1. PHYSICS 111 Rotational Momentum and Conservation of Energy

2. Equations:  KE rotation = 1/ 2 I ω  I = mr 2  Ke rotating and translation = 1/2mv^2 ( 1+ I /mr 2 )  W (work) = t ( torque) (Delta ɵ )  W = τΔθ = Δ KE rotation  ᾠ = V final / r  Total E rotation and translation =.5 mv^2 +.5I ᾠ ^2 + mgh  L ( angular momentum) = I ω = p (translational momentum) * r( perpendicular)  P = mv  If angular momentum is conserved = Ii * ω i = If * ω f

3. Concept Checker:  A cylinder is rolling without slipping down an inclined plane. The friction at the contact point P is  1. Static and points up the inclined plane.  2. Static and points down the inclined plane.  3. Kinetic and points up the inclined plane.  4. Kinetic and points down the inclined plane.  5. Zero because it is rolling without slipping.

4. Answer:  Answer 1. The friction at the contact point P is static and points up the inclined plane. This friction produces a torque about the center of mass that points into the plane of the figure. This torque produces an angular acceleration into the plane, increasing the angular speed as the cylinder rolls down.

5. Last Concept  A long narrow uniform stick lies motionless on ice (assume the ice provides a frictionless surface). The center of mass of the stick is the same as the geometric center (at the midpoint of the stick). A puck (with putty on one side) slides without spinning on the ice toward the stick, hits one end of the stick, and attaches to it. Which quantities are constant?  1. Angular momentum of puck about center of mass of stick.  2. Momentum of stick and ball.  3. Angular momentum of stick and ball about any point.  4. 1 and 3  5. 2 and 3  6. 1 and 2

6. Solution:  (2) and (3) are correct. There are no external forces acting on this system so the momentum of the center of mass is constant (1).  There are no external torques acting on the system so the angular momentum of the system about any point is constant (3).  However there is a collision force acting on the puck, so the torque about the center of the mass of the stick on the puck is non-zero, hence the angular momentum of puck about center of mass of stick is not constant.  The mechanical energy is not constant because the collision between the puck and stick is inelastic.

7. Concept Checker  A certain star, of mass m and radius r, is rotating with a rotational velocity ω. After the star collapses, it has the same mass but with a much smaller radius. Which statement below is true?  A) The star's moment of inertia I has decreased, and its angular momentum L has increased.  B) The star's moment of inertia I has decreased, and its angular velocity ω has decreased.  C) The star's moment of inertia I remains constant, and its angular momentum L has increased.  D) The star's angular momentum L remains constant, and its rotational kinetic energy has decreased.  E) The star's angular momentum L remains constant, and its rotational kinetic energy has increased.

8. Answer:  The correct answer is e.  According to conservation of angular momentum, the angular momentum L of the star remains constant, so when its moment of inertia I decreases (due to the decreased radius), its angular velocity ω goes up proportionally  L initial = L final  I ᾠ final = I ᾠ initial

9. Practice Problem:  A skater spins with an angular speed of 12.0 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia from to.  Calculate her initial and final rotational kinetic energy.  How do you account for the change in kinetic energy?

10. Solution:  Her initial energy is:  KE =.5 I ᾠ ^2 =.5 (41)(12) = 2952 Joules  Her angular momentum is:  L = I ᾠ = 41 (12) = 492 kg* m^2/s  You know that angular momentum is conserved. Meaning that Lfinal and L inital = 492 kg* m^2/s  Then you need to solve for wfinal  L = I ᾠ  492 = 36 ( ᾠ f ) ᾠ = 13.67 rad/s  So her final rotational kinetic energy is:  Ke=.5I ᾠ ^2 =.5 (36)(13.67) = 3364 Joules Why? Because it took her energy to pull her arms in

11. Practice Problem:  A 65.0 kg woman stands at the rim of a horizontal turntable that has a moment of inertia of 400 kg ∙ m² and a radius of 3.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 2.0 m/s relative to the Earth.  A) In what direction and with what angular speed does the turntable rotate?

12. Solution: M= 65 kg I = 400 kg*m^2 R = 3 m V = 2 m/s  Recognize that this problem focuses on the law of conservation of angular momentum. So…We know that initially  L=0 because we are told that the turntable is initially at rest. Letting the subscript w stand for the woman and t for the table, we can write: 0 = Iw * ω w + It * ω t  0 =mr^2(v/r) + Itwt  0 =65 *3^2( -2/3)+400*wt  Wt =.98 rad/ s and since it is positive, means ccw