PHY 001 (Physics I) Lecture 7

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1 PHY 001 (Physics I) Instructor: Dr. Mohamed Fouad Salem

2 Textbook University Physics, 12 th edition, Young and Freedman

3 Course Material Website

4 Chapter 9 Rotation of Rigid Bodies

5 9-4 Rotational Kinetic Energy A set of masses m i uniformly rotating with angular velocity ω about some fixed axis A possesses a kinetic energy defined by K= 1 m 2 i v 2 i = 1 m 2 i r i2 ω 2 i where r i is the distance from the i th mass to the rotation axis. For such a set of mass, or for a continuous body, we define the moment of inertia I about the specified axis A as i I= m i r i 2 Then the rotational kinetic energy can be written as K= 1 2 Iω 2 i

6 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

7 Moment of Inertia for different rotation axes (Example) An engineer is designing a machine part consisting of three heavy disks 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 the diagram? (b) What it is the moment of inertia about an axis through the center of disks B and C? (c) If the body rotates about an axis through A as in (a) with angular speed ω = 4.0 rad/s, what it is the kinetic energy?

8 (a) Moment of Inertia for different rotation axes (Example solution) (b) (c)

9 Moments of inertia of some common bodies

10 Rotational Energy (Example 1) We wrap a light, nonstretching cable around a solid cylinder of mass 50 kg and diameter 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. Find its final angular speed and the final speed of the cable.

11 Rotational Energy (Example 1 solution) The work done on the cylinder is: The moment of inertia is: Conservation of energy gives:

12 Rotational Energy (Example 1 solution continuation) The final tangential speed of the cylinder, and hence the final speed of the cable is:

13 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 stretching or slipping. Find expressions for the speed of the falling block and the angular speed of the cylinder as the block strikes the floor.

14 Rotational Energy (Example 2 solution)

15 Rotational Energy (Example 2 solution continuation) Solving for the linear velocity gives:

16 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 extended body is the same as if all the mass were concentrated at its center of mass: U grav = Mgy cm Where y cm is the y-coordinate of the center of mass.

inertia about the displaced axis is given by where M is the object s mass and I cm is

17 The parallel-axis theorem Given 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 where M is the object s mass and I cm is the moment of inertia measured about the axis that passes through the object s center of mass.

18 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 Use Table 9.2 to get I CM for each part about the 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

19 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 I p = Kg. m 2. What it is the moment of inertia I cm about a parallel axis through the center of the mass?

20 Calculations of Moment of Inertia For 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 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

21 Calculations of Moment of Inertia (Example 1) The figure shows a slender uniform rod with mass M and length L. Compute its moment of inertia about an axis through O, at any arbitrary distance h from the end.

22 Calculations of Moment of Inertia (Example 1 Solution) h = 0 h = L h = L/2

23 Calculations of Moment of Inertia (Example 2) The figure shows a hollow, uniform cylinder with length L, inner radius R 1, and outer radius R 2. Find the moment of inertia about the axis of symmetry of the cylinder.

24 Calculations of Moment of Inertia (Example 2 Solution) 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 R 1 = 0, then I is

25 Next Time Chapter 10

26 Assignment # 5 Section and 57 Section , 61 and 62

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