Problem Statement
Solve the kinematics problem: A rigid body rotates about a fixed axis with angular acceleration $\beta(t)$. Express the linear acceleration of a point at radius $r$ from the axis. A point at perpendicular distance $r$ from the axis has: Tangential acceleration (along the circle): $$w_\tau = \beta r$$ Centripetal (normal) acceler
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
$$w_\tau = \beta r$$
$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$
$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$
Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
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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