1. Rotational Motion
Handout
HW #1 & 2

2. I. Introduction:
A rotating object is one that spins on a fixed axis. The position and direction of the rotation axis will remain constant. The position of some part of the object can be specified with standard cartesian coordinates, (x,y). All objects will be assumed to rotate in a circular path of constant radius. y
Both coordinates, (x, y), change over time as the object rotates.
(x, y) R
The object’s position can also be specified with polar coordinates, (r, q). q x
For this last coordinate system, only the angle changes. The radius stays constant.

3. II. Definitions:
Since the polar coordinates only have one changing variable, the angle, we will use this to simplify analysis of motion.
position
A. The ___________ is defined as where an object is in space. Here, we only need to specify the angle of the object with respect to some origin or reference line.
The position of the object is measured by an angle, q, measured counterclockwise from the positive x – axis.
The angle will be measured in units of ___________ rather than ___________ .
radians q
degrees

4. Radians are defined as the ratio of the length along the arc of a circle to the position of an object divided by the radius of the circle. R s q
s = length along the arc measured counterclockwise from the +x – axis.
R = radius of the circular path.
Since q is the ratio of two lengths, the angle measurement really does not have any units. The term “radians” is just used to specify how the angle is measured.
1 revolution = 360 degrees = 2p radians

5. B. The _____________ of the object is just the difference in its position. We define the ____________________ of the object as the difference in its angular position.
displacement
angular displacement
A positive Dq shows a counterclockwise {ccw} rotation, while a negative Dq shows a clockwise {cw} rotation.
velocity
C. Motion can also be measured through a rate of rotation, a __________.
angular velocity
The ____________________ of an object is defined as the amount of rotation of an object per time. The angular velocity is represented by the greek letter w (lower case omega).
average angular velocity:
t = elapsed time.

6. D. Another measure of motion is the rate of change of velocity, called an
________________.
acceleration
angular acceleration
The ____________________ of an object is defined as the amount of change of the angular velocity of an object per time. The angular acceleration is represented by the greek letter a (lower case alpha).
average angular acceleration:
In general, the motion may be complex, but we will again look at constant angular acceleration cases, exactly the same way as we did back in Ch. 2. The same equations of motion can be derived for circular motion.

7. Ch. 2
Ch. 7
Linear Motion Rotational Motion
This is only true for constant accelerations.
Problem solving is the same as before.

8. E. The motion of an object around a circle can also be represented as actual distances along the circle and speeds tangent to the circle. R vt
tangential speed
The _________________ , vt, of an object is defined as the angular velocity times the radius of the circle.
The tangential speed measures the actual speed of the object as it travels around the circle.
length along arc
vt =
elapsed time

9. tangential acceleration
The _________________________ , at, of an object is defined as the angular acceleration times the radius of the circle.
The tangential acceleration measures how the tangential speed increases or decreases over time.

10. Example #1: A disk rotates from rest to an angular speed of 78
Example #1: A disk rotates from rest to an angular speed of rpm in a time of seconds. a. What is the angular acceleration of the disk?
wo = 0, w = revolutions per minute, t = seconds. a = ?

11. b. Through what angle does the disk turn?

12. c. Through what angle will the disk turn if it were to maintain the same angular acceleration up to rad/s?

13. d. The disk has a diameter of 12. 00 inches
d. The disk has a diameter of inches. What is the tangential speed and acceleration at the edge of the disk the moment the disk reaches rpm?

14. Handout HW #3 Turn in Momentum Lab
Centripetal Force
Handout
HW #3
Turn in Momentum Lab

15. III. Centripetal Acceleration and Force.
When an object moves in a circle, the direction of its velocity is always changing. This means the object is always accelerating! For an object rotating at a constant angular speed, the acceleration of the mass is always towards the center of the motion. This kind of acceleration is called a ___________ {“center seeking”} acceleration, and is represented as ac. The amount of acceleration depends on the radius of the circular path and the speed around the circle.
centripetal
w = angular speed, v = tangential speed, r = radius of circular path.

16. Example #2: A wheel of a car has a diameter of 32. 0 inches
Example #2: A wheel of a car has a diameter of 32.0 inches. A rock is wedged into the grooves of the tire. a. What is the centripetal acceleration on the rock if the wheel turns a rate equal to 70.0 mph?

17. b. What is the rotation rate of the tire, in rpm?

18. If there is a centripetal acceleration making an object move in a circle, then there must be an unbalanced force creating this acceleration.
centripetal
This force is called the _____________ force, and it also points towards the center of the circular motion. This force must be made by real forces acting on an object. The centripetal force will be the sum of the radial components of the forces acting on the object. A radial component points towards the ___________ of the circle.
center

19. Example #3: A 0. 475 kg mass is tied to the end of a 0
Example #3: A kg mass is tied to the end of a meter string and the mass is spun in a horizontal circle. If the mass makes 22.0 revolutions in a time of 2.50 seconds, what is the tension in the string holding the mass to the circular motion?
rotation R T m

20. Example #4: A penny sits at the edge of a 12. 00 inch diameter record
Example #4: A penny sits at the edge of a inch diameter record. If the coefficient of static friction is between the penny and the record, what is the maximum rotation rate of the record that will allow the penny to remain on the record?
rotation n
vertical forces balance. Fs
by definition friction is: mg

21. set the friction force equal to the centripetal force

22. Example #5: A mass m on a frictionless table is attached to a hanging mass M by a cord through a hole in the table. Find the speed which m must move in order for M to stay at rest. Evaluate the speed for m = 2.00 kg,
M = 15.0 kg, and r = m.
Since M is at rest, the tension force lifting it is equal to the weight of M:
This tension is also the centripetal force on the mass m, causing it to spin in a circular path:

23. Set the two equations equal to one another:

24. Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room.

25. definition of force of static friction
Example #6: A common amusement park ride involves a spinning cylinder with a floor that drops away. When a high enough rotation speed is achieved, the people in the ride will stay on the side of the wall. A static friction force holds each person up. Solve for the rotation rate of the room, given the coefficient of friction for the wall and the radius of the room.
set the friction force equal to the weight as the vertical forces balance. Fs
forces balance n m
definition of force of static friction
Horizontal direction: set the normal force equal to the centripetal force. mg

26. Combine all the information to solve for the rotation rate, w.
What would be the rotation rate for a room 3.00 m wide and a carpeted wall with a coefficient of friction of 0.750?

27. component of the normal force that is vertical
Example #7: (Banking Angle) Determine the angle of the roadway necessary for a car to travel around the curve without relying on friction. Assume the speed of the car and the radius of the curve are given.
component of the normal force that is vertical
component of the normal force that is horizontal, also becomes the centripetal force

28. balance the vertical forces:
set the net horizontal force equal to the centripetal force, with towards the center of the circular path as the positive direction:
substitute in:

29. note: divide… Example #8: Conical Pendulum.
A mass of m = 1.5 kg is tied to the end of a cord whose length is L = 1.7 m. The mass whirls around a horizontal circle at a constant speed v. The cord makes an angle q = 36.9o. As the bob swings around in a circle, the cord sweeps out the surface of a cone. Find the speed v and the period of rotation T of the pendulum bob.
note:
divide…

30. simplify…

31. the period, T, is the time for one revolution:

32. Example #9: Another common amusement park ride is a rollercoaster with a loop. Determine the minimum speed at the top of the loop needed to pass through the top of the loop.
There are two forces acting on the car: the weight pulling straight downwards and the normal force pushing perpendicular to the track.
As the car goes through the loop, the normal force always points towards the center of the loop. The radial component of the vector sum of the normal force and the weight equal the centripetal force.
The faster the car goes, the greater the normal force to push the car into a circular path. The minimum speed for the car at the top of the loop is where the normal force goes to zero.

33. The net force for the mass at the top of the loop is:
The faster the car goes, the greater the normal force to push the car into a circular path. The minimum speed for the car at the top of the loop is where the normal force goes to zero. mg r

34. Example #9: (b) Use energy conservation to find the speed of the mass at the bottom of the loop.
Set the two energies equal:
Divide by m and multiply by 2:

35. Example #9: (c) What is the necessary starting height of the mass if sliding from rest down the ramp?
Set the two energies equal:
Divide both sides by m and g:

36. Newton’s Universal Law of Gravity
Handout
HW #4
Turn in Rotary Motion Lab

37. The force is proportional to each mass.
Newton’s Law of Universal Gravity:
Any two objects are attracted to each other through the force of gravity.
The force is proportional to each mass.
The force is inversely proportional to the square of the center to center distance between the masses.
“inverse square law”

38. By Newton’s 3rd Law, forces are equal and opposite!
These two equations can be combined into
a single equation, and a proportionality
constant can be introduced to make
an equality:
By Newton’s 3rd Law, forces are equal and opposite!

39. Example #1: Two students, each with mass 70 kg, sit 80 cm apart.
(a) What is the force of attraction between the two students?

40. (b) How does this compare to the weight of the students?
The force between you and your neighbor is less than 1 part in a billion of your normal body weight.

41. All forces balance out pairwise except for one:
Example #2: What is the direction of the net force on the mass at the center of the image? Why?
All forces balance out pairwise except for one:
Force on central mass points upwards.

42. All forces balance out pairwise except for one:
Example #3: What is the direction of the net force on the mass at the center of the image? Why?
All forces balance out pairwise except for one:
Force on central mass points towards the left.

43. Remember, r is the center to center (of the earth) distance.
II. Compare the force of gravity to weight:
The weight of an object near the surface of the Earth is just the force of gravity exerted on the object by the Earth.
Remember, r is the center to center (of the earth) distance.

44. The mass of the object cancels out, and what is left is a theoretical value for the acceleration due to gravity near the surface of the Earth.
Example #4: Estimate the value of ‘g’ for the Earth.

45. Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.

46. Estimate volume as sphere 1.4 m in diameter.
Example #5: Estimate the value of ‘g’ near the surface of Peter Griffin.
Estimate volume as sphere 1.4 m in diameter.

47. Multiply by density to get mass:
Estimate ‘g’ at surface:

48. III. Satellite Motion.
A satellite is an object that orbits (travels in a circular path around) some gravity source
The force of gravity provides the centripetal force to keep the satellite moving in a circular path.

49. The equation of motion is:
m = satellite mass (divides out…)
M = mass of the gravity source
r = radius of the orbit (center to center distance!)
v = the tangential speed of the orbiting object

50. Example #6: The space shuttle orbits the Earth at an altitude of 400 km above the surface of the Earth. (a) Determine the speed of the space station around the Earth.

51. (b) How much time will it take for the space station to orbit the Earth?

52. Example #7: Write an equation that gives the period (time for one revolution) of an orbit in terms of the radius of the orbit.

53. Example #8: Determine the altitude (height above the surface) for a geosynchronous satellite. The period of the satellite matches the length of the Earth’s day, 24 hours.

54. This is the height from the center of the Earth.
Subtract the radius of the Earth to get the height from the surface.

55. Kepler’s Laws of Satellite Motion
Handout
HW #5

56. IV. Kepler’s Laws.
Johannes Kepler was a German mathematician, astronomer and astrologer. He is best known for his eponymous laws of planetary motion, which he was able to put together by painstakingly analyzing the volumes of data collected by Tycho Brahe of the planets motions in the sky.
Kepler
Brahe

57. Kepler’s 1st Law:
All planets (satellites) orbit the sun (or gravity source) in elliptical orbits.

58. The sun sits at one focus of the ellipse
The sun sits at one focus of the ellipse. The planet will be closer to the sun for part of its “year”, and farther away for the other part of its “year”.
Rp = perihelion = closest distance of planet to sun
Ra = aphelion = farthest distance of planet to sun

59. Kepler's first law. An ellipse is a closed curve such that the sum of the distances from any point P on the curve to two fixed points (called the foci, F1 and F2) remains constant. That is, the sum of the distances, F1P + F2P, is the same for all points on the curve. A circle is a special case of an ellipse in which the two foci coincide, at the center of the circle.

60. Potential energy from gravity still has the same behavior on this scale:
The farther the planet moves from the sun, the higher the potential energy of the planet. But total energy is still constant:
As the planet moves farther from the sun, it slows down. The closer the planet to the sun, the faster it moves in its orbit.

61. Kepler’s 2nd Law:
As planets orbit around the sun, an imaginary line from the planet to the sun will sweep out equal areas in equal times.
Kepler's second law. The two shaded regions have equal areas. The planet moves from point 1 to point 2 in the same time as it takes to move from point 3 to point 4. Planets move fastest in that part of their orbit where they are closest to the Sun. Exaggerated scale.

63. Measurable Application: Earth’s weather is due to the tilt of the Earth’s axis relative to its orbital plane.
Earth at aphelion
Earth at perihelion

64. Since summer (for the northern hemisphere) occurs when Earth is farthest from the sun, the Earth spends slightly more time in this part of the orbit as compared to winter (again, for northern hemisphere). The effect is that summer is about 3 days longer than winter. This is sometimes called the “summer analemma”.

65. The analemma is the location of the sun in the sky at the same time each day of the year plotted on a graph or photograph. The uneven shape is due to the elliptical orbit of the Earth around the sun.

66. Kepler’s 3rd Law:
The square of the period of the orbit of a planet is proportional to the cube of the semimajor axis of the orbit of the planet.
a = length of semimajor axis.
T = period of orbit.
Note: This was derived for circular orbits in example #7. The derivation for elliptical orbits is beyond the scope of the class…

69. Example #9: Earth’s orbital information is sometimes used as a standard: The orbital period is one year and its semimajor axis is one AU (astronomical unit). If the orbital period for Jupiter is 11.9 years, determine the semimajor axis distance for Jupiter.

71. Example #10: Earth’s orbit has a semimajor axis length of 1
Example #10: Earth’s orbit has a semimajor axis length of 1.50×108 km and a period of days. Determine the mass of the sun.

72. http://www. youtube. com/watch

75. Torque and Center of Mass
Handout
HW #6

76. Center of Mass:
The center of mass (or mass center) is the mean location of all the mass in a system.
The motion of an object can be characterized by this point in space. All the mass of the object can be thought of being concentrated at this location. The motion of this point matches the motion of a point particle.

79. Finding the Center of Mass:
Uniform geometric figures have the center of mass located at the geometric center of the object.
Note that the center of mass does not have to be contained inside the volume of the object.

80. Collections of Point Masses:
The center of mass for a collection of point masses is the weighted average of the position of the objects in space.
Each object will have a position in space. The center of mass is found as:

81. Example #1: A 10. 0 kg mass sits at the origin, and a 30
Example #1: A 10.0 kg mass sits at the origin, and a 30.0 kg mass rests at the 12.0 m mark on the x – axis. (a) Find the center of mass for this system.

82. (b) Find the center of mass for this system relative to the mass at the right.
Although numerically different, it is the same point in space relative to the masses…

83. Example #2: A 10. 0 cm long wire has a mass of 4. 00 grams
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

84. Treat as two objects: 6 cm object: 4 cm object:
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.
Treat as two objects:
6 cm object:
4 cm object:

86. Example #3: Determine the center of mass of the following masses, as measured from the left end. Assume the blocks are of the same density.

89. Torque
Torque is the rotational equivalent of force. A torque is the result of a force applied to an object that tries to make the object rotate about some pivot point.

90. Note that torque is maximum when the angle q is 90º.
Equation of Torque:
applied force
distance from pivot to applied force
angle between direction of force and pivot distance.
pivot point
Note that torque is maximum when the angle q is 90º.
The units of torque are Nm or newton · meter

91. The torque is also the product of the distance from the pivot times the component of the force perpendicular to the distance from the pivot.

92. The torque is also the product of the force times the lever arm distance, d.

93. Example #4: Calculate the torque for the force shown below.

94. Example #5: Calculate the total torque about point O on the figure below. Take counterclockwise torques to be positive, and clockwise torques to be negative.

95. Example #6: The forces applied to the cylinder below are F1 = 6
Example #6: The forces applied to the cylinder below are F1 = 6.0 N, F2 = 4.0 N, F3 = 2.0 N, and F4 = 5.0 N. Also, R1 = 5.0 cm and R2 = 12 cm. Determine the net torque on the cylinder.

96. Static Equilibrium: Torque and Center of Mass
Handout
HW #7
Extra credit due Friday

97. Static Equilibrium:
Static equilibrium was touched on in the unit of forces. The condition for static equilibrium is that the object is at rest. Since the object is not moving, it is not accelerating. Thus the net force is zero. Shown at right is a typical example from that unit: Find the force of tension in each rope.
A new condition can now be added into this type of problem: Since the object is at rest, it must not be rotating, as that would also require an acceleration.

98. Answer: The pivot point can be put anyplace you want!
If there is no rotation, there must not be a rotational acceleration. Thus the net torque must be zero. This is an application of Newton’s 2nd law to rotational motion.
If the net torque is zero, then all the counterclockwise (ccw) torques must balance all the clockwise (cw) torques.
If there is no rotation, where is the pivot point for calculating torque?
Answer: The pivot point can be put anyplace you want!
Hint: Put the pivot point at one of the unknowns. This eliminates the unknown from the torque equation.

99. Example #1: A meter stick has a mass of 150 grams and has its center of mass located at the 50.0 cm mark. If the meter stick is supported at each of its ends, then what forces are needed to support it?
Show that the two forces are equal through torque. Put the pivot point at the left end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
Force mmsg makes a cw torque.

100. The other unknown must also equal half the weight, so:
Balance the net ccw and cw torque:
The other unknown must also equal half the weight, so:

101. Example #2: Suppose the meter stick above were supported at the 0 cm mark (on the left) and at the 75 cm mark (on the right). What are the forces of support now?
Find the two unknown forces through torque. Put the pivot point at the left end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
Force mmsg makes a cw torque.

102. Balance the net ccw and cw torque:
If force F2 holds 2/3 the weight, then F1 must hold the remaining 1/3 of the weight.

103. Example #3: A meterstick is found to balance at the 49
Example #3: A meterstick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
The meterstick behaves as if all of its mass was concentrated at its center of mass.
Calculate the torque about the pivot point. The support force of the fulcrum will not contribute to the torque in this case.

104. Force maddedg makes a ccw torque.
Force mmsg makes a cw torque.
Balance the net ccw and cw torque:

105. Example #4: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and is 3.00 m long. What is the tension in each rope when the 700-N worker stands 1.00 m from one end?
Put the pivot point on the left end. The force F1 does not contribute torque. Solve for F2.

106. Solve F1 from Newton’s laws:

107. Put the pivot point at the left end and balance the torques.
Example #5: A cantilever is a beam that extends beyond its supports, as shown below. Assume the beam has a mass of 1,200 kg and that its center of mass is located at its geometric center. (a) Determine the support forces.
Put the pivot point at the left end and balance the torques.

108. The fact FA is negative means that the force really points downwards.
Balance the net ccw and cw torque:
Solve FA from Newton’s laws:
The fact FA is negative means that the force really points downwards.
When the wrong direction is chosen for a force, it just comes out negative at the end.

109. Static Equilibrium: Day #2
Handout
HW #7
Extra credit due Friday

110. Example #6: Calculate (a) the tension force FT in the wire that supports the 27.0 kg beam shown below.
Put the pivot point at the left end. The wall support does not contribute to torque.
Note that q and 40° are supplements, so it does not matter which is used in the sine function.
Beam length is L.

111. Balance the net ccw and cw torque:

112. Balance forces in each component direction.
(b) Determine the x and y components to the force exerted by the wall.
Balance forces in each component direction.

113. Example #7: A shop sign weighing 245 N is supported by a uniform 155 N beam as shown below. Find the tension in the guy wire and the horizontal and vertical forces exerted by the hinge on the beam.
Put the pivot point at the left end. The wall support does not contribute to torque.

114. Balance the net ccw and cw torque:

115. The fact Fy is negative means that the force really points downwards.

116. Example #8: A person bending forward to lift a load “with his back” (see figure below) rather than “with his knees” can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in the figure below of a person bending forward to lift a 200-N object. The spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°. Find the tension in the back muscle and the compressional force in the spine.

117. Put the pivot point at the left end
Put the pivot point at the left end. The hip support does not contribute to torque.

119. Example #9: A person in a wheelchair wishes to roll up over a sidewalk curb by exerting a horizontal force to the top of each of the wheelchair’s main wheels (Fig. P8.81a). The main wheels have radius r and come in contact with a curb of height h (Fig. P8.81b). (a) Assume that each main wheel supports half of the total load, and show that the magnitude of the minimum force necessary to raise the wheelchair from the street is given by
where mg is the combined weight of the wheelchair and person. (b) Estimate the value of F, taking mg = N, r = 30 cm, and h = 10 cm.

121. Example #10: A circular disk 0
Example #10: A circular disk m in diameter, pivoted about a horizontal axis through its center, has a cord wrapped around its rim. The cord passes over a frictionless pulley P and is attached to an object that weighs 240 N. A uniform rod 2.00 m long is fastened to the disk, with one end at the center of the disk. The apparatus is in equilibrium, with the rod horizontal. (a) What is the weight of the rod?

122. (b) What is the new equilibrium direction of the rod when a second object weighing 20.0 N is suspended from the other end of the rod, as shown by the broken line in the image below? That is, what angle does the rod then make with the horizontal?

124. Inertia and Rotary Motion Moment of Inertia
Handout
HW #8

125. Final Schedule: Mosig’s Class
Monday 12/9 Notes
Tuesday 12/10 Finish notes
Wednesday 12/11 Study / Tower practice day
Thursday 12/12 Study / Tower practice day
Friday 12/13 Final Exam {covers current unit}
Monday 12/16 Towers p. 0, 1, & 6
Tuesday 12/17 Towers p. 2 & 3
Wednesday 12/18 Towers p. 4 & 5
Thursday 12/19 Elf Dance / Non – Academic day
Friday 12/20 No School

126. Definition: Inertia is the ability of an object to resist a change in its motion.
Inertia was introduced earlier in the Force unit. For straight line motion, the inertia of an object is measured through mass: The more massive an object, the more it is able to resist changes to its (straight line) motion.
Force and acceleration were related through inertia:
There is an equivalent to inertia in rotary motion. Here, inertia would try to resist a change to the angular motion. This form of inertia will depend on mass, just as before, but it will also depend on the distribution of the mass.
Demonstration:

127. When the mass is distributed close to the center of rotation, the object is relatively easy to turn. When the mass is held much further away, it is more difficult to rotate the object.
For example, if you had a ring (hoop) and a disk with the same radius and same mass, the ring would show more resistance to rotation than would the disk. The disk has mass uniformly distributed across its body, so some of the mass is near the center of rotation. The ring has more mass concentrated at the outside edge of the body than does the disk, so it will show more inertia (even though the mass and radius are the same).

128. The dependence of the inertia on mass and distribution can be built up through the kinetic energy of an object moving through a circular path.
The kinetic energy of a mass m moving at a speed v around the circle is:
Let the circle have a radius r, and let the angular speed of the mass be w. Write the kinetic energy in terms of the angular speed:
Define the moment of inertia I of this point mass to be:
Then: This is now the definition.

129. Now build up to a more complicated object:
The total kinetic energy becomes:
Mass m1 travels in a circle of radius r1, etc. All masses have the same angular velocity, w. Write the kinetic energy in terms of this:

130. In general, the moment of inertia is found as:
The farther the mass is from the center of rotation, the higher the moment of inertia.
Example #1: (a) If m = 2.00 kg and d = m for the image below, determine the moment of inertia of this group of objects.

131. (b) If this apparatus is rotating at 4
(b) If this apparatus is rotating at 4.00 rad/s, what is its kinetic energy?

132. For more complicated objects, the moment of inertia is tabulated below
For more complicated objects, the moment of inertia is tabulated below. The calculations of these are complex and beyond the scope of the class.

133. Example of calculating the rotational inertia of a solid ball
Example of calculating the rotational inertia of a solid ball. You are not responsible for knowing these calculations…

134. Newton’s 2nd Law and Rotational Motion:
There is an equivalent to Newton’s 2nd law for rotational motion.
This can be combined with the kinematics equations from earlier in the unit to solve uniform motion problems;

135. Example #2: A force of N is applied to the edge of a disk that can spin about its center. The disk has a mass of 240 kg and a diameter of 3.20 m.
If the force is applied for 24.0 s, how fast will the disk be turning if it starts from rest?

137. Example #3: Variation of the Atwood’s machine. A 6
Example #3: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm. What is the acceleration of the hanging mass downwards?
The tangential acceleration of the disk is equal to the linear acceleration of the falling mass.
The net torque on the disk gives one equation:

138. Next use Newton’s 2nd law on the falling mass:
Combine the equations:

140. Inertia and Rotary Motion Day #2
Handout
HW #8

141. Rolling and Energy Conservation.
When an object rolls along the ground, the tangential speed of the outside edge of the object is the same as the speed of the center of mass of the object relative to the ground.

142. The rolling object has two parts to its motion
The rolling object has two parts to its motion. First is the motion of the center of mass, and second is the rotation around the center of mass. The total kinetic energy is the sum of the kinetic energy associated with each part.
The kinetic energy associated with the center of mass moving in a straight line is given by the term:
The portion associated with rotating around the center of mass is:
Icm is the moment of inertia of the object about its center of mass. Refer to the given table for values.

143. Example #4: A solid ball of radius 10. 0 cm and mass 10
Example #4: A solid ball of radius 10.0 cm and mass 10.0 kg rolls at a given speed vo of 5.00 m/s. (a) What is the total kinetic energy of this rolling ball?
For a rolling ball: &

144. (b) What percentage of the total kinetic energy is rolling?

145. Example #5: Four different objects are placed at the top on an incline, as shown below. A point particle can slide down without friction. The other three objects will roll down the incline. In what order will the objects reach the bottom, from fastest to slowest?
(a) What is the speed of the sliding point particle when it reaches the bottom?
Energy conservation!

146. (b) Solve for the speed of the sphere (solid ball) at the bottom.
Energy conservation!
Note that there is a fixed starting energy, and this is split between linear motion and rotation. The rolling objects are slower!

148. (c) Solve for the speed of the hoop at the bottom.
Energy conservation!
The hoop has the slowest speed, and thus takes the longest to reach the bottom. The disk will be between the ball and the hoop.

150. Example #6: Variation of the Atwood’s machine. A 6
Example #6: Variation of the Atwood’s machine. A 6.00 kg mass is tied via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm.
(a) How fast will the mass be traveling after it falls a distance of h = 4.00 m downwards?
Energy conservation!
Speed of falling mass equals tangential speed of disk:

152. (b) Solve for the acceleration of the hanging mass:
Shortcut: Remember the kinematics equation…
Same result as before!