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NEET
Exam
NEET
Subject
Physics
Chapter
Rotational Motion

NEET · Physics · Rotational Motion

An electric fan with blades of length 0.4 m starts from rest. It...

An electric fan with blades of length 0.4 m0.4 \text{ m} starts from rest. It reaches a steady rotational speed of 300 rpm300 \text{ rpm} in 5 s5 \text{ s} with uniform angular acceleration. The tangential acceleration of a point at the tip of the blade is:

Options

  1. A0.8π m s−20.8\pi \text{ m s}^{-2}
  2. B2π m s−22\pi \text{ m s}^{-2}
  3. C4π m s−24\pi \text{ m s}^{-2}
  4. D24 m s−224 \text{ m s}^{-2}

Correct answer

A. 0.8π m s−20.8\pi \text{ m s}^{-2}

Step-by-step solution

Given, initial angular speed ω0=0\omega_0 = 0
Final angular speed ω=300 rpm\omega = 300 \text{ rpm}
Time t=5 st = 5 \text{ s}
Radius of the circular path of the tip R=0.4 mR = 0.4 \text{ m}

First, convert the final angular speed into rad s−1\text{rad s}^{-1}:
ω=300×2π60 rad s−1=10π rad s−1\omega = 300 \times \dfrac{2\pi}{60} \text{ rad s}^{-1} = 10\pi \text{ rad s}^{-1}

Calculate the angular acceleration α\alpha using the first equation of rotational kinematics:
α=ω−ω0t=10π−05=2π rad s−2\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{10\pi - 0}{5} = 2\pi \text{ rad s}^{-2}

The tangential acceleration ata_t is related to angular acceleration by the formula:
at=αRa_t = \alpha R
at=2π×0.4=0.8π m s−2a_t = 2\pi \times 0.4 = 0.8\pi \text{ m s}^{-2}

Answer: 0.8π m s−20.8\pi \text{ m s}^{-2}
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