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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 uniform rod having a mass of 300 g and a length of 50 cm rotates...

A uniform rod having a mass of 300 g300 \text{ g} and a length of 50 cm50 \text{ cm} rotates at a uniform angular speed of 2 rad/s2 \text{ rad/s} about an axis perpendicular to the rod through one end. Calculate the angular momentum of the rod about the axis of rotation.

Options

  1. A0.15 kg-m2/s0.15 \text{ kg-m}^2/\text{s}
  2. B0.025 kg-m2/s0.025 \text{ kg-m}^2/\text{s}
  3. C0.0125 kg-m2/s0.0125 \text{ kg-m}^2/\text{s}
  4. D0.05 kg-m2/s0.05 \text{ kg-m}^2/\text{s}

Correct answer

D. 0.05 kg-m2/s0.05 \text{ kg-m}^2/\text{s}

Step-by-step solution

Mass of the rod, m=300 g=0.3 kgm = 300 \text{ g} = 0.3 \text{ kg}

Length of the rod, l=50 cm=0.5 ml = 50 \text{ cm} = 0.5 \text{ m}

Angular speed, ω=2 rad/s\omega = 2 \text{ rad/s}

The moment of inertia of a uniform rod about an axis passing through one end and perpendicular to its length is:

I=ml23I = \dfrac{ml^2}{3}

Substituting the given values:

I=0.3×(0.5)23=0.1×0.25=0.025 kg-m2I = \dfrac{0.3 \times (0.5)^2}{3} = 0.1 \times 0.25 = 0.025 \text{ kg-m}^2

The angular momentum LL is given by:

L=IωL = I\omega

Substituting the values:

L=0.025×2=0.05 kg-m2/sL = 0.025 \times 2 = 0.05 \text{ kg-m}^2/\text{s}
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