1. Chapter 8
Rotational Motion

2. Introduction Until now, objects have been treated as point particles
The point particle treatment is the correct way to describe translational motion
The point particle approximation does not allow us to analyze rotational motion
Need to learn how to use Newton’s Laws to describe and analyze rotational motion
The size and shape of the object will have to be taken into account
Introduction

3. Describing Rotational Motion
To deal with rotational motion, rotational quantities need to be defined
Angular position
Angular velocity
Angular acceleration
These will be analogous to the translational (linear) quantities of position, velocity and acceleration
Section 8.1

4. Coordinate System Need to identify the rotation axis
In A, the rod is fixed at one end and the rotational axis is the z- axis
The angle θ is measured with respect to the x-axis
For the CD in B, the z-axis is the axis of rotation
The angular position, θ, is specified by the angle the reference line makes with the x-axis
Section 8.1

5. Radian The end of the rod sweeps out a circle of radius r
Assume the end of the rod travels a distance s along the circular path
At the same time, the rod sweeps out an angle θ
Section 8.1

6. Radian, cont. The distance s and angle θ are related by
θ is measured in radians
Angles can also be measured in degrees
360°= 2 π rad
Both measure one complete circle
Section 8.1

7. Angular Velocity
Angular velocity describes how the angular position is changing with time
Denoted by ω
For some time interval, Δt, the average angular velocity is
Section 8.1

8. Instantaneous Angular Velocity
The instantaneous angular velocity is
The instantaneous angular velocity equals the average angular velocity when ω is constant
SI unit is rad/s
May also see rpm (revolutions / minute)
Section 8.1

9. Angular Velocity, Direction
Since angular velocity is a vector quantity, it must have a direction
If θ increases with time, then ω is positive
Therefore, a counterclockwise rotation corresponds to a positive angular velocity
Clockwise would be negative
Section 8.1

10. Angular Acceleration
Angular acceleration is the rate of change of the angular velocity
Denoted by α
The average angular acceleration is
The instantaneous angular acceleration is
SI units is rad/s²
The instantaneous angular acceleration equals the average angular acceleration when α is constant
Section 8.1

11. Angular Acceleration and Centripetal Acceleration
Angular acceleration and centripetal acceleration are different
As an example, assume a particle is moving in a circle with a constant linear velocity
The particle’s angular position increases at a constant rate, therefore its angular velocity is constant
Its angular acceleration is 0
Since it is moving in a circle, it experiences a centripetal acceleration of ac = v2/ r
This is not zero, even though the angular acceleration is zero
The centripetal acceleration refers to the linear motion of the particle
The angular acceleration is concerned with the related angular motion
Section 8.1

12. Angular and Linear Velocities Compared
When an object is rotating, all the points on the object have the same angular velocity
Makes ω a useful quantity for describing the motion
The linear velocity is not the same for all points
It depends on the distance from the rotational axis
Section 8.1

13. Period of Rotational Motion
One revolution of an object corresponds to 2 π radians
The object will move through ω / 2π complete revolutions each second
The time required to complete one revolution is the period of the motion
Denoted by T with
Section 8.1

14. Connection Between Linear and Angular Velocities
The linear velocity of any point on a rotating object is related to its angular velocity by v = ωr
r is the distance from the rotational axis to the point
Technically, this is the relationship between the speeds, since direction has not been taken into account
When a point is farther from the axis of rotation (rA > rB , for example), then its linear velocity will be greater
vA > vB
The angular velocities of both points are the same
Section 8.1

15. Connection Between Linear and Angular Accelerations
The relationship between linear and angular velocities can be used to determine the relationship between linear and angular accelerations
Analysis indicates the angular acceleration and the linear acceleration of a point a distance r from the rotation axis is given by a = α r
Section 8.1

16. Torque and Newton’s Laws for Rotational Motion
A connection between force and rotational motion is needed
Specifically, how forces give rise to angular accelerations
The approach will be similar to looking at forces and linear motion
Section 8.2

17. Torque
Torque is the product of an applied force and the distance it is applied from the support point
Denoted τ
The point P is called the pivot point
Since the object can rotate around that point
Section 8.2

18. Lever Arm and Torque
The lever arm is the distance between the pivot point and where the force acts
When the force is perpendicular to a line connecting its point of application to the pivot point, the torque is given by τ = F r
Torque in rotational motion is analogous to force in translational motion
Torque is the product of the force and the distance to the pivot point
SI unit is N. m
Section 8.2

19. Torque and Angular Acceleration
Only forces with a component perpendicular to the rod can contribute to the angular acceleration
Newton’s Second Law for rotational motion is written as Στ = I α
I is called the moment of inertia of the object
Section 8.2

20. Newton’s Second Law – Rotational
The Στ is the vector sum of all the torques acting on the object
The moment of inertia, I, plays the same role that mass did in translational motion
For a ball and massless hinged rod, I = m r2
Section 8.2

21. Analogy with Translational Motion
Torque plays the role of force in rotational motion
Torque depends on the magnitude of the force and where the force is applied relative to the pivot point
There may be multiple forces acting on the system, all involving a single pivot point
Section 8.2

22. Analogy with Translational Motion, 2
The motion of inertia enters into rotational motion in the same way that mass enters into translational motion
For an object composed of many pieces of mass located at various distances from the pivot point, the moment of inertia of the object is
The moment of inertia depends on the mass and on how that mass is distributed relative to the axis of rotation
Section 8.2

23. Analogy with Translational Motion, 3
Newton’s Second Law for translational motion can be used to derive Newton’s Second Law for rotational motion
For rotational motion, Newton’s Second Law becomes Στ = I α
Section 8.2

24. Linear and Rotational Comparison
Section 8.2

25. Torque and Lever Arm: Generalized
In general, the force that produces a torque does not have to be applied in a perpendicular direction
Assume the force acts at an angle ϕ with respect to the rod
Only the perpendicular component contributes to the torque
Section 8.2

26. Torque and Lever Arm: Generalized, cont.
The perpendicular component of the applied force is Fapplied sin ϕ
The torque is Fapplied r sin ϕ
This is a general definition of torque
It can be used when the force is not directed perpendicular to the lever arm
When ϕ is 90°, sin ϕ = 1 and Fapplied = F|
When ϕ is 0°, sin ϕ = 0 and Fapplied = 0
A force applied parallel to the lever arm cannot cause an object to rotate
Section 8.2

27. One Way to Think About Torque
Since the perpendicular component of the force is Fapplied sin ϕ, the torque can be expressed as Fapplied r sin ϕ
The angle can be found by extending the radius line beyond the point where the force acts
The angle is between the force and this radius line
Section 8.2

28. Another Way to Think About Torque
You can also use the perpendicular distance from the pivot point to where the force is acting to calculate the torque
rperpendicular = r sin ϕ
This gives the general expression for the lever arm
Therefore,
τ = Fapplied r sin ϕ
Section 8.2

29. Ways to Think About Torques, Final
Using the idea of the perpendicular force or the perpendicular distance (lever arm) will give the same results
For example:
ϕ = 90° gives maximum torque
ϕ = 0° give zero torque
Section 8.2

30. Torque and Direction – Mass at End
For a single rotation axis, the direction of the torque is specified by its sign
A positive torque is one that would produce a counterclockwise rotation
A negative torque would produce a clockwise rotation
Section 8.2

31. Torque and Direction – Distributed Mass
We could imagine breaking the clock hand up into many infinitesimally small pieces and finding the torques on each piece
A more convenient approach is to use the center of gravity of the hand
Section 8.2

32. Center of Gravity
For the purposes of calculating the torque due to the gravitational force, you can assume all the force acts at a single location
The location is called the center of gravity of the object
The center of gravity and the center of mass of an object are usually the same point
Section 8.2

33. Rotational Equilibrium
Equilibrium may include rotational equilibrium
An object can be in equilibrium with regard to both its translation and its rotational motion
Its linear acceleration must be zero and its angular acceleration must be zero
The total force being zero is not sufficient to ensure both accelerations are zero
Section 8.3

34. Example: Equilibrium
The applied forces are equal in magnitude, but opposite in direction
Therefore, ΣF = 0
However, the object is not in equilibrium
The forces produce a net torque on the object
There will be an angular acceleration in the clockwise direction
Section 8.3

35. Rotational Equilibrium, cont.
For an object to be in complete equilibrium, the angular acceleration is required to be zero
Στ = 0
This is a necessary condition for rotational equilibrium
All the torques will be considered to refer to a single axis of rotation
The same ideas can also be applied to multiple axes
Section 8.3

36. Problem Solving Strategy – Equilibrium
Recognize the principle
The object is in translational static equilibrium if its linear acceleration and linear velocity are both zero
The object is in rotational static equilibrium if its angular acceleration and angular velocity are both zero
Sketch the problem
Show the object of interest along with all the forces that act on it
Include a set of coordinate axes
Section 8.3

37. Problem Solving Strategy, 2
Identify the relationships
Find the rotation axis and the pivot point
Calculate the torque from each force
Determine the lever arm for each force
Calculate the magnitude of the force using τ = F r sin ϕ
Determine the sign of the torque
If the force acting alone would produce a counterclockwise rotation, the torque is positive
If the force acting alone would produce a clockwise rotation, the torque is negative
Add the torques from each force to get the total force
Be sure to include the sign of each torque
Section 8.3

38. Problem Solving Strategy, 3
Solve
Solve for the unknowns by applying the condition for rotational equilibrium
Στ = 0
If necessary, apply the condition for translational equilibrium
ΣF = 0
Check
Consider what your answer means
Check that your answer makes sense
Section 8.3

39. Problem Solving Reminders
There are some new considerations to remember
Draw the entire object to show where the forces are acting on it
You can no longer draw the object as a point
Decide where to put the pivot point
There is often a natural choice for the pivot point
Sometimes there can be more than one plausible choice
For an object in rotational equilibrium, any spot may be chosen to be the pivot point without affecting the final answer
To simplify the equations, remember that forces whose lever arms are zero will not contribute to the torque
Section 8.3

40. Rotational Equilibrium Example: Lever
Use rotational equilibrium to find the force needed to just lift the rock
We can assume that the acceleration and the angular acceleration are zero
Also ignore the mass of the lever
mlever << mrock
Section 8.3

41. Lever Example, cont. Three forces are present
The force from the rock on the lever
The force the person applied to the lever
The force the of support on the lever
Choosing this point to be the pivot point produces a zero torque from this force
The force exerted by the person can be less than the weight of the rock
If Lperson > Lrock
The lever will amplify the force exerted by the person
Section 8.3

42. Amplification of Forces in the Ear
The incus is supported by a hinge that acts like a lever
The ratio of the lever arms is about 3
The lever amplifies the forces associated with a sound vibration by the same factor
Section 8.3

43. Example: Tipping a Crate
We can calculate the force that will just cause the crate to tip
When on the verge of tipping, static equilibrium applies
If the person can exert about half the weight of the crate, it will tip
Section 8.3

44. Moment of Inertia
The moment of inertia of an object composed of many pieces of mass is
The moment of inertia of an object depends on its mass and on how this mass is distributed with respect to the rotation axis
The definition can be applied to find the moment of inertia of various objects for any rotational axis
SI unit of moment of inertia is kg · m2
Section 8.4

45. Moment of Inertia, cont.
The value of I depends on the choice of rotation axis
In the two examples, m and L are the same
Their moments of inertia are different due to the difference in rotation axes
Section 8.4

46. Various Moments of Inertia
Section 8.4

47. Rotational Dynamics Newton’s Second Law for a rotating system states
Once the total torque and moment of inertia are found, the angular acceleration can be calculated
Then rotational motion equations can be applied
For constant angular acceleration:
Section 8.5

48. Kinematic Relationships
Section 8.5

49. Example: Real Pulley with Mass
Up to now, we have assumed a massless pulley
Using rotational dynamics, we can deal with real pulleys
The torque on the pulley is due to the tension in the rope
Apply Newton’s Second Laws for translational motion and for rotational motion
Section 8.5

50. Problem Solving Strategy – Rotational Dynamics
Recognize the principle
Find the total torque
Find the moment of inertia
Use Newton’s Second Law to find the angular acceleration
Sketch the problem
Show all the objects of interest
Include all the forces that act on the objects
Include coordinate axes for translational motion
Section 8.5

51. Problem Solving Strategy – Rotational Dynamics, 2
Identify the relationships
Determine the rotation axis and the pivot point for calculating torques on any object that rotates
Find the total torque on the objects that are undergoing rotational motion
These torques will be used in Newton’s Second Law: Στ = Iα
Calculate the sum of the forces acting on the objects that are undergoing linear motion
These will be used in Newton’s Second Law: ΣF = ma
Check for the relationship between linear and rotational accelerations
Section 8.5

52. Problem Solving Strategy, 3
Solve
Use both forms of Newton’s Second Law to solve for the unknown(s)
Check
Consider what your answer means
Be sure your answer makes sense
Section 8.5

53. Example: Motion of a Crate
The crate undergoes translational motion
The pulley undergoes rotational motion
For the pulley:
The tension in the rope supplies the torque
The pulley rotates around its center, so that is a logical axis of rotation
Section 8.5

54. Motion of a Crate Example, cont.
Pulley equation
Στ = - T Rpulley = Ipulley α
The pulley is a disc, so I = ½ mpulley R²pulley
For the crate
Take the +y direction as +
Equation: ΣF = T – mcrate g = mcrate a
Relating the accelerations
a = α Rpulley
Combine the equations and solve
Section 8.5

55. Combined Motions
Many common situations include a combination of rotational and translational motion
Two examples are
A rolling wheel
The motion of a baseball bat
Section 8.6

56. Rolling Wheel
Rolling motion combines translational and rotational motions
Assume the center of the wheel is moving at a constant linear speed v
The point on the edge of the wheel does not move with constant velocity
The point of the wheel in contact with the ground is at rest during the instant it is in contact

57. Rolling Wheel, 2
The wheel undergoes rotational motion about an axis through its center
The axle
The rotational motion is described by an angular velocity, ω
The wheel starts with point 1 in contact with the ground
Section 8.6

58. Rolling Wheel, cont.
When it completes one rotation it has traveled a distance along the ground equal to the circumference of the wheel
Combining gives
v = ω R
Section 8.6

59. Rolling Wheel, final For the accelerations, a = α R
Follows the same argument as for the velocity
Friction is essential for rolling motion
Usually, the wheels do not slip so static friction is involved
Section 8.6

60. Sweet Spot of a Baseball Bat
Newton’s Second Law for the linear motion is applied to the center of mass of the bat
As the batter swings, the bat undergoes rotational motion about the batter’s hands
We will approximate P as being fixed
Section 8.6

61. Sweet Spot of a Baseball Bat
If the force acts at the sweet spot, there is no recoil at your hands
If the bat was uniform, the sweet spot would be L/6
Bats are not actually uniform, but the sweet spot of a real bat can be found in a similar way
Section 8.6