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Physics 111: Lecture 17 Today’s Agenda

Physics 111: Lecture 17 Today’s Agenda. Rotational Kinematics Analogy with one-dimensional kinematics Kinetic energy of a rotating system Moment of inertia Discrete particles Continuous solid objects Parallel axis theorem. Rotation.

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Physics 111: Lecture 17 Today’s Agenda

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  1. Physics 111: Lecture 17Today’s Agenda • Rotational Kinematics • Analogy with one-dimensional kinematics • Kinetic energy of a rotating system • Moment of inertia • Discrete particles • Continuous solid objects • Parallel axis theorem

  2. Rotation • Up until now we have gracefully avoided dealing with the rotation of objects. • We have studied objects that slide, not roll. • We have assumed pulleys are without mass. • Rotation is extremely important, however, and we need to understand it! • Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics.

  3. Lecture 17, Act 1Rotations • Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. • Klyde’s angular velocity is: (a)the same as Bonnie’s (b)twice Bonnie’s (c)half Bonnie’s

  4. Lecture 17, Act 1Rotations • The angular velocity w of any point on a solid object rotating about a fixed axis is the same. • Both Bonnie & Klyde go around once (2p radians) every two seconds. (Their “linear” speed v will be different since v = wr). w

  5. Rotational Variables. Spin round blackboard • Rotation about a fixed axis: • Consider a disk rotating aboutan axis through its center: • First, recall what we learned aboutUniform Circular Motion: (Analogous to )  

  6. constant Rotational Variables... • Now suppose  can change as a function of time: • We define the angular acceleration:  • Consider the case when is constant. • We can integrate this to find  and  as a function of time:  

  7. Rotational Variables... constant v • Recall also that for a point at a distance R away from the axis of rotation: • x = R • v = R And taking the derivative of this we find: • a = R x R   

  8. Summary (with comparison to 1-D kinematics) Angular Linear And for a point at a distance R from the rotation axis: • x = Rv = R a = R

  9. a R Example: Wheel And Rope • A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

  10. = 0 + 0(10) + (10)(10)2 = 500 rad Wheel And Rope... • Use a = Rto find : = a / R = 4 m/s2 / 0.4 m = 10 rad/s2 • Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. a  R

  11. m4 m1 r1  r4 m3 r2 r3 m2 Rotation & Kinetic Energy • Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). • The kinetic energy of this system will be the sum of the kinetic energy of each piece:

  12. which we write as: v1 m4 v4 m1 r1  r4 v2 m3 r2 r3 m2 Define the moment of inertia about the rotation axis v3 Rotation & Kinetic Energy... • So: butvi = ri I has units of kg m2.

  13. Rotation & Kinetic Energy... • The kinetic energy of a rotating system looks similar to that of a point particle:Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

  14. Moment of Inertia Inertia Rods • So where • Notice that the moment of inertia I depends on the distribution of mass in the system. • The further the mass is from the rotation axis, the bigger the moment of inertia. • For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). • We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics!

  15. Calculating Moment of Inertia • We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: m m L m m

  16. I = 2mL2 Calculating Moment of Inertia... • The squared distance from each point mass to the axis is: Using the Pythagorean Theorem so L/2 m m r L m m

  17. m m I = mL2 m m Calculating Moment of Inertia... • Now calculate I for the same object about an axis through the center, parallel to the plane (as shown): r L

  18. I = 2mL2 Calculating Moment of Inertia... • Finally, calculate I for the same object about an axis along one side (as shown): r m m L m m

  19. Calculating Moment of Inertia... • For a single object, I clearly depends on the rotation axis!! I = 2mL2 I = mL2 I = 2mL2 m m L m m

  20. Lecture 17, Act 2Moment of Inertia • A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively. • Which of the following is correct: a (a)Ia > Ib > Ic (b)Ia > Ic > Ib (c)Ib > Ia > Ic b c

  21. Calculate moments of inerta: Lecture 17, Act 2Moment of Inertia • Label masses and lengths: m a L b So(b) is correct:Ia > Ic > Ib L c m m

  22. dm r Calculating Moment of Inertia... • For a discrete collection of point masses we found: • For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm. • We have to do anintegral to find I :

  23. R Thin hoop of mass M and radius R, about an axis through a diameter. Moments of Inertia Hoop • Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop. R

  24. Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. R Moments of Inertia... Sphere and disk • Some examples of I for solid objects: Solid sphere of mass M and radius R, about an axis through its center. R

  25. Lecture 17, Act 3Moment of Inertia • Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. • Which one has the biggest moment of inertia about an axis through its center? (a) solid aluminum (b) hollow gold (c) same hollow solid same mass & radius

  26. Lecture 17, Act 3Moment of Inertia • Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center. • The spherical shell (gold) will have a bigger moment of inertia. ISOLID < ISHELL hollow solid same mass & radius

  27. Thin rod of mass M and length L, about a perpendicular axis through its end. L Moments of Inertia... Rod • Some examples of I for solid objects (see also Tipler, Table 9-1): Thin rod of mass M and length L, about a perpendicular axis through its center. L

  28. Parallel Axis Theorem • Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known. • The moment of inertia about an axis parallel to this axis but a distance D away is given by: IPARALLEL = ICM + MD2 • So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.

  29. We know So which agrees with the result on a previous slide. Parallel Axis Theorem: Example • Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod. IPARALLEL = ICM + MD2 D=L/2 M CM x L ICM IEND

  30. Connection with CM motion • Recall what we found out about the kinetic energy of a system of particles in Lecture 15: KREL KCM • For a solid object rotating about its center of mass, we now see that the first term becomes: Substituting but

  31. Connection with CM motion... • So for a solid object which rotates about its center or mass and whose CM is moving: VCM  We will use this formula more in coming lectures.

  32. Recap of today’s lecture • Rotational Kinematics (Text: 9-1) • Analogy with one-dimensional kinematics • Kinetic energy of a rotating system • Moment of inertia (Text: 9-2, 9-3, Table 9-1) • Discrete particles (Text: 9-3) • Continuous solid objects (Text: 9-3) • Parallel axis theorem (Text: 9-3) • Look at textbook problems Chapter 9: # 7, 11, 27, 31, 33, 37

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