Rotational equilibrium & dynamics

1. Rotational equilibrium & dynamics PHY231

2. Extended objects • Until now we have treated (almost) all the objects as point-like objects • For point-like objects, only the magnitude and direction of the force matters • For extended objects, the place where the force is applied also matters!! PHY231

3. Book on a table • Lay a book on a table, push on its sides with your hands • Net force from your hands is Fnet=F-F=0 • Case1: Press in the middle • Case2: Press at extremities F F -F -F • Book rotates !? • No motion • is F=ma wrong ??

4. Were Newton’s laws wrong? • Newton’s laws are still correct • They describe the translational motion of the object’s center of mass (also called center of gravity) • We will see next how to describe rotational motion for objects ! PHY231

5. Center of mass (or gravity) • The (2-D) position of the object’s center of mass is defined as • With • It is different than the object’s geometric center • If you hang the object at rCM it won’t move ! (demo) • Applying Newton’s laws on the object as a whole, seems like all the mass is concentrated at rCM PHY231

6. example • Two masses are connected by a very light rod. The mass m2 is much heavier than m1. Where is the center of mass located? m1 m2 • In the middle of the rod • Closer to m1 • Closer to m2 • Impossible to say PHY231

7. example • Two masses are connected by a very light rod. The mass m2 is much heavier than m1. Where is the center of mass located? L/2 -L/2 0 x m1 m2 • In the middle of he rod • Closer to m1 • Closer to m2 • Impossible to say PHY231

8. y Tray • A waiter is ready to lift his tray with three bottles on it (1, 2 and 3) • The positions (in inch) and masses of the bottles are: • At what position should he put his hand to lift the tray? 1 x 2 3 A) B) C) PHY231

9. y • We apply Definition for CM 1 x 2 3 A) B) C) PHY231

10. Torque • Forces cause acceleration • Torques cause angular acceleration • The torque is linked to forces and position were they are applied • SI unit: N.m • Torque is calculated with respect to a reference axis (Here axis normal to the slide and passing through O) • The net torque is the sum of the torques from all forces acting on the object PHY231

11. Torque • Only the component of the force perpendicular to the position vector and to the axis of rotation contributes to the torque • Torque is positive if “motion” going from r to F is counterclockwise (as in example above) • Torque is negative if it is clockwise This component of F produces all the torque This component of F doesn’t do anything PHY231

12. Recipe for lots of torque • For lots of torque you need • Apply the force far from the pivot point • Apply a large force • Apply the force perpendicularly • As an example • Applying a 200 N force 1 m away from the pivot is the same than applying 100 N but 2 m away PHY231

13. example • You push up on the door with a 100 N force at an angle of q=30º and 2.0 m from the hinges PHY231

14. Zero net torque • Jane tries to push the door down toward Kate by applying a 200 N force at 30º and 1.0 m away from the hinges • How much force does Kate need to apply to cancel the torque from Jane if she pushes up, 2.0 m away from the hinges and perpendicularly to the door? • 50 N • 100 N • 200 N • 400 N 2.0m 30º 1.0m 200N PHY231

15. Zero net torque • zero net torque means • 50 N • 100 N • 200 N • 400 N 2.0m 30º 1.0m 200N PHY231

16. Torque from an object’s own weight xj • Torque for a rod made of smaller rods • Everything look like the entire weigh is at xCM!! O tj=xj*mjg O xCM tj=xCM*Mg

17. Choice of axis • The torque for any force considered depends on the axis that is chosen as reference. BE CAREFUL ! • For example, the torques from gravity on a uniform rod of length L for two different axis of rotations (indicated by a black dot ) O O xCM=L/2 Fg Fg PHY231

18. Balancing act • Three masses are hanging on a swing. You have a 100 g in your hand. Can you hang it somewhere to balance the situation? • A • B • C • Impossible A C B 1000g 500g 100g PHY231

19. A • B • C • Impossible A C B 1000g 500g 100g PHY231

20. Balancing act • At what distance x from the center should the man M=75 kg seat so there is no net torque on the plank ? (woman has m=55 kg) • 0.37 m • 0.75 m • 1.0 m • 1.5 m PHY231

21. Forces on plank: • Weight from woman • Weight from man • Weight from plank • Reaction force at pivot O • 0.37 m • 0.75 m • 1.0 m • 1.5 m PHY231

22. Mechanical equilibrium for an object • An object is in mechanical equilibrium if: • The net external force is zero • The net external torque is zero • The first condition means the translational acceleration is zero • The second condition means the rotational acceleration is zero PHY231

23. Mechanical equilibrium • A 50 kg uniform beam is attached to a wall with a hinge and supported by a string connecting its extremity to the wall • The angle between the string and the beam is q=40º • What is the tension in the string and reaction force from the wall if the system is in mechanical equilibrium? y x • 221 N • 381 N • 490 N • Can’t tell PHY231

24. 221 N • 381 N • 490 N • Can’t tell L R T G

25. Angular acceleration • If the net torque is not equal to zero, the object experiences some angular acceleration. Let’s go back to a simple mass rotating: • Angular acceleration is proportional to the torque • Acceleration inversely proportional to the moment of inertia = mr2 PHY231

26. Moment of inertia - I • Let’s now consider a rigidobject made of a many masses (a disk for example) • All the points have the same angular velocity and angular acceleration (rigid object) • The net torque is given by the sum on all the masses • I is called moment of inertia and depends on the choice of axis of rotation ! Be careful !! PHY231

27. Parallel notions • There is again a direct parallel between governing equation for translational motion (for center of mass) and for rotational motion • Acceleration proportional to force • The larger the mass, the smaller the acceleration • Angular acceleration proportional to torque • The larger the moment of inertia, the smaller the angular acceleration PHY231

28. Blue VS red • Two batons (B: Blue, R:Red) • Same total mass • Same size • Hold it in the baton in the middle • Seems like twisting the red baton is much easier, why? PHY231

29. Blue VS red • Two batons (B: Blue, R:Red) • Same total mass • Same size (length L) • Not the same moment of inertia I !! • Batons’ mass distributions are different • Blue baton, most of the mass is at extremities • Red baton, most mass is at the center m m m m • Hard to rotate • Easy to rotate

30. Partial summary • Center of mass • Torque depends on rot. axis • Moment of inertia depends on rot. axis • Torque/Angular acceleration • Mechanical equilibirum

31. I for some objects with uniform composition PHY231

32. Pivoting rods • We have three rods (A, B and C) of uniform composition • A has mass M and length L • B has mass 3M and length L • C has mass 2M and length 2L • All are hold at angle of q at rest and released at the same time, which one has the smallest amplitude of acceleration just after release? • Rod A • Rod B • Rod C q PHY231

33. Pivoting rods • A has mass M and length L • B has mass 3M and length L • C has mass 2M and length 2L • Rod A • Rod B • Rod C Mg*cosq q Mg q

34. Rotational kinetic energy • A particle of mass m moving through space with speed v has kinetic energy • A rigid object in rotation is made of many particles moving with same angular velocity w and the rotational energy is • Notice (again) that it is easily obtained by replacing • velocity v by angular velocity w • Mass m by moment of inertia I

35. Rotational kinetic energy - proof • KEr is given by summing the KE for all the particles composing the rigid body PHY231

36. KEr - example • A solid cylinder of uniform composition is rotating at a constant angular velocity of 10 rad/s around its axis. The cylinder has 10 cm radius and is weighting 40 kg. • What is the KEr of the cylinder? • 10 J • 400 J • 4000 J • Wha ? PHY231

37. KEr - example • A solid cylinder of uniform composition is rotating at a constant angular velocity of 10 rad/s around its axis. The cylinder has 10 cm radius and is weighting 40 kg. • What is the KEr of the cylinder? • 10 J • 400 J • 4000 J • Wha ? PHY231

38. Mechanical energy • Since we have a new form of energy, we should revisit the work-energy theorem to include it • Where we distinguish between KE of translation and rotation. In absence of non-conservative forces, the total mechanical energy is conserved

39. Rolling down • A solid sphere and a hollow cylinder of same mass and same radius are initially at rest and released from the top of an incline simultaneously • They are both rolling without slipping • Which one as the largest KEr at the bottom? • Sphere • Cylinder • Same KEr

40. Rolling down • Sphere • Cylinder • Same KEr PHY231

41. The pulling mass • A string that is wrapped around a solid cylinder (R=0.50 m and M=20 kg) and a mass m=10 kg is attached at its extremity. • The cylinder is attached and can rotate freely around its axis. The system is initially at rest and the mass is released and pulls down on the string. • The string has exactly 2.0 m wrapped around the axis and the friction with the axis of rotation is negligible. • What is the rotational kinetic energy of the cylinder after all the string has been unwrapped?

42. We will solve this problem using two different methods and arrive at the same result • Force-Torque + kinematics • Energy conservation • Method 1: Force-Torque + kinematics >0 -T T >0 -mg

43. Method 1: Force-Torque + kinematics (2nd part) PHY231

44. Method 2: Energy conservation KEr,i=0 KEr,f KEt,i=0 GPEi=mgh KEt,f GPEf=0

45. Angular momentum • As for translational motion, there is a notion of momentum for rotational motion • We found the notion of angular momentum ! PHY231

46. Conservation of L • If there is no net external torque acting on a system we have • In absence of net external torque, the angular momentum of a system is conserved PHY231

47. The spinning stool • A man is spinning at wi=2.0*π rad/s on a frictionless stool with his arms extended and holding weights. • He can be seen as a uniform cylinder of M=64 kg and radius R=50 cm. The weights (m=4.0kg each) can be considered point-like and rotating at r=1.0 m radius. • The man tucks his arms in so that they are very close to the axis of rotation, what is the angular velocity of the man then? • Smaller w • Same w • Larger w PHY231

48. Smaller w • Same w • Larger w PHY231

49. Rotational motion • Ang. displacement • Ang. velocity • Ang. accele. • Moment of Inertia • Torque • Rotational KE • Ang. momentum • Translational motion • Displacement • Velocity • Acceleration • Mass • Force • Kinetic energy • Momentum