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MCE263-DYNAMICS

MASS MOMENT OF INERTIA

Let’s start with Newton’s Law for linear motion

𝑭 = 𝒎a; where a the linear acceleration of the mass

If we want to apply this law to a rotating mass at some radius we can multiply both sides of the equation by the radius. Then

𝒍𝑭 = 𝒎𝒍a→ Torque=𝒎𝒍(𝒍𝜶) → 𝑻𝒂𝒍𝒒𝒂𝒍 = (𝒎𝒍𝟐)𝜶

The term 𝒎𝒍𝟐 𝒍𝑴 𝒅𝒍𝒐𝒍𝒍𝒍𝒅 𝒍𝑴 𝒎𝒍𝑴𝑴 𝒎𝒂𝒎𝒍𝒍𝒂 𝒂𝒐 𝒍𝒍𝒍𝒍𝒂𝒍𝒍, 𝑰;

𝑻𝑻𝒍𝒍 𝒂𝑻𝒍 𝒎𝒍𝑴𝑴 𝒎𝒂𝒎𝒍𝒍𝒂 𝒂𝒐 𝒍𝒍𝒍𝒍𝒂𝒍𝒍 𝒍𝒃𝒂𝒂𝒂

𝒂𝑻𝒍 𝒛 − 𝒍𝒙𝒍𝑴 at point A is given as follows:

𝐚 = 𝐫𝛂 ≡ 𝒍𝒍𝒍𝒍𝒍𝒍 𝒍𝒂𝒂𝒍𝒍𝒍𝒍𝒍𝒂𝒍𝒂𝒍

𝒎

𝒍

𝒚

𝒙

𝜶 ≡ 𝒍𝒍𝒂𝒂𝒍𝒍𝒍 𝒍𝒂𝒂𝒍𝒍𝒍𝒍𝒍𝒂𝒍𝒂𝒍

𝑭

𝑻𝒂𝒍𝒒𝒂𝒍 = 𝑰𝜶 𝒂𝒍

�𝑴���⃗ 𝑨𝒛 = 𝑰𝑨𝒛 𝜶

𝑰𝑨𝒛 = 𝒎𝒍𝟐 and Newton’s law for a rotating mass becomes

Where ∑𝑴���⃗ 𝑨 ≡ 𝑺𝒂𝒎𝒎𝒍𝒂𝒍𝒂𝒍 𝒂𝒐 𝒎𝒂𝒎𝒍𝒍𝒂𝑴 𝒍𝒃𝒂𝒂𝒂 𝒑𝒂𝒍𝒍𝒂 𝑨

2

𝒚

𝒅𝒎

𝒍

𝒙

In the most general case, given an arbitrary rigid body rotating about z-axis, integration is needed to calculate the mass moment of inertia of the body

𝑰𝒛 = �𝒍𝟐 𝒅𝒎

This can then be written as an

Integral over the volume V, as follows:

𝑰𝒛 = ∫𝒍𝟐 𝝆 𝒅𝑽

and if the

density 𝝆 is constant it can

be brought outside the integral

to obtain a simpler volume integral

𝑰𝒛 = 𝝆∫𝒍𝟐 𝒅𝑽𝒂𝒍 𝑰𝒛 = 𝝆∭𝒍𝟐 𝒅𝒙 𝒅𝒚 𝒅𝒛

𝑰𝒍𝒂𝒍𝒂𝒍𝒍𝒂𝒍𝒂𝒍 𝒂𝒈𝒍𝒍 𝒍𝒍𝒂𝒍𝒍𝒍 𝒃𝒂𝒅𝒚

𝒛

𝑹𝑬𝑻𝑨𝑿‼!𝒀𝑶𝑼 𝑾𝑰𝑻𝑻 𝑵𝑶𝑻 𝑯𝑨𝑽𝑬 𝑻𝑶 𝑫𝑶 𝑰𝑵𝑻𝑬𝑮𝑹𝑨𝑻𝑰𝑶𝑵

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𝒙

𝒚

𝒛

𝒅𝒍 𝒍

𝒛

𝒙

𝒚

𝑹

h/2

h/2

UNITS: kg.m2 or slug.ft2

𝑴𝒍𝑴𝑴 𝒎𝒂𝒎𝒍𝒍𝒂 𝒂𝒐 𝑰𝒍𝒍𝒍𝒂𝒍𝒍 𝒂𝒐 𝒍 𝑴𝒂𝒍𝒍𝒅 𝒂𝒚𝒍𝒍𝒍𝒅𝒍𝒍 𝒍𝑴 𝒍𝑴 𝒐𝒂𝒍𝒍𝒂𝒇: 𝑰𝒛 =𝟏𝟐𝑴𝑹

𝟐

EXAMPLE 1 – MASS MOMENT OF INERTIA OF A SOLID CYLINDER

BY INTEGRATION

Since all points of the cylindrical shell are at the same distance from the Z-axis,

the inertia of the cylindrical shell about the Z-axis can be written as follows:

d𝑰𝒛 = [𝒅𝒎] 𝒍𝟐 = [𝝆(𝟐𝝅𝒍)𝑻 𝒅𝒍]𝒍𝟐 → d𝑰𝒛 = [(𝟐𝝅𝝆𝑻)]𝒍𝟑𝒅𝒍 →

𝑰𝒛 = [(𝟐𝝅𝝆𝑻)]� 𝒍𝟑𝒅𝒍𝑹

𝟎→ 𝑰𝒛 =

[(𝟐𝝅𝝆𝑻)]𝑹𝟒

𝟒→ 𝑰𝒛 =

[(𝝅𝝆𝑻)]𝑹𝟒

𝟐

But total mass M = 𝝆[𝝅𝑹𝟐]𝑻 →

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5

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The table gives equations for mass the mass moment of inertia, I, about certain axes but we may need the mass

moment of inertia about some arbitrary axis; The Parallel Axis Theorem accomplishes that task

𝑷𝒍𝒍𝒍𝒍𝒍𝒍𝒍 𝑨𝒙𝒍𝑴 𝑻𝑻𝒍𝒂𝒍𝒍𝒎 𝑰𝒛 = 𝑰𝑮𝒛′ + 𝒎 𝒅𝟐 𝒇𝑻𝒍𝒍𝒍 𝒎 ≡ 𝒎𝒍𝑴𝑴 𝒂𝒐 𝒂𝑻𝒍 𝒃𝒂𝒅𝒚

NOTES:

1. Mass moment of inertia

𝑰𝒛 𝒍𝑴 𝒍𝒃𝒂𝒂𝒂 𝒍𝒍 𝒍𝒙𝒍𝑴 𝒛 𝒑𝒍𝒍𝒍𝒍𝒍𝒍𝒍 𝒂𝒂 𝒂𝑻𝒍 𝒍𝒙𝒍𝑴 𝒛′ 𝒂𝑻𝒍𝒂𝒂𝒂𝑻 𝒂𝑻𝒍 𝒂𝒍𝒍𝒂𝒍𝒍 𝒂𝒐 𝒂𝒍𝒍𝒈𝒍𝒂𝒚 𝑮 and through an arbitrary point

2. Mass moment of inertia 𝑰𝑮𝒛′ 𝒍𝑴 𝒍𝒃𝒂𝒂𝒂 𝒍𝒍 𝒍𝒙𝒍𝑴 𝒛′ 𝒂𝑻𝒍𝒂𝒂𝒂𝑻 𝒂𝑻𝒍 𝒂𝒍𝒍𝒂𝒍𝒍 𝒂𝒐 𝒂𝒍𝒍𝒈𝒍𝒂𝒚 𝑮

3. Distance d is the perpendicular distance between the two parallel axes 𝒛 𝒍𝒍𝒅 𝒛′ 4. The equation is valid only as shown above; any point on the left of the

equation; the center of gravity on the right side of the equation 5. 𝑰𝑮𝒛′ = 𝑰𝒛 + 𝒎 𝒅𝟐

𝑨𝑵𝑶𝑻𝑯𝑬𝑹 𝑫𝑬𝑭𝑰𝑵𝑰𝑻𝑰𝑶𝑵

𝑰𝒛 = 𝒎 𝒌𝒛𝟐

𝒍𝒂′𝑴 𝑴𝒍𝒎𝒍𝒍𝒍𝒍 𝒂𝒂 𝑰𝒛 = 𝒎 𝒍𝟐

RADIUS OF GYRATION 𝒌 𝒛 ; 𝒍𝒂′𝑴 𝒍𝒂𝒂 𝒑𝒍𝒍𝒂 𝒂𝒐 𝒂𝑻𝒍 𝒃𝒂𝒅𝒚; 𝒍𝒂′𝑴 𝒍𝒒𝒂𝒍𝒈𝒍𝒍𝒍𝒍𝒂 𝒂𝒂 𝑴𝒂𝒎𝒍 𝒍𝒍𝒅𝒍𝒂𝑴 𝒍

Significance of 𝒌 𝒛; 𝒍𝒍 𝒂𝒂𝒎𝒑𝒍𝒍𝒙 𝒃𝒂𝒅𝒍𝒍𝑴 𝒂𝒂𝑻𝒍𝒍𝑴 𝑻𝒍𝒈𝒍 𝒂𝒍𝒍𝒂𝒂𝒍𝒍𝒂𝒍𝒅 𝒂𝑻𝒍 𝒍𝒍𝒅𝒍𝒂𝑴 𝒂𝒐 𝒂𝒚𝒍𝒍𝒂𝒍𝒂𝒍 which can be used to get the mass moment of inertia

Example 1: For a body with mass m and given 𝒌𝑮𝒛, 𝒂𝒍𝒍𝒂𝒂𝒍𝒍𝒂𝒍 𝑰𝑮𝒛; 𝑰𝑮𝒛 = 𝒎 𝒌𝑮𝒛𝟐

Example 2: For a body with mass m and given 𝒌𝑨𝒛, 𝒂𝒍𝒍𝒂𝒂𝒍𝒍𝒂𝒍 𝑰𝑨𝒛; 𝑰𝑨𝒛 = 𝒎 𝒌𝑨𝒛𝟐

And so on!!

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Calculation of mass moment of inertia of composite bodies by adding or subtracting mass moments of inertia of various segments

EXAMPLE 1 – COMPOSITE BODIES – MASS MOMENT OF INERTIA CALCULATION

The pendulum shown below consists of two thin rods, each weighing 10 lb with mass (10/32.2)=0.31 slug. Determine the mass moment of inertia of the following composite body (pendulum) about an axis perpendicular to the page passing through

a. Point of suspension O b. Center of Gravity G of the entire system

SOLUTION about point of suspension O:

1. Bar OA

𝑰𝑶(𝑶𝑨) =𝟏𝟑𝒎 𝒍𝟐 =

𝟏𝟑∗ 𝟎.𝟑𝟏 ∗ 𝟐𝟐 →

2. Bar BC; use Parallel Axis Theorem

𝑰𝑶(𝑩𝑪) = 𝑰𝑮(𝑩𝑪) + 𝒎𝒅𝟏𝟐 →𝟏𝟏𝟐

𝒎𝒍𝟐 + 𝒎𝒅𝟏𝟐 𝒂𝒍

𝑰𝑶(𝑩𝑪) = 𝟏𝟏𝟐∗ 𝟎.𝟑𝟏 ∗ 𝟐𝟐 + 𝟎.𝟑𝟏 ∗ 𝟐𝟐

𝑰𝑶(𝑻𝑶𝑻𝑨𝑻) = 𝑰𝑶(𝑶𝑨) + 𝑰𝑶(𝑩𝑪) = 𝟎.𝟒𝟏𝟒 + 𝟏.𝟑𝟒𝟕 →

SOLUTION about point of Gsystem: Must first calculate 𝒚� = 𝟏.𝟐 𝒐𝒍𝒍𝒂 Then use Parallel Axis Theorem

𝑰𝑶(𝑻𝑶𝑻𝑨𝑻) = 𝑰𝑮𝑴𝒚𝑴𝒂𝒍𝒎 + 𝒎𝒂𝒂𝒂𝒍𝒍𝒚�𝟐 → 𝟏.𝟕𝟕 = 𝑰𝑮𝑴𝒚𝑴𝒂𝒍𝒎 + (𝟐 ∗ 𝟎.𝟑𝟏) ∗ 𝟏.𝟐𝟐

𝟏 𝒐𝒂𝒂𝒂 𝟏 𝒐𝒂𝒂𝒂

𝒚�

𝑮𝑴𝒚𝑴𝒂𝒍𝒎

𝑰𝑶(𝑶𝑨) = 𝟎.𝟒𝟏𝟒 𝑴𝒍𝒂𝒂.𝒐𝒂𝟐

𝑰𝑶(𝑩𝑪) = 𝟏.𝟑𝟒𝟕 𝑴𝒍𝒂𝒂.𝒐𝒂𝟐

𝑶

𝑩 𝑪 𝑨

𝒅𝟏 = 𝟐 𝒐𝒍𝒍𝒂

𝑮𝑩𝑪

𝑰𝑶(𝑻𝑶𝑻𝑨𝑻) = 𝟏.𝟕𝟕 𝑴𝒍𝒂𝒂.𝒐𝒂𝟐

𝑰𝑮𝑴𝒚𝑴𝒂𝒍𝒎 = 𝟎.𝟑𝟕𝟐 𝑴𝒍𝒂𝒂.𝒐𝒂𝟐

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𝑹 = 𝟎.𝟐𝟐𝟎 𝒎𝒎

𝑶

𝒍 = 𝟎.𝟏𝟐𝟐 𝒎𝒎

𝒚

𝒙

𝑮

EXAMPLE 2 – COMPOSITE BODIES – MASS MOMENT OF INERTIA CALCULATION

The ring shown below is 10 mm thick and its density is 8,000 kg/m3. Determine the mass moment of inertia about an axis perpendicular to the page through a point on the outer circumference, as shown, Point O, IZO.

SOLUTION:

𝑻𝑻𝒍 𝒎𝒍𝑴𝑴 𝒂𝒐 𝒂𝑻𝒍 𝒍𝒍𝒍𝒂𝒍 𝒅𝒍𝑴𝒌,𝒎𝒍𝒍𝒍𝒂𝒍 = 𝒈𝒂𝒍𝒂𝒎𝒍 ∗ 𝒅𝒍𝒍𝑴𝒍𝒂𝒚 =

�𝝅 ∗ 𝟎.𝟐𝟐𝟎𝟐 ∗.𝟎𝟏𝟎� ∗ 𝟖,𝟎𝟎𝟎 → 𝒎𝒍𝒍𝒍𝒂𝒍 = 𝟏𝟐.𝟕𝟏 𝒌𝒂

𝑻𝑻𝒍 𝒎𝒍𝑴𝑴 𝒂𝒐 𝒂𝑻𝒍 𝑴𝒎𝒍𝒍𝒍 𝒅𝒍𝑴𝒌,𝒎𝑴𝒎𝒍𝒍𝒍=

= �𝝅 ∗ 𝟎.𝟏𝟐𝟐𝟎𝟐 ∗.𝟎𝟏𝟎� ∗ 𝟖,𝟎𝟎𝟎 → 𝒎𝑴𝒎𝒍𝒍𝒍 = 𝟑.𝟗𝟑 𝒌𝒂

𝟏. Find the mass moment of inertia of the large circle about point O

2. Find the mass moment of inertia of the small circle about point O 3. Subtract result from Step 2 from the result of Step 1 1. Use Parallel Axis Theorem

𝑰𝑶(𝒍𝒍𝒍𝒂𝒍) = 𝑰𝑮(𝒍𝒍𝒍𝒂𝒍) + 𝒎𝒍𝒍𝒍𝒂𝒍𝒅𝟐 → 𝑰𝑶(𝒍𝒍𝒍𝒂𝒍) = 𝟏𝟐𝒎𝒍𝒍𝒍𝒂𝒍𝑹𝟐 + 𝒎𝒍𝒍𝒍𝒂𝒍(𝟎.𝟐𝟐𝟎)𝟐

𝑰𝑶(𝒍𝒍𝒍𝒂𝒍) = 𝟎.𝟐 ∗ 𝟏𝟐.𝟕𝟏 ∗ 𝟎.𝟐𝟐𝟎𝟐 + 𝟏𝟐.𝟕𝟏 ∗ 𝟎.𝟐𝟐𝟎𝟐 →

2. Use Parallel Axis Theorem

𝑰𝑶(𝑴𝒎𝒍𝒍𝒍) = 𝑰𝑮(𝑴𝒎𝒍𝒍𝒍) + 𝒎𝑴𝒎𝒍𝒍𝒍𝒅𝟐 → 𝑰𝑶(𝑴𝒎𝒍𝒍𝒍) =𝟏𝟐𝒎𝑴𝒎𝒍𝒍𝒍𝒍𝟐 + 𝒎𝑴𝒎𝒍𝒍𝒍(𝟎.𝟐𝟐𝟎)𝟐

𝑰𝑶(𝑴𝒎𝒍𝒍𝒍) = 𝟎.𝟐 ∗ 𝟑.𝟗𝟑 ∗ 𝟎.𝟏𝟐𝟐𝟐 + 𝟑.𝟗𝟑 ∗ 𝟎.𝟐𝟐𝟎𝟐 →

3. Subtract Eqn. (2) from Eqn. (1) 𝑰𝑶(𝒍𝒂𝒂𝒂𝒍𝒍 𝑴𝒚𝑴𝒂𝒍𝒎) = 𝑰𝑶(𝒍𝒍𝒍𝒂𝒍) − 𝑰𝑶(𝑴𝒎𝒍𝒍𝒍) = 𝟏.𝟒𝟕𝟑 − 𝟎.𝟐𝟕𝟕

𝑰𝑶(𝒍𝒍𝒍𝒂𝒍) = 𝟏.𝟒𝟕𝟑 𝒌𝒂.𝒎𝟐 (1)

𝑰𝑶(𝑴𝒎𝒍𝒍𝒍) = 𝟎.𝟐𝟕𝟕 𝒌𝒂.𝒎𝟐 (2)

𝑰𝑶(𝒍𝒂𝒂𝒂𝒍𝒍 𝑴𝒚𝑴𝒂𝒍𝒎) = 𝟏.𝟐𝟎 𝒌𝒂.𝒎𝟐