1. 1/11/16Oregon State University PH 212, Class 41 Motion and preferred (SI) units for a rotating (rigid-body) object Entire rotating object A point on the object at radius r (m) from center Angular measure Translational measure ang. position:  (rad) arc position: s =  r (m) ang. displacement:  (rad)arc displacement:  s = (  )r (m) ang. velocity:  (rad/s)velocity: v =  X r = (r  ) T (m/s) ang. acceleration:  (rad/s 2 )tangential accel.: a T =  X r = (r  ) T (m/s 2 ) radial accel: a R =  X v = (r  2 ) R (m/s 2 ) = (v 2 /r) R (m/s 2 ) total acceleration: a = a T + a R (m/s 2 )

2. 1/11/16Oregon State University PH 212, Class 42 Here again are some of the direct analogies between translational and rotational motion: Quantity or PrincipleLinearRotation Displacement x  Velocity v  Acceleration a  Inertia (resistance to mass (m)moment of acceleration) inertia (I) Moment of inertia (I) is different than the simple measure of mass (m). As we have observed with baseball bats, for example, an object’s resistance to angular acceleration depends not only on how much mass it has but where that mass is located with respect to the axis of rotation.

3. Center of Mass To consider how the distribution of mass in an object affects its motion, start with the center of mass (c.m.). The is the point in an object which it’s often accurate to model as the location of all its mass. The mass is not really all located there, but in many ways, it behaves as if it is. For example: The c.m. is the “balance” point of the object, at which you can rest the entire object upon a fulcrum—or hang it by a thread—and gravity will not rotate the object. Thus the c.m. is also known as the center of gravity (c.g.)—the point at which the entire F G acting on the object is effectively exerted.) The c.m. is also the point about which an unconstrained rotating object will rotate. Throw a hammer across the room…. What point does it rotate around? The same point it “balances” on! We will soon see how/why this must be true. 1/11/163Oregon State University PH 212, Class 4

4. Calculating the Center of Mass For any uniform object, (symmetric, with uniform mass distribu- tion), the center of mass is the geometric center of the object. For any continuous distribution of mass, the c.m. is found by x cm = 1/M  x dm y cm = 1/M  y dm Essentially here, you’re multiplying each bit of mass by its distance from the edge of the object, summing all those contri- butions, then dividing by the total mass to find the “average” location of mass from the edge of the object. (And you do this for each axial direction.) That’s why the c.m. is the balance point. 1/11/164Oregon State University PH 212, Class 4

5. A straight, rigid, mass-less rod of length 1.00 m connects the point masses shown. Assigning x = 0 at the location of the 10-kg mass, where must you locate a fulcrum (support point) so that the object balances (does not tip) with gravity? In other words, where is the c.m. of this rod? 1/11/165Oregon State University PH 212, Class 4 1.x = 0.333 m 2.x = 0.500 m 3.x = 0.667 m 4.x = 0.750 m 5.None of the above. 10.0 kg30.0 kg

6. 1/11/16Oregon State University PH 212, Class 46 Moment of Inertia (I) So how do we understand the baseball bat situation—where the bat’s resistance to angular acceleration depends on where the axis of rotation is located? Is an object’s “angular inertia” (called its moment of inertia) also the sum of each bit of mass times its distance from the rotation axis? In other words, does an object double its resistance to angular acceleration when you double the average distance of all its mass from the point you’re trying to rotate it around? No. The resistance quadruples. 10.0 kg

7. 1/11/16Oregon State University PH 212, Class 47 The location dependency is the square of the distance. For discrete point masses: I =  1 n  (m i ·r i 2 ) For most objects—anything with a continuous mass distribution—I is an integral I =  1 n  r 2 dm See page 318 in your textbook for the I values of common object shapes and axes of rotation.

8. 1/11/16Oregon State University PH 212, Class 48 Again, the 1.00-m rod here is massless. What’s the moment of inertia of this object for an axis of rotation (out of the board) located at the rod’s midpoint? 10.0 kg30.0 kg

9. 1/11/16Oregon State University PH 212, Class 49 You have an ordinary meter stick. Now suppose you attach one more bit of mass (a “point mass”), in one of the following configurations. Which will result in the greatest total I about the left (x = 0.00) end of the stick? 1.1 kg at x = 1.00 2.2 kg at x = 0.75 3.3 kg at x = 0.50 4.None of the above.

10. 1/11/16Oregon State University PH 212, Class 410 The Parallel Axis Theorem There is a fairly simple way to get many values of I. If you know the I value as measured around an axis through the object’s c.m., then the I of that object as measured around any other axis parallel to the c.m. axis is given by: I p.axis = I c.m. + Md 2 where M is object’s entire mass, and d is the straight-line distance between the two axes. Use a uniform object—such as a simple meter stick—for an example… Notice what this tells us: The moment of inertia about an axis through the c.m. is the minimum I value for any rotation axis parallel to that axis. The “balancing” nature of the c.m. does come into play with I (just not with the same calculation as for the c.m. itself).