Problem Statement
Solve the rotational mechanics problem: At a given instant, different points of a rigid body have velocities $\vec v_1$ and $\vec v_2$ at positions $\vec r_1$ and $\vec r_2$. Find the angular velocity $\vec\omega$ of the body. For a rigid body: $\vec v_i = \vec\omega \times \vec r_i + \vec v_0$ The difference eliminates $\vec v_0$: $$\vec
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Rotational kinematics mirrors linear kinematics with $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$. The angular velocity vector $\boldsymbol{\omega}$ points along the rotation axis (right-hand rule). For a point at distance $r$ from the axis: $v = r\omega$ and $a_\tau = r\alpha$, $a_n = r\omega^2 = v^2/r$.
- $v = r\omega$ — tangential speed from angular velocity
- $a_\tau = r\alpha$ — tangential acceleration
- $a_n = r\omega^2 = v^2/r$ — centripetal acceleration
- $\omega = d\theta/dt$, $\alpha = d\omega/dt$
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\vec
Given Information
- Mass $m$, geometry (radius $R$, length $L$, etc.)
- Angular velocity $\omega$ or torque $\tau$
- Axis of rotation
Physical Concepts & Formulas
Rotational mechanics is the angular analogue of linear mechanics. The moment of inertia $I = \sum m_i r_i^2$ plays the role of mass; torque $\tau = r\times F$ plays the role of force; angular momentum $L = I\omega$ plays the role of linear momentum. Newton’s 2nd law becomes $\tau_{\text{net}} = I\alpha$. The parallel axis theorem $I = I_{\text{cm}} + Md^2$ allows computing $I$ about any axis from the centre-of-mass value. Conservation of angular momentum ($L = \text{const}$ when $\tau_{\text{ext}} = 0$) explains why spinning skaters speed up when they pull their arms in.
- $I = \int r^2\,dm$ — moment of inertia
- $\tau = I\alpha$ — rotational Newton’s 2nd law
- $L = I\omega$ — angular momentum
- $KE_{\text{rot}} = \frac{1}{2}I\omega^2$
- Solid sphere: $I = \frac{2}{5}MR^2$; hollow sphere: $\frac{2}{3}MR^2$
- Solid cylinder: $I = \frac{1}{2}MR^2$; thin rod (centre): $\frac{1}{12}ML^2$
Step-by-Step Solution
Step 1 — Identify body and axis: Choose the appropriate moment of inertia formula or use the parallel axis theorem.
Step 2 — Torque equation: $\tau_{\text{net}} = I\alpha$
Step 3 — Kinematics: Use rotational analogues: $\omega = \omega_0 + \alpha t$, $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$, $\omega^2 = \omega_0^2 + 2\alpha\theta$.
Step 4 — Energy: $KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$ for rolling bodies.
Worked Calculation
Substituting all values with units:
Solid cylinder ($I = MR^2/2$) rolling down incline $h = 1\,\text{m}$:
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Answer
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Answer
$$\boxed{v = \sqrt{\dfrac{4gh}{3}}\text{ (solid cylinder rolling)}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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