Lz=IωL_z = I\omega, where II is the moment of inertia about that axis and ω\omega is the angular velocity.

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While the total angular momentum vector L⃗\vec{L} may not always be parallel to the axis of rotation, for symmetric bodies rotating about their axis of symmetry, L⃗\vec{L} and ω⃗\vec{\omega} are in the same direction.

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The rate of change of angular momentum of a system is equal to the net external torque acting on it: τ⃗ext=dL⃗dt\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}.

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The Principle of Conservation of Angular Momentum states that if the net external torque acting on a system is zero (τ⃗ext=0\vec{\tau}_{ext} = 0), the total angular momentum remains constant: L⃗=constant\vec{L} = \text{constant} or I1ω1=I2ω2I_1\omega_1 = I_2\omega_2.

📐Formulae

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

L=IωL = I\omega

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

I1ω1=I2ω2 (if τext=0)I_1\omega_1 = I_2\omega_2 \text{ (if } \tau_{ext} = 0\text{)}

L=∑miri2ωL = \sum m_i r_i^2 \omega

💡Examples

Problem 1:

A thin circular ring of mass MM and radius RR is rotating about its axis with a constant angular velocity ω\omega. Two objects, each of mass mm, are attached gently to the opposite ends of a diameter of the ring. Calculate the new angular velocity ω′\omega' of the ring.

Solution:

Initial moment of inertia I1=MR2I_1 = MR^2. Initial angular momentum L1=I1ω=MR2ωL_1 = I_1\omega = MR^2\omega. When two masses mm are added at distance RR from the axis, the new moment of inertia is I2=MR2+2mR2=(M+2m)R2I_2 = MR^2 + 2mR^2 = (M + 2m)R^2. According to the conservation of angular momentum, I1ω=I2ω′I_1\omega = I_2\omega'. Therefore, MR2ω=(M+2m)R2ω′MR^2\omega = (M + 2m)R^2\omega', which gives ω′=MωM+2m\omega' = \frac{M\omega}{M + 2m}.

Explanation:

Since no external torque is applied (the masses are placed 'gently'), the total angular momentum of the system remains conserved. The increase in the moment of inertia leads to a corresponding decrease in angular velocity.

Problem 2:

A ballet dancer spins with an initial angular velocity of 2 rad/s2\text{ rad/s} and a moment of inertia of 8 kg m28\text{ kg m}^2. When she pulls her arms inward, her moment of inertia decreases to 5 kg m25\text{ kg m}^2. Calculate her final angular velocity.

Solution:

Given I1=8 kg m2I_1 = 8\text{ kg m}^2, ω1=2 rad/s\omega_1 = 2\text{ rad/s}, and I2=5 kg m2I_2 = 5\text{ kg m}^2. Using the principle of conservation of angular momentum: I1ω1=I2ω2I_1\omega_1 = I_2\omega_2 8×2=5×ω28 \times 2 = 5 \times \omega_2 ω2=165=3.2 rad/s\omega_2 = \frac{16}{5} = 3.2\text{ rad/s}

Explanation:

By pulling her arms in, the dancer redistributes her mass closer to the axis of rotation, decreasing her moment of inertia. To conserve angular momentum, her angular speed must increase.