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Q6.13: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.13 Solution: A rope of negligible mass is wound round a hollow cylinder of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the solid‑cylinder formula $I=\frac12 MR^2$ instead of the correct thin‑walled (hollow) cylinder inertia $I=MR^2$, which halves the computed angular acceleration. Remember to match the geometry of the body to its moment‑of‑inertia expression before using $alpha=\tau/I$.

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

Step-by-Step Verified Solution

**Answer:** α=25\alpha = 25 rad/s_, a=10a = 10 m/s_. **Step-by-Step Explanation:** Torque τ=F×R=30×0.4=12\tau = F \times R = 30 \times 0.4 = 12 N m. Moment of inertia (hollow cylinder): I=MR2=3×(0.4)2=0.48 kg m2I = MR^2 = 3 \times (0.4)^2 = 0.48 \text{ kg m}^2. Angular acceleration: α=τ/I=12/0.48=25 rad/s2\alpha = \tau/I = 12/0.48 = 25 \text{ rad/s}^2. Linear acceleration: a=αR=25×0.4=10 m/s2a = \alpha R = 25 \times 0.4 = 10 \text{ m/s}^2 [citation:28].

All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

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