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System of Particles and Rotational Motion - Torque and Angular Momentum

Grade 11CBSEPhysics

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

🔑Concepts

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Torque (τ⃗\vec{\tau}), or the moment of force, is the rotational equivalent of force. It is defined as the cross product of the position vector r⃗\vec{r} and the force vector F⃗\vec{F}, given by τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}.

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The magnitude of torque is τ=rFsin⁡θ\tau = rF \sin \theta, where θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}. It is measured in Newton-meters (N⋅mN \cdot m).

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Angular Momentum (L⃗\vec{L}) is the rotational analog of linear momentum. For a particle, it is the moment of its linear momentum: L⃗=r⃗×p⃗\vec{L} = \vec{r} \times \vec{p}.

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The Principle of Moments for a rigid body states that for rotational equilibrium, the sum of the anti-clockwise torques must equal the sum of the clockwise torques about a pivot.

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The Newton's Second Law for rotation states that the torque acting on a body is equal to the rate of change of its angular momentum: τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}.

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For a rigid body rotating about a fixed axis, the angular momentum is related to the moment of inertia (II) and angular velocity (ω\omega) by the relation L=IωL = I\omega.

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Conservation of Angular Momentum: If the net external torque acting on a system is zero (τ⃗ext=0\vec{\tau}_{ext} = 0), the total angular momentum of the system remains constant, meaning I1ω1=I2ω2I_1\omega_1 = I_2\omega_2.

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A body is in complete mechanical equilibrium if it satisfies both translational equilibrium (∑F⃗=0\sum \vec{F} = 0) and rotational equilibrium (∑τ⃗=0\sum \vec{\tau} = 0).

📐Formulae

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

τ=rFsin⁡θ\tau = rF \sin \theta

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

L=mvrsin⁡θL = mvr \sin \theta

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

L=IωL = I\omega

τ=Iα\tau = I\alpha

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

💡Examples

Problem 1:

A force F⃗=2i^−3j^+4k^\vec{F} = 2\hat{i} - 3\hat{j} + 4\hat{k} N acts at a point whose position vector is r⃗=3i^+2j^+3k^\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k} m. Calculate the torque about the origin.

Solution:

Torque is given by τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. Using the determinant method: τ⃗=∣i^j^k^3232−34∣\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} τ⃗=i^(2×4−(−3)×3)−j^(3×4−2×3)+k^(3×(−3)−2×2)\vec{\tau} = \hat{i}(2 \times 4 - (-3) \times 3) - \hat{j}(3 \times 4 - 2 \times 3) + \hat{k}(3 \times (-3) - 2 \times 2) τ⃗=i^(8+9)−j^(12−6)+k^(−9−4)\vec{\tau} = \hat{i}(8 + 9) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) τ⃗=17i^−6j^−13k^ N m\vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \text{ N m}

Explanation:

The torque is found by taking the vector cross product of the position vector and the force vector.

Problem 2:

A dancer spins with an initial angular velocity ω1=4 rad/s\omega_1 = 4 \text{ rad/s} with her arms extended, having a moment of inertia I1=2.5 kg m2I_1 = 2.5 \text{ kg m}^2. When she pulls her arms in, her moment of inertia decreases to I2=1.0 kg m2I_2 = 1.0 \text{ kg m}^2. Calculate her new angular velocity.

Solution:

By the Law of Conservation of Angular Momentum (I1ω1=I2ω2I_1\omega_1 = I_2\omega_2): 2.5×4=1.0×ω22.5 \times 4 = 1.0 \times \omega_2 10=1.0×ω210 = 1.0 \times \omega_2 ω2=10 rad/s\omega_2 = 10 \text{ rad/s}

Explanation:

Since no external torque is acting on the dancer, the angular momentum is conserved. Reducing the moment of inertia results in an increase in angular velocity.

Torque and Angular Momentum - Revision Notes & Key Formulas | CBSE Class 11 Physics