Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

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2 Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection

3 Translational and Rotational Kinematics Table 10.1, p.97

4 In Sum.

5 Uniform Circular Motion Motion in a circle at constant angular speed. ω: angular velocity (rad/s)

6 Rotation Angle The rotation angle is the ratio of arc length to radius of curvature. For a given angle, the greater the radius, the greater the arc length. θ Δ Δs r Δθ rotation angle Δs arc length r radius

7 How many radians in 360 0? Ans: π radians Consider a circle with radius r. θ s/r θ πr /r θ π r How many degrees in 1 radian? s? s C πr π rad rad / π

8 θ r 4 m s A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of What is the distance, d, that the box moves? d 43 o 43 o x1 43 o x.75rad π 360 o First: How many radians is 43 0?

9 θ r 4 m rad s d A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of.75 rad. What is the distance, d, that the box moves? s r θ d r θ d 4m (0.75 rad) d 3 m rad d 3 m Radian is dimensionless and is dropped!

10 Angular Velocity ω: angular velocity (rad/s) θ ω d dt ( / ) ds r dt 1 ds rdt Δt v v ωr r

11 Angular & Tangential Velocity ω: angular velocity (rad/s) v: tangential velocity (m/s) θ d ω dt v r v ωr ω is the same for every point & v varies with r.

12 Problem Calculate the angular velocity of a m radius car tire when the car travels at a constant speed of 5.0 m/s. ω v r 5 m/ s.5m 50 rad / s Insert rad for angular speed. Keep it to define units.

13 Angular & Tangential Velocity ω: angular velocity (rad/s) v: tangential velocity (m/s) θ d ω dt v r v ωr ω is the same for every point & v varies with r.

14 Tangential and Angular a t dv dt Acceleration ω α d dt ( ω ) d r dt d r dt ω at αr

15 Centripetal and Tangential Accelerations at αr a c v r ( ωr) r ac ω r (r constant)

16 Total Acceleration r t a a + a a-ar+aθˆ ˆ r θ a a ω r v / r r c a a αr dv/ dt θ t

17 Angular Tangential Bike A bike wheel with a radius of 0.5 m undergoes a constant angular acceleration of.50 rad/s. The initial angular speed of the wheel is 5.00 rad/s. After 4.00 s a) What angle has the wheel turned through? Givens: b) What is the final angular speed? ω0 c) What is the final tangential speed of the bike? α d) How far did the bike travel? 1 1 Δ θ ω0t+ αt 5 rad / s4s +.5 rad / s (4 s) 40rad ω ω + αt 5 rad / s +.5 rad / s 4s 15 rad / s f 0 v ωr 15 rad/ s.5m 3.75 m/ s d Δ sδ θ r 40 rad.5m 10m 5 rad / s.5 rad / s t 4s

18 NEW: Rotational Inertia The resistance of an object to rotate. The further away the mass is from the axis of rotation, the greater the rotational inertia.

19 Rotational Inertia Depends on the axis.

20 Which has greater Rotational Inertia? (Both have Same m, R) ICM mr WHY? Mass further from axis of rotation. ICM 1 mr

21 Calculate Rotational Inertia For a system of point particles: I mr I i i i Where r is the distance to the axis of rotation. SI units are kg. m

22 Calculate: Moment of Inertia of a Rigid Object I rdm R dm The trick is to write dm in terms of the density: dm ρ dv Divide the cylinder into concentric shells with radius r, thickness dr and length L: R 1 4 I rdm r ( πρlrdr) πρlr 0 But ρ M / π R L dv π rl dr I z 1 MR

23 Moments of Inertia of Various Rigid Objects

24 Superposition of Inertia The Parallel Axis Th m Superposition: Inertia ADD I I i The theorem states I I CM + MD I is about any axis parallel to the axis through the center of mass of the object I CM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis

25 Moment of Inertia for a Rod Rotating Around One End The moment of inertia of the rod about its center is ICM 1 1 ML The position of the CM is D½ L I ICM + MD Therefore, L 1 1 I ML + M ML 1 3 It is easier to rotate a rod about its center than about an end.

26 Rotational Inertia Which reaches the bottom first? (Same mass and radius)

27 Why Solid Cylinder? Less Rotational Inertia! ICM mr ICM 1 mr

28 Rolling The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid The green line shows the path of the center of mass of the object which moves in linear motion.

29 Rolling without Slipping All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity v CM, and the point P moves with a velocity v CM. (Why?) Fig. 10.8, p.317

30 For pure rolling motion, (no slipping) as the cylinder rotates through an angle θ, its center moves a linear distance s Rθ with speed v CM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.

31 The motion of a rolling object can be modeled as a combination of pure translation and pure rotation. Translation: CM moves with v CM. Rotation: All points rotate about P with angular speed ω. Fig. 10.9, p.318

32 Rolling Without Slipping Kinetic Energy K 1 P I ω Use Parallel-axis Th : I I + MR m P CM 1 K ( ICM + MR ) ω 1 1 K I ω + mv CM CM The total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.

33 Rolling Without Slipping Friction If an object rolls without slipping then the frictional force that causes the rotation is a static force. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. No energy is dissipated and Mechanical Energy is Conserved. If the object slips, then friction is not static, does work on the object and dissipates energy.

34 Rolling Without Slipping Conservation of Energy vcm 10 7 gh E i E f 1 1 Mgh MvCM + ICMω 1 1 vcm MvCM + ICM ( ) R 1 1 v ( )( CM Mv + MR ) 5 R MvCM + MvCM Mv 5 10 CM CM

35 Rolling Without Slipping Conservation of Energy Compare to Slipping Only: What about little ball big ball? vcm 10 7 gh E i Mgh vcm E 1 f Mv gh Some of the gravitational potential energy goes into rolling the sphere so the translational velocity of the cm is less when rolling!

36 Does ω change while in flight? Rolling Without Slipping Conservation of Energy Compare to Slipping Only: Rolling down: vcm 10 7 gh E i Mgh vcm E 1 f Mv gh What about the acceleration down the incline?

37 10 Pre Lab

38 Total Kinetic Energy A 1.0-kg wheel in the form of a solid disk rolls without slipping along a horizontal surface with a speed of 6.0 m/s. What is the total kinetic energy of the wheel? 1 1 Ktotal Ktrans + Krot mv + Iω v ( )( ) mv + mr r 1 1 mv + mv (1 )(6 / ) 4 mv 4 kg m s 7J What I to use?

39 Why Solid Cylinder? PROVE IT! ICM mr ICM 1 mr

40 Total Energy E KE + KE + PE trans rot 1 1 mgh mv + Iω v mv + I ( ) r v mgh 0 m+ I / r

41 General equation for the total final velocity at the end of the ramp: v mgh m+ I / r v gh 0 v 4 3 gh 0 Solid Disk has the greatest velocity at the bottom of the ramp! Note: the velocity is independent of the radius!

42 Torque: Causes Rotations τ Frsinφ Fd lever arm: d rsinφ The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force The horizontal component of F (F cos φ) has no tendency to produce a rotation

43 Torque: Causes Rotations τ Frsinφ Fd The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.

44 Torque Is there a difference in torque? (Ignore the mass of the rope) NO! In either case, the lever arm is the same! What is it? 3m

45 Newton s 1 st Law for Rotation If the sum of the torques is zero, the system is in rotational equilibrium. τ boy 500N 1.5m 750Nm τ 0 τ girl 50N 3m + 750Nm

46 Newton s nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ I α Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

47 Net Torque m 4.3 a) τ? b) α? kg R.314m τ Fr+ Fr 1 ( + 90N 15 N).314m 11Nm τ τ α I 1 mr 11Nm 1/ 4.3kg(.314m) 9. rad / s

48 Net Torque: You Try It! τ Frsinφ Fd τ Fd 1 1+ Fd ( 0 N)(.5 m) + (35 N)(1.10msin60) OR ( 0 N)(.5 m) + (35Ncos30)(1.10 m) F and d must be mutually perpendicular! Nm CCW

49 A 50 N m torque acts on a wheel of moment of inertia 150 kg m. If the wheel starts from rest, how long will it take the wheel to make a quarter turn (90 degrees)? 1 Use : Δ θ ω0t + αt τ Iα α τ 1 Δθ Δθ Δ θ ω0t+ αt t α τ / I I t π /rad 50 N m/150kg m 3.1s

50 Rolling Ball Problem A tennis ball starts from rest and rolls without slipping down the hill. Treat the ball as a solid sphere. Find the range x. R v t vcosθ t 1 1 x mgh mvcm + ICMω 1 1 ( v 10 )( ) mv + Mr v gh 5 r v sinθ Δ y vt 0 vsinθ t gt t y + at y g v R vt x vcosθ t sinθ cosθ (10 / 7 gh)sinθ cosθ g g 10h sin θ R.5m 7g Does ω change while in flight?

51 A meter stick is attached at one end (the zero mark) and is free to rotate on a horizontal, frictionless table. A particle of mass kg is shot at the meter stick with initial speed 3.00 m/s, as shown. The particle strikes and sticks to the meter stick at the 75.0-cm mark. The meter stick has a mass kg. a) Calculate the rotational inertia and the center of mass of the stick/particle system. b) Calculate the angular speed of the system just after the particle hits and sticks to the meter stick. c) How long does it take for the system to make one complete revolution?

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