HC Verma Chapter 7 Problem 28 — Relation between angular and linear acceleration

Problem Statement

Solve the kinematics problem: Solve the kinematics problem: A flywheel of radius 0.3 m has angular acceleration 5 rad/s². A point on the rim has what linear (tangential) acceleration? $a_t = r\alpha$ Step 1: $a_t = r\alpha = 0.3 \times 5 = 1.5$ m/s². $$\boxed{a_t = 1.5\text{ m/s}^2}$$ Initial velocity $u$ (or $v_0$) Acceleration

Given Information

  • $\boxed{a_t = 1.5\text{ m/s}$

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

  • See the step-by-step solution for the specific equations applied.
  • All quantities are in SI units unless otherwise stated.

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

$$\boxed{a_t = 1.5\text{ m/s}^2}$$

$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$

$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$

Answer

$$\boxed{a_t = 1.5\text{ m/s}^2}$$

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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