1. Rigid Body Dynamics chapter 10 continues
around and around we go …

2. Rigid Body Rotation
Rotation Axis
Moment of Inertia:

3. CORRESPONDENCE:

4. Rotational Kinetic Energy
We there have an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ Iw2)
Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object
The units of rotational kinetic energy are also Joules (J)

5. Important Concept: Moment of Inertia
The definition of moment of inertia is
The dimensions of moment of inertia are ML2 and its SI units are kg.m2
We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass Dmi

6. Moment of Inertia, cont
We can rewrite the expression for I in terms of Dm
With the small volume segment assumption,
If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

7. Question – WHAT IS THE MOMENT OF INERTIA OF THIS OBJECT??
WHICH AXIS ?

8. Let’s Look at the possibilities
c d

9. Two balls with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure P For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is
I = L2, where = mM/(m + M).

10. Remember the Various Densities
Volumetric Mass Density –> mass per unit volume: r = m / V
Face Mass Density –> mass per unit thickness of a sheet of uniform thickness, t : s = rt
Linear Mass Density –> mass per unit length of a rod of uniform cross-sectional area: l = m / L = rA

11. Moment of Inertia of a Uniform Thin Hoop
Since this is a thin hoop, all mass elements are the same distance from the center

12. Moment of Inertia of a Uniform Rigid Rod
The shaded area has a mass
dm = l dx
Then the moment of inertia is

13. Moment of Inertia of a Uniform Solid Cylinder
Divide the cylinder into concentric shells with radius r, thickness dr and length L
Then for I
dV=L(2prdr)

14. Moments of Inertia of Various Rigid Objects

15. Parallel-Axis Theorem
In the previous examples, the axis of rotation coincided with the axis of symmetry of the object
For an arbitrary axis, the parallel-axis theorem often simplifies calculations
The theorem states I = ICM + MD 2
I is about any axis parallel to the axis through the center of mass of the object
ICM is about the axis through the center of mass
D is the distance from the center of mass axis to the arbitrary axis

16. Howcome?? The new axis is parallel to the old axis of rotation.
Assume that the object rotates about an axis parallel to the z axis.
The new axis is parallel to the original axis.  mi L ri
Not in
same
plane

17. From the top:
 
NEW OLD
 mi L ri
Center of Mass

18. Remember the Center of Mass?
Since for our problem the sum is ABOUT the center of
mass, rCM must be zero

19. So:
ZERO
Inew = ICM + ML2
QED

20. Parallel-Axis Theorem Example
The axis of rotation goes through O
The axis through the center of mass is shown
The moment of inertia about the axis through O would be IO = ICM + MD 2

21. Moment of Inertia for a Rod Rotating Around One End
The moment of inertia of the rod about its center is
D is ½ L
Therefore,

22. Many machines employ cams for various purposes, such as opening and closing valves. In Figure P10.29, the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed about the axis of the

23. Torque (Another Vector): tF

24. Torque
Torque, t, is the tendency of a force to rotate an object about some axis
Torque is a vector
t = r F sin f = Fd = rXF
F is the force
f is the angle the force makes with the horizontal
d is the moment arm (or lever arm)

25. More Torqueing
The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force
d = r sin Φ

26. Torque
The horizontal component of F (F cos f) has no tendency to produce a rotation
Torque will have direction
If the turning tendency of the force is counterclockwise, the torque will be positive
If the turning tendency is clockwise, the torque will be negative
Right Hand Screw Rule

27. Net Torque
The force F1 will tend to cause a counterclockwise rotation about O
The force F2 will tend to cause a clockwise rotation about O
St = t1 + t2 = F1d1 – F2d2

28. Torque vs. Force Forces can cause a change in linear motion
Described by Newton’s Second Law
Forces can cause a change in rotational motion
The effectiveness of this change depends on the force and the moment arm
The change in rotational motion depends on the torque

29. Torque Units The SI units of torque are N.m
Although torque is a force multiplied by a distance, it is very different from work and energy
The units for torque are reported in N.m and not changed to Joules

30. Torque and Angular Acceleration
Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force Ft
The tangential force provides a tangential acceleration:
Ft = mat
Ftr=matr=m(ar)r=mr2a
t=Ia

31. More Associations:
How about that?

32. SO?
Worry about concepts.
Don’t worry about too many new “formulas”.

33. Torque and Angular Acceleration, Extended
Consider the object consists of an infinite number of mass elements dm of infinitesimal size
Each mass element rotates in a circle about the origin, O
Each mass element has a tangential acceleration

34. Torque and Angular Acceleration, Extended cont.
From Newton’s Second Law
dFt = (dm) at
The torque associated with the force and using the angular acceleration gives
dt = r dFt = atr dm = ar 2 dm
Finding the net torque
This becomes St = Ia

35. Torque and Angular Acceleration, Extended final
This is the same relationship that applied to a particle
The result also applies when the forces have radial components
The line of action of the radial component must pass through the axis of rotation
These components will produce zero torque about the axis

36. Torque and Angular Acceleration, Wheel Example
The wheel is rotating and so we apply St = Ia
The tension supplies the tangential force
The mass is moving in a straight line, so apply Newton’s Second Law
SFy = may = mg - T

37. Problem
Find the net torque on the wheel in Figure P10.31 about the axle through O if a = 10.0 cm and b = 25.0 cm.

38. Anudder one
An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel, as shown in Figure P The flywheel is a solid disk with a mass of 80.0 kg and a diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of m. If the tension in the upper (taut) segment of the belt is 135 N and the flywheel has a clockwise angular acceleration of 1.67 rad/s2, find the tension in the lower (slack) segment of the belt.

39. Torque and Angular Acceleration, Multi-body Ex., 1
Both masses move in linear directions, so apply Newton’s Second Law
Both pulleys rotate, so apply the torque equation

40. Torque and Angular Acceleration, Multi-body Ex., 2
The mg and n forces on each pulley act at the axis of rotation and so supply no torque
Apply the appropriate signs for clockwise and counterclockwise rotations in the torque equations

41. 32. The tires of a 1 500-kg car are 0
32. The tires of a kg car are 0.600 m in diameter and the coefficients of friction with the road surface are s = and k = Assuming that the weight is evenly distributed on the four wheels, calculate the maximum torque that can be exerted by the engine on a driving wheel, without spinning the wheel. If you wish, you may assume the car is at rest.

42. Work in Rotational Motion
Find the work done by F on the object as it rotates through an infinitesimal distance ds = r dq
dW = F . d s
= (F sin f) r dq
dW = t dq
The radial component of F
does no work because it is
perpendicular to the
displacement

43. Power in Rotational Motion
The rate at which work is being done in a time interval dt is
This is analogous to P = Fv in a linear system

44. Work-Kinetic Energy Theorem in Rotational Motion
The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object’s rotational kinetic energy

45. Work-Kinetic Energy Theorem, General
The rotational form can be combined with the linear form which indicates the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies

46. Energy in an Atwood Machine, Example
The blocks undergo changes in translational kinetic energy and gravitational potential energy
The pulley undergoes a change in rotational kinetic energy

47. Summary of Useful Equations

48. Rolling Object
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

49. Pure Rolling Motion
In pure rolling motion, an object rolls without slipping
In such a case, there is a simple relationship between its rotational and translational motions

50. Rolling Object, Center of Mass
The velocity of the center of mass is
The acceleration of the center of mass is

51. Rolling Object, Other Points
A point on the rim, P, rotates to various positions such as Q and P ’
At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs

52. Rolling Motion Cont.
Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion

53. Total Kinetic Energy of a Rolling Object
The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass
K = ½ ICM w2 + ½ MvCM2

54. Total Kinetic Energy, Example
Accelerated rolling motion is possible only if friction is present between the sphere and the incline
The friction produces the net torque required for rotation

55. Total Kinetic Energy, Example cont
Despite the friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant
Let U = 0 at the bottom of the plane
Kf + U f = Ki + Ui
Kf = ½ (ICM / R 2) vCM2 + ½ MvCM2
Ui = Mgh
Uf = Ki = 0