Understanding Torque and Rotational Dynamics Lecture

1. Lecture 18 Goals: • Define and analyze torque • Introduce the cross product • Relate rotational dynamics to torque • Discuss work and work energy theorem with respect to rotational motion • Specify rolling motion (center of mass velocity to angular velocity • So what causes rotation?

2. So what forces make things rotate? F F F • Which of these scenarios will cause the bar to spin? Fixed rotation axis. A F B C F D F F F E

3. Net external torques cause objects to spin • An external force (or forces) properly placed induced changes in the angular velocity. This action is defined to be a “torque” • A force vector or a component of a force vector whose “line of action” passes through the axis of rotation provides no torque • A force vector or the component of a force vector whose “line of action” does NOT pass through the axis of rotation provides torque • The exact position where the force is applied matters. Always make sure the force vector’s line of action contacts the point at which the force is applied.

4. Net external torques cause objects to spin q F r • Torque increases proportionately with increasing force • Only components perpendicular to r vector yields torque • Torque increases proportionately with increasing distance from the axis of rotation. • This is the magnitude of the torque • q is the angle between the radius and the force vector • Use Right Hand Rule for sign

5. Force vector line of action must pass through contact point • Force vector cannot be moved anywhere • Just along line of action q F r r line of action q F

6. Resolving force vector into components is also valid F = Fr + Ft F t F r F t = Fsin q • Key point: Vector line of action must pass through contact point (point to which force is applied) r line of action q Fr F

7. Exercise Torque L F F L axis case 1 case 2 • Case 1 • Case 2 • Same • In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. • Remember torque requires F, rand sin q or the tangential force component times perpendicular distance

8. Torque is constant along the line of action q L p-q F F L axis p-q case 1 case 2 • Even though case 2 has a much larger radius vector the torque remains constant. • Case 1 : t = L F • Case 2 : t = r F sin q = F r sin q • Notice that the sin q = sin (p-q) = - cos p sin q = sin q • Case 2 : t = r F sin q = F r sin q = F r sin (p-q) = F L Here q = 135° and |r| = 2½ L so r sin q = 2-1/2 (2½ L ) = L

9. Torque, like w, is a vector quantity r sin q Ftangential line of action r q F F F Fradial • Magnitude is given by (1) |r| |F| sin q (2) |Ftangential | |r| (3) |F| |rperpendicular to line of action | • Direction is parallel to the axis of rotation with respect to the “right hand rule” • Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m

10. Torque can also be calculated with the vector cross product • The vector cross product is just a definition

11. Torque and Newton’s 2nd Law • Applying Newton’s second law

12. Example: Wheel And Rope • A solid 16.0 kg wheel with radius r = 0.50 mrotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest, the rope is pulled such that it has a constant tangential force of F = 8 N. How many revolutions has the wheel made after 10 seconds? F r

13. Example: Wheel And Rope F r • m = 16.0 kg radius r = 0.50 m • F = 8 N for 10 seconds Constant F  Constant t  constant a Isolid disk = ½ mr2 = 2 kg m2 I a = t = r F  a = 4 Nm/ 2 kg m2 = 2 rad/s2 q = q0 + w0 Dt + ½ a Dt2 q - q0 = w0 Dt + ½ a Dt2 Rev = (q - q0) / 2p =( 0 + ½ a Dt2 )/ 2p Rev = (0.5 x 2 x 100) / 6.28 = 16

14. Work F  axis of rotation R dr =Rd d • Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an angular displacement d then ds = r d dW = FTangential dr = Ft ds dW = (Ft r)d • dW =  d (and with a constant torque) • We can integrate this to find: W =  = t (qf - qi)

15. 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.

16. Work & Kinetic Energy: • Recall the Work Kinetic-Energy Theorem: K = WNETor WEXT • This is true in general, and hence applies to rotational motion as well as linear motion. • So for an object that rotates about a fixed axis:

17. Example: Wheel And Rope • A solid 16 kg wheel with radius r = 0.50 mrotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest, the rope is pulled such that it has a constant tangential force of F = 8 N. What is the angular velocity after 16 revolutions ? F r

18. Example: Wheel And Rope • Mass 16 kg radius r = 0.50 m Isolid disk=½mr2= 2 kg m2 • Constant tangential force of F = 8 N. • Angular velocity after you pull for 32p rad? W = F (xf - xi) = r F Dq =0.5 x 8.0 x 32p J = 402 J DK = (Kf - Ki) = Kf = ½ I w2 = 402 J w = 20 rad/s F r

19. Exercise Work & Energy • Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. • Disk 1, on the left, has a bigger radius, but both have the same mass. Both disks rotate freely around axes though their centers, and start at rest. • Which disk has the biggest angular velocity after the pull? w2 w1 W =  =F d = ½ I w2 Smaller I bigger w (A)Disk 1 (B)Disk 2 (C)Same F F start d finish

20. Home Example: Rotating Rod L m • A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? (B) its initial angular acceleration ? (C) initial linear acceleration of its free end ?

21. Example: Rotating Rod L m • A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (B) its initial angular acceleration ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! S F = 0 occurs only at the hinge buttz = I az = r F sin 90° at the center of mass and (ICM + m(L/2)2) az = (L/2) mg and solve foraz mg

22. Example: Rotating Rod L m • A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (C)initial linear acceleration of its free end ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! a = a L mg

23. Example: Rotating Rod • A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and use the Work-Energy Theorem 2. The hinge changes everything! L W = mgh = ½ I w2 m W = mgL/2 = ½ (ICM + m (L/2)2) w2 and solve forw mg L/2 mg

24. Connection with CM motion • If an object of mass M is moving linearly at velocity VCMwithout rotating then its kinetic energy is • If an object of moment of inertia ICMis rotating in place about its center of mass at angular velocityw then its kinetic energy is • What if the object is both moving linearly and rotating?

25. Rolling Motion • Now consider a cylinder rolling at a constant speed. VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

26. Rolling Motion • Again consider a cylinder rolling at a constant speed. 2VCM CM VCM

27. Rolling Motion • Now consider a cylinder rolling at a constant speed. VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

28. Motion Both with |vt| = |vCM | Rotation only vt = wR Sliding only 2VCM VCM CM CM CM VCM • Again consider a cylinder rolling at a constant speed.

29. For Tuesday • Read though 11.3