PHY 001 (Physics I)

Lecture 7

Instructor: Dr. Mohamed Fouad Salem

mohamed.salem@gmail.com

TextbookUniversity Physics, 12th edition,

Young and Freedman

Lecture 7

Course Material Website

http://meryesk.wordpress.com/phy001/

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Chapter 9

Rotation of Rigid Bodies

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9-4 Rotational Kinetic Energy A set of masses mi uniformly rotating with angular velocity ω about some fixed axis A possesses a kinetic energy defined by

where ri is the distance from the ith mass to the rotation axis.For such a set of mass, or for a continuous body, we

K = 12 miv i

2 = 12 mi

i

∑ ri2ωωωω 2

i

∑

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For such a set of mass, or for a continuous body, we define the moment of inertia I about the specified axis Aas

Then the rotational kinetic energy can be written as

I = miri2

i

∑

K = 12 Iω

2

Moment of Inertia The greater the moment of inertia of a rigid body, the more difficult to make it rotate if it is at rest or to stop it if it started rotating

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Moment of Inertia for different rotation axes(Example)

An engineer is designing a machine part consisting of three heavydisks linked by lightweight struts as shown(a) What it is the moment of inertia of this body about an axis

through the center of disk A, perpendicular to the plane of thediagram?

(b) What it is the moment of inertia about an axis through the center of disks B and C?

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center of disks B and C?(c) If the body rotates about

an axis through A as in (a) with angular speedω = 4.0 rad/s, what it is the kinetic energy?

Moment of Inertia for different rotation axes(Example solution)

(a)

(b)

(c)

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Moments of inertia of some common bodies

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Rotational Energy (Example 1)

• We wrap a light, nonstretching cable around a solid cylinder of mass 50 kg and diameter 0.120 m, which rotates in frictionless bearings about a stationary axis. We pull the free end of the cable with a constant 9.0 N force for a distance of 2.0 m; it turns the cylinder as it unwinds without slipping. The cylinder is initially at rest.

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cylinder is initially at rest. Find its final angularspeed and the finalspeed of the cable.

The work done on the cylinder is:

The moment of inertia is:

Rotational Energy (Example 1 solution)

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Conservation of energygives:

Rotational Energy (Example 1 solution continuation)

The final tangential speed of the cylinder, and hence

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The final tangential speed of the cylinder, and hence the final speed of the cable is:

Rotational Energy (Example 2) We wrap a light, nonstretching cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a block of mass m and release the block from rest at a distance h above the floor. As the block falls,the cable unwinds without stretchingor slipping. Find expressions

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for the speed of the falling blockand the angular speed of thecylinder as the block strikesthe floor.

Rotational Energy (Example 2 solution)

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Rotational Energy (Example 2 solution continuation)

Solving for the linear velocity gives:

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Gravitational potential energy of an extended body

• In the previous example if the cable were to have considerable mass not negligible as assumed, we need to calculate gravitational potential energy for it.

• The gravitational potential energy of an

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• The gravitational potential energy of an extended body is the same as if all the mass were concentrated at its center of mass:

Ugrav = Mgycm

• Where ycm is the y-coordinate of the center of mass.

The parallel-axis theoremGiven two parallel axes (lines), one passing through an object’s center of mass and the other displaced by a distance d, the object’s moment of inertia about the displaced axis is given by

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where M is the object’s mass and Icm is the moment of inertia measured about the axis that passes through the object’s center of mass.

Calculation of Moment of Inertia for Complex Object

1. To find I for a complex object, split it into simple geometrical shapes that can be found in Table 9.2

2. Use Table 9.2 to get ICM for each part about the axis parallel to the axis of rotation and going

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axis parallel to the axis of rotation and going through the center-of-mass

3. If needed use parallel-axis theorem to get I for each part about the axis of rotation

4. Add up moments of inertia of all parts

The parallel-axis theorem (Example)

A part of mechanical linkage is shown has a mass of 3.6 Kg. We measure its moment of inertia about an axis 0.15 m from the center of mass to be Ip = 0.132 Kg. m2. What it is the moment of inertia Icmabout a parallel axis through the center of the mass?

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Calculations of Moment of InertiaFor a rigid body with a total mass M, divide the body into very small elements of mass dm and assume that each element has a distance r from the axis of rotation, then the moment of inertia is

but the density of mass is

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but the density of mass is

So,

For a constant density of mass

Finally, we need to express the volume element dv in terms of the differentials of integration variables, then we integrate to get I

Calculations of Moment of Inertia (Example 1)

The figure shows a slender uniform rod with massM and length L. Compute its moment of inertia about an axis through O, at any arbitrary distance h from the end.

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Calculations of Moment of Inertia (Example 1 Solution)

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h = 0

h = L

h = L/2

The figure shows a hollow, uniform cylinder with length L,inner radius R1, and outer radius R2. Find the moment ofinertia about the axis of symmetry of the cylinder.

Calculations of Moment of Inertia (Example 2)

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Calculations of Moment of Inertia (Example 2 Solution)

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But the total volume of the cylinder is given by

Hence, the total mass is

The moment of inertia of the hollow cylinder

For a solid cylinder R1 = 0, then I is

Next Time

• Chapter 10

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Assignment # 5

• Section 9-554 and 57

• Section 9-660, 61 and 62

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