1. Chapter 10 Chapter 10 Rotational motion Rotational motion Part 2 Part 2

2. Parallel axis theorem Proof: Rotational inertia about a given axis = Rotational Inertia about a parallel axis that extends trough body’s Center of Mass + Mh 2 h = perpendicular distance between the given axis and axis through COM. R

3. Units: Nm Tangential component, F t : does cause rotation  pulling a door perpendicular to its plane. F t = F sinφ Radial component, F r : does not cause rotation  pulling a door parallel to door’s plane.TorqueTorque: Twist  “Turning action of force F ”.

4. r ┴ : Moment arm of F r : Moment arm of F t Sign: Torque >0 if body rotates counterclockwise. Torque <0 if clockwise rotation. Torque <0 if clockwise rotation. Superposition principle: When several torques act on a body, the net torque is the sum of the individual torques Vector quantity

5. Newton’s second law for rotation Proof: Particle can move only along the circular path  only the tangential component of the force F t (tangent to the circular path) can accelerate the particle along the path.

6. Kinetic energy of rotation Reminder: Angular velocity, ω is the same for all particles within the rotating body. Linear velocity, v of a particle within the rigid body depends on the particle’s distance to the rotation axis (r). Moment of Inertia

7. Rotational Kinetic Energy We must rewrite our statements of conservation of mechanical energy to include KE r Must now allow that (in general): ½ mv 2 +mgh+ ½ I  2 = constant Could also add in e.g. spring PE

8. VII. Work and Rotational kinetic energy TranslationRotation Work-kinetic energy Theorem Work, rotation about fixed axis Work, constant torque Power, rotation about fixed axis Proof:

9. Example - Rotational KE  What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30 o incline? mgh+ ½ mv 2 + ½ I  2 = constant  Is there enough information? 2 m

10. Example - Rotational KE So we have that The velocity of the centre of mass and the tangential speed of the sphere are the same, so we can say that: Rearranging for v f :

11. Example: Conservation of KE r A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he falls off? Roundabout is a disk: Boy has

12. During a certain period of time, the angular position of a swinging door is described by During a certain period of time, the angular position of a swinging door is described by θ= 5.00 + 10.0t + 2.00t 2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s. Solution: (a)

13. During a certain period of time, the angular position of a swinging door is described by θ= 5.00 + 10.0t + 2.00t 2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s. Solution: Solution: (b)

14. A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel? and are two equations in two unknowns

15. The four particles are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. If the system rotates in the xy plane about the z axis with an angular speed of 6.00 rad/s, calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system.

16. (a)

17. In this case, (b)

18. Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm. The thirty-degree angle is unnecessary information.

19. A block of mass m 1 = 2.00 kg and a block of mass m 2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0  as in the Figure above. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

20. For m 1 : For pulley,

21. For m 2 :

22. (a) Add equations for m 1, m 2, and for the pulley:

23. (b)