Torque & Angular Momentum – Friendly Notes

1. Why Torque Matters

Push a door near the hinge and nothing happens; push the outer edge and it swings open. That everyday difference introduces torque – the twist that forces create when they act away from the pivot. :contentReference[oaicite:0]{index=0}

2. Defining Torque (Moment of Force)

• Vector form: \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \) – “pivot-to-point” vector \( \mathbf{r} \) crossed with the applied force \( \mathbf{F} \). :contentReference[oaicite:1]{index=1}
• Size (magnitude): \( \tau = r F \sin\theta = r_{\perp} F = r F_{\perp} \). Here \( r_{\perp} \) is the perpendicular distance (lever arm) and \( F_{\perp} \) is the force part that’s perpendicular to \( \mathbf{r} \). :contentReference[oaicite:2]{index=2}
• Torque drops to zero when the force line passes through the pivot or when the force itself is zero. :contentReference[oaicite:3]{index=3}
• Direction follows the right-hand rule (curl fingers from \( \mathbf{r} \) to \( \mathbf{F} \); thumb shows \( \boldsymbol{\tau} \)). :contentReference[oaicite:4]{index=4}
• SI unit: newton-metre (N m). :contentReference[oaicite:5]{index=5}

3. Factors that Boost Torque

• Bigger force \( F \)
• Larger lever arm \( r_{\perp} \)
• Force applied at a right angle (\( \theta = 90^\circ \) makes \( \sin\theta = 1 \))

4. Angular Momentum of a Particle

• Vector form: \( \mathbf{l} = \mathbf{r} \times \mathbf{p} \) with \( \mathbf{p}=m\mathbf{v} \). :contentReference[oaicite:6]{index=6}
• Size: \( l = r p \sin\theta = r_{\perp} p = r p_{\perp} \). :contentReference[oaicite:7]{index=7}
• Angular momentum vanishes if the mass sits at the pivot, has no linear momentum, or its motion line cuts through the pivot. :contentReference[oaicite:8]{index=8}

5. Torque–Angular Momentum Connection

A changing spin needs a twist: \( \displaystyle \frac{d\mathbf{l}}{dt} = \boldsymbol{\tau} \). This is the rotational twin of Newton’s \( \mathbf{F}=d\mathbf{p}/dt \). :contentReference[oaicite:9]{index=9}

6. Systems of Particles (including Rigid Bodies)

• Total angular momentum: \( \mathbf{L}= \sum_i \mathbf{r}_i \times \mathbf{p}_i \). :contentReference[oaicite:10]{index=10}
• Total external torque changes that total spin: \( \displaystyle \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}} \). :contentReference[oaicite:11]{index=11}
• Internal forces (action–reaction pairs) create no net external torque, so they don’t affect \( \mathbf{L} \). :contentReference[oaicite:12]{index=12}

7. Conservation of Angular Momentum

When \( \boldsymbol{\tau}_{\text{ext}} = 0 \), the total angular momentum stays fixed: \( \mathbf{L} = \text{constant} \). Each component \( L_x,\;L_y,\;L_z \) keeps its individual value. :contentReference[oaicite:13]{index=13}

8. Mechanical Equilibrium of a Rigid Body

A body enjoys complete balance only when both conditions hold: :contentReference[oaicite:14]{index=14}

• No net force: \( \sum \mathbf{F}_i = \mathbf{0} \)  ➔ no linear acceleration.
• No net torque: \( \sum \boldsymbol{\tau}_i = \mathbf{0} \)  ➔ no angular acceleration.

Those two vector equations unfold into six scalar ones (three for forces, three for torques) that must all be satisfied. :contentReference[oaicite:15]{index=15}

High-Yield Ideas for NEET

• The cross-product definitions \( \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \) and \( \mathbf{l} = \mathbf{r}\times\mathbf{p} \) – memorize the forms and right-hand directions.
• Lever-arm approach: \( \tau = r_{\perp}F \) is the fastest way to solve door-style and balancing-rod questions.
• Conservation of angular momentum: expect problems with zero external torque, such as spinning skaters or collapsing nebulae.
• Relation \( \boldsymbol{\tau} = d\mathbf{l}/dt \): essential for calculating angular acceleration from a given torque.
• Equilibrium twin rules (net force = 0, net torque = 0): crucial for ladder, beam, and bridge support problems.

Keep practicing with everyday examples—twisting a screwdriver, swinging a bat, balancing a seesaw—and these ideas will stick!