Figure 10-1. Caption: Looking at a wheel that is rotating counterclockwise about an axis through the wheel’s center at O (axis perpendicular to the page). Each point, such as point P, moves in a circular path; l is the distance P travels as the wheel rotates through the angle θ .
Figure 10-3. Caption: (a) Example 10–1. (b) For small angles, arc length and the chord length (straight line) are nearly equal. For an angle as large as 15°, the error in making this estimate is only 1%. For larger angles the error increases rapidly. Solution: a. 0.017 ° b. 3 cm (assuming the arc length and the chord length are the same)
Figure 10-4. Caption: A wheel rotates from (a) initial position θ 1 to (b) final position θ 2 . The angular displacement is Δθ = θ 2 – θ 1 .
Figure 10-5. Caption: A point P on a rotating wheel has a linear velocity v at any moment.
Answer: The horse has a greater linear velocity; the angular velocities are the same.
Figure 10-6. Caption: A wheel rotating uniformly counterclockwise. Two points on the wheel, at distances R A and R B from the center, have the same angular velocity ω because they travel through the same angle θ in the same time interval. But the two points have different linear velocities because they travel different distances in the same time interval. Since R B > R A , then v B > v A (because v = R ω ).
Figure 10-7. Caption: On a rotating wheel whose angular speed is increasing, a point P has both tangential and radial (centripetal) components of linear acceleration. (See also Chapter 5.)
Figure 10-8. Caption: Example 10–3. The total acceleration vector a = a tan + a R , at t = 8.0 s. Solution: a. The angular velocity increases linearly; at 8.0 s it is 0.48 rad/s. b. The linear velocity is 1.2 m/s. c. The tangential acceleration is 0.15 m/s 2 . d. The centripetal acceleration at 8.0 s is 0.58 m/s 2 . e. The total acceleration is 0.60 m/s 2 , at an angle of 15 ° to the radius.
Solution: a. F = 120 Hz, so the angular velocity is 754 rad/s. b. V = 22.6 m/s. c. 45 megabits/s
Solution: a. The angular acceleration is the derivative of the angular velocity: 1.2 rad/s 2 . b. V = 12.0 m/s; a tan = 3.6 m/s 2 ; a R = 48 m/s 2 .
Figure 10-9. Caption: (a) Rotating wheel. (b) Right-hand rule for obtaining direction of ω .
Solution: a. The final angular velocity is 2100 rad/s, so the acceleration is 70 rad/s 2 . b. The total angle is 3.15 x 10 4 rad, which is 5000 rev.
Figure 10-11. Caption: Top view of a door. Applying the same force with different lever arms, R A and R B . If R A = 3R B , then to create the same effect (angular acceleration), F B needs to be three times F A , or F A = 1/3 F B .
Figure 10-12. Caption: (a) A tire iron too can have a long lever arm. (b) A plumber can exert greater torque using a wrench with a long lever arm.
Figure 10-13. Caption: (a) Forces acting at different angles at the doorknob. (b) The lever arm is defined as the perpendicular distance from the axis of rotation (the hinge) to the line of action of the force.
Figure 10-14. Caption: Torque = R ┴ F = RF ┴ .
Figure 10-15. Caption: Example 10–7.The torque due to F A tends to accelerate the wheel counterclockwise, whereas the torque due to F B tends to accelerate the wheel clockwise. Solution: The net torque is R A F A – R B F B sin 60 ° = -6.7 m·N.
Figure 10-17. Caption: A mass m rotating in a circle of radius R about a fixed point.
Figure 10-18. Caption: A large-diameter cylinder has greater rotational inertia than one of equal mass but smaller diameter.
Figure 11-1: A skater doing a spin on ice, illustrating conservation of angular momentum: (a) I is large and ω is small; (b) I is smaller so ω is larger. Figure 11-2: A diver rotates faster when arms and legs are tucked in than when they are outstretched. Angular momentum is conserved.
Solution: Conservation of angular momentum gives the speed as 4.0 m/s.
Solution: a. The angular momentum is 7.8 kg ·m 2 /s. b. The torque is the change in angular momentum divided by the time, 3.9 m·N. c. Angular momentum is conserved (this is a rotational collision), so the new angular speed is 2.9 rad/s.
Solution: Angular momentum is conserved; the rotation rate would be about 600 rev/s.
Figure 11-5. (a) A person on a circular platform, both initially at rest, begins walking along the edge at speed v. The platform, assumed to be mounted on friction-free bearings, begins rotating in the opposite direction, so that the total angular momentum remains zero, as shown in (b).
Solution: Angular momentum is conserved, so the angular velocity is 0.42 rad/s.
Angular momentum is conserved, so the teacher will start spinning in the direction the wheel was spinning originally.
Angular Quantities Rigid object – shape doesn’t change In purely rotational motion – an object move in rotation motion with (“ O ”) as the axis of rotational Point P - distance R from the center - move in a circle (same as all other point in the object)
<ul><li>All points on a straight line drawn through the axis move through the same angle in the same time. </li></ul><ul><li>X-axis as a reference frame - can indicate the angular position of point P - θ usually in degree - radian is use for mathematical purposes </li></ul><ul><li>The angle θ in radians is defined: </li></ul><ul><li>[dimensionless] </li></ul><ul><li>l - is the arc length. R - distance from the axis of rotational </li></ul>
<ul><li>Example 10-1: Birds of prey—in radians. </li></ul><ul><li>A particular bird’s eye can just distinguish objects that subtend an angle no smaller than about 3 x 10 -4 rad. </li></ul><ul><li>How many degrees is this? </li></ul><ul><li>How small an object can the bird just distinguish when flying at a height of 100 m? </li></ul>
<ul><li>To define linear motion of an object we use quantities such as displacement, velocity and acceleration of the object </li></ul><ul><li>For rotational motion, we will make use angular quantities such as angular displacement, angular velocity and angular acceleration </li></ul><ul><li>The concept is similar to linear motion </li></ul>
A wheel rotating – some point initially specified by θ 1 move to point θ 2 Angular displacement: [rad] The average angular velocity (omega, ω ) is defined as the total angular displacement divided by time : [rad/s] The instantaneous angular velocity: [rad/s]
<ul><li>It is not only the point (we measure) move in that angular velocity, all point in the object rotate with the same angular velocity – every position in the object move through the same angle in the same time interval </li></ul><ul><li>By convention – object moving counterclockwise has + value – object moving clockwise has - value </li></ul>
The angular acceleration (alpha, α ) is the rate at which the angular velocity changes with time : [rad/s 2 ] The instantaneous acceleration: [rad/s 2 ] All points in the object have the same angular acceleration
Every point on a rotating body has, at any instant a linear velocity v and a linear acceleration. We can relate the linear quantities ( v and a ) to the angular quantities ( ω and α ) Linear velocity and angular velocity are related:
Conceptual Example 10-2: Is the lion faster than the horse? On a rotating carousel or merry-go-round, one child sits on a horse near the outer edge and another child sits on a lion halfway out from the center. (a) Which child has the greater linear velocity? (b) Which child has the greater angular velocity?
Points farther from the axis of rotation will move faster (linear velocity) but the angular velocity for all points is the same.
If the angular velocity of a rotating object changes, it has a tangential acceleration: Even if the angular velocity is constant, each point on the object has a centripetal acceleration:
<ul><li>Total acceleration of the object: </li></ul>
Here is the correspondence between linear and rotational quantities:
<ul><li>Example 10-3: Angular and linear velocities and accelerations. </li></ul><ul><li>A carousel is initially at rest. At t = 0 it is given a constant angular acceleration α = 0.060 rad/s 2 , which increases its angular velocity for 8.0 s. At t = 8.0 s, determine the magnitude of the following quantities: </li></ul><ul><li>the angular velocity of the carousel; </li></ul><ul><li>the linear velocity of a child located 2.5 m from the center; </li></ul><ul><li>(c) the tangential (linear) acceleration of that child; </li></ul><ul><li>(d) the centripetal acceleration of the child; and </li></ul><ul><li>(e) the total linear acceleration of the child. </li></ul>
The frequency is the number of complete revolutions per second : Frequencies are measured in hertz: The period is the time one revolution takes:
<ul><li>Example 10-4: Hard drive. </li></ul><ul><li>The platter of the hard drive of a computer rotates at 7200 rpm (rpm = revolutions per minute = rev/min). </li></ul><ul><li>What is the angular velocity (rad/s) of the platter? </li></ul><ul><li>If the reading head of the drive is located 3.00 cm from the rotation axis, what is the linear speed of the point on the platter just below it? </li></ul><ul><li>If a single bit requires 0.50 μ m of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis? </li></ul>
<ul><li>Example 10-5: Given ω as function of time. </li></ul><ul><li>A disk of radius R = 3.0 m rotates at an angular velocity ω = (1.6 + 1.2 t ) rad/s, where t is in seconds. At the instant t = 2.0 s, determine </li></ul><ul><li>the angular acceleration, and </li></ul><ul><li>the speed v and the components of the acceleration a of a point on the edge of the disk. </li></ul>
Vector Nature of Angular Quantities The angular velocity vector points along the axis of rotation, with the direction by convention is given by the right-hand rule . If the direction of the rotation axis does not change, the angular acceleration vector points along it as well.
Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones.
<ul><li>Example 10-6: Centrifuge acceleration. </li></ul><ul><li>A centrifuge rotor is accelerated from rest to 20,000 rpm in 30 s. </li></ul><ul><li>What is its average angular acceleration? </li></ul><ul><li>Through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration? </li></ul>
Torque To make an object start rotating, a force is needed; the position and direction of the force matter as well. The greater the F A , the more quickly the door will open. Same force apply close to the hinge (F B ), door will not open so quickly. The angular acceleration of the door is proportional not only to the magnitude of the force , but also directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm or moment arm .
A longer lever arm is very helpful in rotating objects.
<ul><li>If R A is three times larger than R B (R A =3R B ), with a same force, the angular acceleration will be three times as great at F A than F B </li></ul><ul><li>To get the same acceleration, F B had to be three times of F A (F B =3F A ) </li></ul><ul><li>Angular acceleration is proportional to the product of force times lever arm call moment of force or torque ( τ ) </li></ul>
Lever arm is define as the perpendicular distance from the axis of rotation to the line of action of the force Here, the lever arm for F A is the distance from the knob to the hinge; the lever arm for F D is zero; and the lever arm for F C is as shown.
The torque is defined as: Also: [Unit: m ּ N ]
Two thin disk-shaped wheels, of radii R A = 30 cm and R B = 50 cm, are attached to each other on an axle that passes through the center of each, as shown. Calculate the net torque on this compound wheel due to the two forces shown, each of magnitude 50 N. Example 10-7: Torque on a compound wheel.
<ul><li>The angular acceleration, α of a rotating object is proportional to the net torque, τ applied to it </li></ul><ul><li>Correspond to translational motion </li></ul><ul><li>a is also inversely proportional to the inertia of the object </li></ul>Rotational Dynamics; Torque and Rotational Inertia
Consider particle mass m rotating with radius R , a single force F acts on m , tangent to the circle Knowing that , we see that This is for a single point mass; what about an extended object? As the angular acceleration is the same for the whole object, we can write: R
The total summation of is known as the moment of inertia ( rotational inertia ), I of an object. [unit: kg∙m 2 ] Total torque:
<ul><li>The distribution of mass matters here—these two objects have the same mass, but the one on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation. </li></ul>
<ul><li>Example 10-8: Two weights on a bar: different axis, different I </li></ul><ul><li>Two small “weights,” of mass 5.0 kg and 7.0 kg, are mounted 4.0 m apart on a light rod (whose mass can be ignored) as shown. Calculate the moment of inertia of the system </li></ul><ul><li>(a) when rotated halfway between the weight </li></ul><ul><li>(b) when rotated about an axis 0.50 m to the left of the 5.0 kg mass </li></ul>
The calculation for moment of inertia for most ordinary object can be difficult. Moment of regularly shape object can be work out using calculus (10-7) The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation—compare (f) and (g), for example.
<ul><li>Example 10-9: A heavy pulley </li></ul><ul><li>A 15.0 N force ( F T ) is applied to a cord wrapped around a pulley of mass M = 4.00 kg and radius R O = 33.0 cm. The pulley accelerates uniformly from rest to an angular speed of 30.0 rad/s in 3.00 s. If there is a friction torque τ fr = 1.10 m∙N at the axle, determine the moment of inertia of the pulley. The pulley rotates about its center. </li></ul>
<ul><li>Example 10-10: Pulley and Bucket </li></ul><ul><li>Consider again the pulley in example 10-9 with the same friction. But this time, instead of a constant 15.0 N force being exerted to the cord, we now have a bucket of weight w = 15.0 N (mass m = 1.53 kg) hanging from the cord. We assume the cord has negligible mass and does not stretch or slip on the pulley. </li></ul><ul><li>(a) Calculate the angular acceleration of the pulley and the linear acceleration of the bucket </li></ul><ul><li>(b) Determine the angular velocity of the pulley and linear velocity of the bucket at t = 3.00 s if the pulley (and bucket) start from rest at t = 0. </li></ul>
<ul><li>Example 10-11: Rotating Rod </li></ul><ul><li>A uniform rod of mass M and length l can pivot freely about a hinge or pin attached to the case of a large machine. The rod is held horizontally and then released. At the moment of release, determine </li></ul><ul><li>(a) the angular acceleration of the rod </li></ul><ul><li>(b) the linear acceleration of the tip of the rod. </li></ul><ul><li>(assume the force of gravity acts at the center of mass of the rod, as shown. Take I =1/3 Ml 2 ) </li></ul>
Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written:
Work The work done on an object rotating about a fixed axis can be written in term of angular quantities The torque does work as it moves the wheel through an angle θ :
Work-Energy Principle The work-energy principle holds for rotation of rigid object about a fixed axis, we find that the work done here is: Power
<ul><li>Example 10-14: Flywheel </li></ul><ul><li>Flywheel, which are simply large rotating disks, have been suggested as a means of storing energy for solar-powered generating system. Estimate the kinetic energy can be stored in an 80000 kg flywheel with a diameter of 10 m Assume it could hold together at 100 rpm. </li></ul><ul><li>( I = ½ MR 0 2 ) </li></ul>
<ul><li>Example 10-15: Rotating Rod </li></ul><ul><li>A rod of mass M is pivoting on a frictionless hinge at one end. The rod is held at rest horizontally and then released. Determine the angular velocity of the rod when it reach the vertical position, and the speed of the rod’s tip at this moment. </li></ul><ul><li>( I = 1/3 Ml 2 ) </li></ul>
Rolling Motion: Rotational Plus Translational Motion <ul><li>Rolling without slipping </li></ul><ul><li>Depend on static friction between rolling object and the ground </li></ul><ul><li>Involves both rotation and translational motion </li></ul>
<ul><li>(a) Relative to the ground </li></ul><ul><li>Point C (center of mass), the axle of the wheel has the velocity, v </li></ul><ul><li>(b)Relative to the axle </li></ul><ul><li>The wheel has an angular velocity ω </li></ul><ul><li>The velocity of the ground is the same magnitude with point P linear velocity = R ω </li></ul>
<ul><li>Instantaneous Axis </li></ul><ul><li>When a wheel roll without slipping, the point of contact of the wheel with the ground is instantaneously at rest </li></ul><ul><li>This point is called the instantaneous axis as all other point is rotating around that point at that moment </li></ul>
<ul><li>The Parallel-axis Theorem </li></ul><ul><li>Relate the moment of inertia, I CM about an axis passing through the center of mass of total mass M , and a moment of inertia, I at any axis of the object with the axis parallel to the I CM </li></ul><ul><li>The parallel-axis theorem gives the moment of inertia, I about any axis that is parallel to an axis that goes through the center of mass of an object: </li></ul>
Example 10-13: Parallel axis. Determine the moment of inertia of a solid cylinder of radius R 0 and mass M about an axis tangent to its edge and parallel to its symmetry axis. (I = ½ MR 0 2 )
<ul><li>Total Kinetic Energy </li></ul><ul><li>Rolling object: rotates while its center of mass undergoes translational motion </li></ul><ul><li>A object that both translational and rotational motion also has both translational and rotational kinetic energy: </li></ul>
<ul><li>Example 10-16: Sphere rolling down an incline </li></ul><ul><li>What will be the speed of a solid sphere of mass M and radius r o when it reaches the bottom of an incline if it start from rest at a vertical height H and rolls without slipping? </li></ul><ul><li>(I = 2/5 MR o 2 ) </li></ul>
<ul><li>Example 10-18 Analysis of sphere on an incline using forces </li></ul><ul><li>Analyze the rolling sphere of example 10-16 in term of forces and torque. In particular, find the velocity v and the magnitude of the friction force, F fr . </li></ul>
Angular Momentum The rotational analog of linear momentum is angular momentum, L : [unit: kg m 2 /s] Then the rotational analog of Newton’s second law is: This form of Newton’s second law is valid even if I is not constant.
Conservation of Angular Momentum In the absence of an external torque, angular momentum is conserved: This is the law of conservation of angular momentum, the total angular momentum of a rotating object remains constant if the net external torque acting on it is zero.
This means, the value of angular momentum is conserved: Where I o – moment of inertia at (t=0) ω o – angular velocity at (t=0) Therefore, if an object’s moment of inertia changes, its angular speed changes as well.
<ul><li>Object change their shape – alter their moment of inertia ( I ), angular velocity ( ω ) change as well so I ω remain constant </li></ul>Skater start spinning at low angular velocity with arm outstretched Bring her arm close to her body, decreasing her moment of inertia For I ω to remain constant, ω increases She will spin more faster
<ul><li>Diver leaves the board with an angular momentum, L = I ω , with small angular velocity. </li></ul><ul><li>Curls herself, rotates more quickly </li></ul><ul><li>Stretch out again, moment of inertia increase, angular velocity decrease </li></ul><ul><li>Enter the water with small angular velocity, ω </li></ul>
Example 11-1: Object rotating on a string of changing length. A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v 1 = 2.4 m/s in a circle of radius R 1 = 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to R 2 = 0.48 m. What is the speed, v 2 , of the mass now?
Example 11-2: Clutch. A simple clutch consists of two cylindrical plates that can be pressed together to connect two sections of an axle, as needed, in a piece of machinery. The two plates have masses M A = 6.0 kg and M B = 9.0 kg, with equal radii R 0 = 0.60 m. They are initially separated. Plate M A is accelerated from rest to an angular velocity ω 1 = 7.2 rad/s in time Δ t = 2.0 s. Calculate (a) the angular momentum of M A . (b) the torque required to have accelerated M A from rest to ω 1 . (c) Next, plate M B , initially at rest but free to rotate without friction, is placed in firm contact with freely rotating plate M A , and the two plates both rotate at a constant angular velocity ω 2 , which is considerably less than ω 1 . Why does this happen, and what is ω 2 ?
Example 11-3: Neutron star. Astronomers detect stars that are rotating extremely rapidly, known as neutron stars. A neutron star is believed to form from the inner core of a larger star that collapsed, under its own gravitation, to a star of very small radius and very high density. Before collapse, suppose the core of such a star is the size of our Sun ( r ≈ 7 x 10 5 km) with mass 2.0 times as great as the Sun, and is rotating at a frequency of 1.0 revolution every 100 days. If it were to undergo gravitational collapse to a neutron star of radius 10 km, what would its rotation frequency be? Assume the star is a uniform sphere at all times, and loses no mass. (I = 2/5 MR 2 )
Direction Nature of Angular Momentum Angular momentum is a vector; the direction is in the same direction as the angular velocity vector. The person start to walk counterclockwise, the angular momentum point upward The platform start to move clockwise, direction of angular momentum is downward This happen to balance out the upward angular momentum so the total angular momentum remain L = 0
Example 11-4: Running on a circular platform. Suppose a 60-kg person stands at the edge of a 6.0-m-diameter circular platform, which is mounted on frictionless bearings and has a moment of inertia of 1800 kg·m 2 . The platform is at rest initially, but when the person begins running at a speed of 4.2 m/s (with respect to the Earth) around its edge, the platform begins to rotate in the opposite direction. Calculate the angular velocity of the platform.
Conceptual Example 11-5: Spinning bicycle wheel. Your physics teacher is holding a spinning bicycle wheel while he stands on a stationary frictionless turntable. What will happen if the teacher suddenly flips the bicycle wheel over so that it is spinning in the opposite direction?