Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

The rotor of an electric motor has rotational inertia $I_{m} = 2 .0 \times 10^{- 3} kg m^{2}$ about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia role="math" localid="1660985808865" $I_{p} = 12 kg . m^{2}$ about this axis. Calculate the number of revolutions of the rotor required to turn the probe through $30 °$ about its central axis.

The number of revolutions of the rotor required to turn the probe through $30 °$ about its central axis are $5.0 \times 10^{2}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The rotational inertia of electric motor is, $I_{m} = 2.0 \times 10^{- 3} kg . m^{2}$
The rotational inertia of probe is, $I_{p} = 12 kg . m^{2} .$
The angle through which the probe is rotated is, $θ_{p} = 30 °$

Step 2: To understand the concept

Using the conservation law of the angular momentum we can find the angle through which the motor is rotated. As in one rotation the motor is rotated through $360$ °, we can find the number of revolutions for the angle through which the motor is rotated.

Formula:

The law of conservation of angular momentum, $L_{i} = L_{f}$

$Iθ = \int Iω$

Step 3: Calculate the number of revolutions

The law of conservation of angular momentum gives

$\begin{matrix} L_{i} = L_{f} \\ I_{m} ω_{m} = I_{p} ω_{p} \\ I_{m} θ_{m} = I_{p} θ_{p} \end{matrix}$

$(As I_{m} θ_{m} = \int I_{m} ω_{m} dt and I_{p} θ_{p} = \int I_{p} ω_{p} dt)$

$θ_{m} = \frac{I_{p} θ_{p}}{I_{m}}$

$\begin{array}{l} θ_{m} = \frac{12 \times 30}{2 \times 10^{- 3}} \\ θ_{m} = 180000 ° \end{array}$

So, the no. of revolutions of the rotor is

$\begin{array}{l} N = θ_{m} / (360 ° / rev) \\ N = 180000 ° / (360 ° / rev) \\ N = 500 rev \end{array}$

Therefore, the no. of revolutions of the rotor is $5.0 \times 10^{2}$ .

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: To understand the concept

Step 3: Calculate the number of revolutions

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: To understand the concept

Step 3: Calculate the number of revolutions

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Famous Physicists

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics