Problem Statement
Solve the kinematics problem: Point $A$ moves on a circle of radius $R_1$ and point $B$ on a circle of radius $R_2$ ($R_2>R_1$), both centered at $O$, at constant angular velocities $\omega_1$ and $\omega_2$. Find the angular velocity of $B$ as seen from $A$. Position vectors: $$\vec r_A = R_1(\cos\omega_1 t, \sin\omega_1 t),\qu
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Rotational kinematics mirrors linear kinematics with $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$. The angular velocity vector $\boldsymbol{\omega}$ points along the rotation axis (right-hand rule). For a point at distance $r$ from the axis: $v = r\omega$ and $a_\tau = r\alpha$, $a_n = r\omega^2 = v^2/r$.
- $v = r\omega$ — tangential speed from angular velocity
- $a_\tau = r\alpha$ — tangential acceleration
- $a_n = r\omega^2 = v^2/r$ — centripetal acceleration
- $\omega = d\theta/dt$, $\alpha = d\omega/dt$
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\vec r_A = R_1(\cos\omega_1 t, \sin\omega_1 t),\qu
Given Information
- Initial velocity $u$ (or $v_0$)
- Acceleration $a$ (constant unless stated otherwise)
- Time $t$ or distance $s$ as given
Physical Concepts & Formulas
Kinematics describes motion without reference to its cause. For constant acceleration, the four SUVAT equations are sufficient to solve any problem. They follow directly from the definitions of velocity ($v = ds/dt$) and acceleration ($a = dv/dt$). For 2D problems (projectile motion), the horizontal and vertical motions are independent — horizontal: constant velocity; vertical: constant acceleration $g$ downward. Relative motion problems require defining a reference frame explicitly and using vector subtraction.
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- $s = \tfrac{1}{2}(u+v)t$
- Range of projectile: $R = \dfrac{u^2\sin 2\theta}{g}$
- Max height: $H = \dfrac{u^2\sin^2\theta}{2g}$
Step-by-Step Solution
Step 1 — List knowns and unknown: $u$, $v$, $a$, $s$, $t$ — identify which three are known.
Step 2 — Choose the SUVAT equation that contains the unknown and all three known quantities.
Step 3 — Substitute and solve algebraically.
Step 4 — For 2D: Decompose $\vec{u}$ into $u_x = u\cos\theta$, $u_y = u\sin\theta$. Solve $x$ and $y$ separately.
Worked Calculation
Substituting all values with units:
Projectile at $u = 20\,\text{m/s}$, $\theta = 30°$:
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Answer
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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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