HC Verma Chapter 7 Problem 33 — Particle moving in circle: find speed from angular velocity

Problem Statement

Solve the rotational mechanics problem: Solve the kinematics problem: A particle moves in a circle of radius 0.8 m at 3 rev/s. Find its speed and centripetal acceleration. $v = \omega r$; $a_c = v^2/r = \omega^2 r$ Step 1: $\omega = 3 \times 2\pi = 6\pi$ rad/s. Step 2: $v = 6\pi \times 0.8 = 4.8\pi \approx 15.08$ m/s. Step 3: $a_c = v^2/r

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

Full substitution shown in the steps above.

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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