A wheel revolves about its axis with a uniform angular acceleration. Beginning from rest, it attains 100 rev/sec in 4 seconds. Determine the angular acceleration and the angle rotated during these four seconds.

Correct answer: A. 25 rev/s^2 and 400π rad Initial angular velocity, ω_0 = 0 Final angular velocity, ω = 100 rev/s Time, t = 4 s Using the first equation of rotational motion, ω = ω_0 + α t 100 = 0 + α (4) α = 25 rev/s^2 Using the second equation of rotational motion, θ = ω_0 t + (1)/(2)α t^2 θ = 0 + (1)/(2)(25)(4)^2 = 200 rev Converting revolutions to radians: θ = 200 × 2π = 400π rad

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rotational Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rotational Mechanics

A wheel revolves about its axis with a uniform angular...

A wheel revolves about its axis with a uniform angular acceleration. Beginning from rest, it attains

100 \text{ rev/sec}

4

seconds. Determine the angular acceleration and the angle rotated during these four seconds.

Options

A $25 \text{ rev/s}^2$ and $400\pi \text{ rad}$
B $50 \text{ rev/s}^2$ and $800\pi \text{ rad}$
C $25 \text{ rev/s}^2$ and $200\pi \text{ rad}$
D $50 \text{ rev/s}^2$ and $400\pi \text{ rad}$

Correct answer

A. $25 \text{ rev/s}^2$ and $400\pi \text{ rad}$

Step-by-step solution

Initial angular velocity,

\omega_0 = 0

Final angular velocity,

\omega = 100 \text{ rev/s}

Time,

t = 4 \text{ s}

Using the first equation of rotational motion,

\omega = \omega_0 + \alpha t

100 = 0 + \alpha (4) \Rightarrow \alpha = 25 \text{ rev/s}^2

Using the second equation of rotational motion,

\theta = \omega_0 t + \dfrac{1}{2}\alpha t^2

\theta = 0 + \dfrac{1}{2}(25)(4)^2 = 200 \text{ rev}

Converting revolutions to radians:

\theta = 200 \times 2\pi = 400\pi \text{ rad}

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