As with conservation of energy and conservation of … · Web viewTitle As with conservation of energy and conservation of linear momentum, conservation of angular momentum is one

Topic 6 – Angular MomentumChapter 10

I. Introduction.a. As with conservation of energy and conservation of linear

momentum, conservation of angular momentum is one of the basic principles of physics.

b. Experimental evidence shows that angular momentum is never created nor destroyed.

c. We extend our study of rotational motion to situations in which the direction of the axis of rotation may change. Angular velocity, angular acceleration, and torque are presented in Chapter 9. Here we begin by introducing the vector nature of these quantities and of angular momentum, which is the rotational analog of linear momentum. We then show that the net torque acting on a system equals the time rate of change of its angular momentum. Angular momentum is conserved in systems that have zero net external torque. Like conservation of linear momentum, conservation of angular momentum is a fundamental law of nature, relating to even atoms, molecules, subatomic particles, and photons.

II. The Vector Nature of Rotation.a. In Chapter 9, we indicated the direction of rotation about an

axis by assigning plus and minus signs to indicate the direction of the angular velocity, just as in Chapter 2 we used plus and minus signs to indicate the direction of the velocity in one-dimensional motion.

b. However, plus and minus signs are not adequate to specify the direction of the angular velocity if the direction of the rotation axis is not fixed in space.

c. This inadequacy is overcome by treating the angular velocity as a vector quantity directed along the rotation axis.

d. If the rotation is directed as shown, is directed as shown; if the

rotation direction is reversed, so is the direction of .

e. The convention relating the direction of with the direction of rotation is specified by a convention called the right-hand rule.

f. You can obtain the direction of by curling the fingers of your right hand in the direction of rotation ( Figure 10-2); your

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thumb then points along the rotation axis in the direction of .

g. In Chapter 9, we indicated the direction of torque about an axis by assigning plus and minus signs to indicate the direction of the torque. In this chapter, we define the torque about a point as a vector quantity, and, as with , the direction of is specified by a right hand rule.

i. Figure 10-3 shows a force acting on a particle at some position relative to the origin .

ii. The torque about exerted by

this force is defined as a vector that is perpendicular to both and and has magnitude , where is the angle between the directions of and .

iii. If and are both perpendicular to the z axis, as in Figure 10-3, the torque vector is parallel to the z axis.

iv. If is applied to the rim of a disk of radius r as shown in Figure 10-4, the torque vector has the magnitude and is along the axis of rotation in the direction shown.

h. The Vector Product.i. Review.

1. The Cross Product Method.

2. The cross product method produces a vector when two vectors are multiplied together.

3. The magnitude is the area of a parallelogram formed by two vectors.

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4. The resultant direction is defined as perpendicular to the other two and is determined by the right-hand rule when two vectors are placed tail to tail.

a. Align your right hand along the first vector so that the palm of your hand is at the tail of the vector and your fingertips are pointing toward the head.

b. Then, curl your fingers via the small angle toward the second vector.

c. The direction your thumb points in making the angle from the first to the second vector is the direction of the resultant.

5. Two vectors that are parallel or antiparallel will have a cross product of 0.

ii. Torque is expressed mathematically as the vector product of and .

iii. Because of the used to indicate this type of multiplication, the vector product is also called the cross product.

iv. The vector product of two vectors and is defined to be a vector

whose magnitude equals the area of the parallelogram formed by and .

v. The vector is perpendicular to both and in the direction of the thumb of your right hand if you curl your fingers from the direction of toward the direction of .

vi. If is the angle between and , * and is a unit vector that is perpendicular to both and in the direction of

, the vector product of and is:

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vii. Note that the order in which two vectors are multiplied in a vector product makes a difference.

i. Finding the Vector Product of Two Vectors.1. PICTURE

a. At times it is easier to find a vector product of two vectors by using the equation

. b. At other times it is easier to find the vector

product using the Cartesian components of the two vectors.

2. SOLVEa. The vector product obeys a distributive

law under addition:

b. If and are functions of some variable such as t, the derivative of follows the usual product rule for derivatives:

c. The unit vectors , , and , which are mutually perpendicular, have vector products given by

i. Reversing the order of multiplication gives and

.ii. A tool for remembering this is shown

in Figure 10 - 8. iii. Furthermore,

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d. CHECK. i. Make sure that your vector products

make sense. ii. For example, the vector product of

two vectors is a vector and is perpendicular to each of the two vectors.

iii. In addition, check your work to make certain you did not inadvertently reverse the order of the two vectors being multiplied, and thus create a sign error.

e. TAKING IT FURTHER. i. Any coordinate system for which the

equations above hold is called a right-handed coordinate system.

ii. Only right-handed coordinate systems are used in this book.

j. Example 10 – 1: Vector and Dot Products - If and = 12 find .

III. Torque and Angular Momentum.a. The linear momentum of the particle is . The angular

momentum of the particle relative to the origin is defined to be the vector product of and :

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b. If and are both perpendicular to the z axis, as in Figure 10-10, is parallel to the z axis and is given by

. Like torque, angular momentum is defined relative to a point in space; in this case the angular momentum is defined about the origin.

c. Figure 10-11 shows a particle of mass m attached to a circular disk of negligible mass moving in a circle in the xy plane with its center at the origin. The disk is spinning about the z axis with angular speed ω The speed of the particle and its angular speed are related by .

d. The angular momentum of the particle relative to the center of the disk is

e. Note: In this example the angular momentum vector is in the same direction as the angular velocity vector.

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f. The angular momentum of this particle about a general point on the z axis is not parallel to the angular velocity vector.

i. Figure 10-12 shows the angular momentum vector for the same particle attached to the same disk, but with computed about a point on the z axis that is not at the center of the circle. In this case, the angular momentum is not parallel to the angular velocity vector

, which is parallel to the z axis.

ii. In Figure 10-13, we attach a second particle of equal mass to the spinning disk at a point diametrically opposite this first particle.

1. The angular-momentum vectors and are shown relative to the same point .

2. The total angular momentum of the two-particle system is again parallel to the angular velocity vector .

3. In this case, the axis of rotation, the z axis, passes through the center of mass of the two-particle system, and the mass distribution is symmetric about this axis.

4. Such an axis is called a symmetry axis. 5. For any system of particles that rotates about a

symmetry axis, the total angular momentum (which is the sum of the angular momenta of the individual particles) is parallel to the angular velocity and is given by

g. Example 10 – 2: Angular Momentum About the Origin. Find the angular momentum about the origin for the following situations. (a) A car of mass 1200 kg moves in a circle of radius 20 m with a speed of 15 m/s. The circle is in the xy plane, centered at the origin. When viewed from a point on the positive z axis, the car moves counterclockwise. (b) The same car moves in the xy plane with velocity along the line y = y0

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= 20 m parallel to the x axis. (c) A uniform disk in the xy plane of radius 20 m and mass 1200 kg rotates at 0.75 rad/s about its axis, which is also the z axis. When viewed from a point on the positive z axis, the disk rotates counterclockwise. Model the car as a point particle and the disk as a uniform disk.

h. There are several additional results concerning torque and angular momentum for a system of particles. The first of these is

i. In Equation 10-10, the net external torque about the point is the vector sum of the external torques about that point acting on the system. Integrating both sides of this equation with respect to time gives

j. Equation 10-11 is the rotational analog of

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k. It is often useful to split the total angular momentum of a system about an arbitrary point O into orbital angular momentum and spin angular momentum.

l. Earth has spin angular momentum due to its spinning motion about its rotational axis, and it has orbital angular momentum about the center of the Sun due to its orbital motion around the Sun (Figure 10-17).

i. The total angular momentum of Earth relative to the center of the Sun is the vector sum of the spin and

orbital angular momenta. is the angular momentum of a system about its center of mass, and

is the angular momentum that a point particle of mass M located at the center of mass and moving at the velocity of the center of mass would have. That is,

m. In

Chapter 9, torques are computed about axes instead of about points.

i. The relation between the torque about an axis and the torque about a point is straightforward.

ii. If point O is the origin and if force exerts torque about O, then (the z component of ) is the torque of

about the z axis.iii. Do not confuse torque about a point with torque about

an axis. The torque of a force about the z axis is the z

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component of the torque of the force about any point on the z axis.

iv. Taking components of vector products requires some care. If then

v. (see Figure 10-18) are the vector components of , , and . The vector component in a given direction is the scalar component in that direction times the unit vector in that direction. For example,

Here, is the vector component of in the positive radial direction (directly away from the z axis),

and is the component of perpendicular to the z axis, and thus parallel to the xy plane .

n. The relation between angular momentum about an axis and angular momentum about a point is also straightforward.

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i. If the angular momentum of a point particle about the origin is then the angular momentum of the particle about the z axis is

where is the component of the linear momentum

perpendicular to the z axis ii. Taking the z vector components of both sides of

Equation 10-10 gives

For a symmetric rigid object rotating about the z axis,

where is the moment of inertia about the z axis.

iii. For a system of particles, the total angular momentum about the z axis equals the sum of the angular momenta about the z axis. In addition, the total torque about the z axis is the sum of the external torques about the z axis acting on the system.

iv. Example 10 – 3: The Atwood‘s Machine Revisited - An Atwood‘s machine has two blocks with masses m1 and m2 (m1 > m2) connected by a string of negligible mass that passes over a pulley with frictionless bearings. The pulley is a uniform disk of mass M and radius R. The string does not slip on the pulley. Apply Equation 10-16 to the system consisting of the two blocks, the

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string, and the pulley, to find the angular acceleration of the pulley and the linear acceleration of the blocks.

IV. Conservation of Angular Momentum.a. When the net external torque acting on a system about some

point remains zero, we have

b. Equation 10-20 is a statement of the law of conservation of angular momentum.

c. This law is the rotational analog of the law of conservation of linear momentum.

i. If a system is isolated from its surroundings, so that there are no external forces or torques acting on it, three quantities are conserved: energy, linear momentum, and angular momentum.

ii. The law of conservation of angular momentum is a fundamental law of nature.

d. Although conservation of angular momentum is a law, independent of Newton‘s laws of motion, the fact that the

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internal torques of a system cancel is suggested by Newton‘s third law.

e. Consider the two particles shown in Figure 10-26. i. Let be the force exerted by particle 1 on particle 2

and be the force exerted by particle 2 on particle 1.

ii. By Newton‘s third law, . The sum of the torques exerted by these forces about the origin O is

iii. The vector is along the line joining the two particles.

iv. If acts parallel to the line joining m1 and m2, and are either parallel or antiparallel and

v. If this is true for all the internal forces, the internal torques cancel in pairs.

f. Example 10 – 4: A Rotating Disk - Disk 1 is rotating freely and has an angular velocity about a vertical axis that coincides with its symmetry axis, as shown in Figure 10-27. Its moment of inertia about this axis is I1. It drops onto disk 2, of moment of inertia I2, that is initially at rest. Disk 2 is centered on the same axis as disk 1 and is free to rotate

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about that axis. Because of kinetic friction, the two disks eventually attain a common angular velocity . Find .

g. In the collision of the two disks in Example 10-4, mechanical energy is dissipated.

i. We can see this by writing the energy in terms of the angular momentum.

ii. An object rotating with an angular velocity ω has kinetic energy.

iii. This result is analogous with that for linear motion Equation 8-25.

1. The initial kinetic energy in Example 10-4 is

2. The final kinetic energy is

3. Because the ratio of the final to the initial kinetic energy is

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which is less than one.

4. This interaction of the disks is analogous to a one-dimensional perfectly inelastic collision of two objects.

iv. Example 10 – 5: Mud in Your Eye - You and three of your friends have been bullied for many years by Gene, who has avoided taking physics classes. So you and your three friends who are taking advanced-placement physics decide to teach him a lesson using conservation of angular momentum. Here is your plan. A nearby park has a small merry-go-round (Figure 10-28) with a 3.0-m-diameter turntable that has a 130-kg · m2 moment of inertia. You initially get all five of you to stand on the merry-go-round next to the rim while the merry-go-round is rotating at a modest 20 rev/min. When the signal is given, you and your friends will quickly walk to the center of the merry-go-round, leaving Gene near the rim. The merry-go-round will speed up, throwing Gene off and into the mud. (You plan to do this after a heavy rain.) Gene is very quick and very strong, so throwing him off will require that the centripetal acceleration of the rim be at least 4.0 gs. Will this plan work? (Assume that each person has a mass of 60 kg.)

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1. The moment of inertia of the students–merry-go-round system decreases as the friends walk toward the center.

2. Thus, the system‘s moment of inertia decreases while its angular momentum remains constant.

3. As a result, we can see from Equation 10-21 that the kinetic energy of the students–merry-go-round system increases.

4. The energy for this kinetic energy increase comes from the internal energy of the friends. Walking radially inward, like walking up a steep incline, requires the expenditure of internal energy.

v. Example 10 – 6: Another Ride on the Merry-Go-Round - A 25-kg child in a playground runs with an initial speed of 2.5 m/s along a path tangent to the rim of a merry-go-round, whose radius is 2.0 m. The merry-go-round, which is initially at rest, has a moment of inertia of 500 kg · m2 The child then jumps on (Figure 10-29). Find the final angular velocity of the child and the merry-go-round together.

vi. PRACTICE PROBLEM 10-1 Calculate the initial and final kinetic energies of the child–merry-go-round system.

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vii. Example 10 – 7: Spinning the Wheel Conceptual - You are sitting on a stool on a frictionless turntable holding a bicycle wheel ( Figure 10-30). Initially, neither the wheel nor the turntable is spinning. Following instructions from your teacher, you hold the spin axis of the wheel vertical with one hand, and with your other hand you set the wheel spinning counterclockwise (as viewed from above). Surprise! When you start the wheel spinning one way, the turntable, the stool, you, and the axis of the wheel start rotating in the opposite direction. After a few seconds, you use your free hand to brake the spinning motion of the wheel. You are surprised again when you, the stool, and the wheel axis cease rotating as the wheel ceases spinning. Explain.

viii. Example 10 – 8: Pulling Through a Hole - A particle of mass m moves with speed in a circle of radius on a frictionless tabletop. The particle is attached to a string that passes through a hole in the table, as shown in Figure 10-31. The string is slowly pulled downward until the particle is a distance from the hole, after which the particle moves in a circle of radius . (a) Find the final velocity in terms of , and . (b) Find the tension when the particle is moving in a circle of radius r in terms of m, r, and the angular momentum . (c) Calculate the work done on the particle by the tension force by integrating . Express your answer in terms of r and .

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ix. In Figure 10-33 a puck on a frictionless plane is given an initial speed The puck is attached to a string that wraps around a vertical post.

1. This situation looks similar to Example 10-8, but it is not the same.

2. There is no agent that does work on the puck, nor is there any mechanism for energy dissipation.

3. Thus, mechanical energy must be conserved. 4. Because , where L is the magnitude

of the angular momentum about the axis of the post, is constant and I decreases as decreases, L must also decrease.

a. Note that the tension force does not act toward the axis of the post.

b. The tension force on the puck produces a torque vector about the axis of the post in the downward direction, which reduces the puck‘s angular-momentum vector about the axis, which is in the upward direction.

x. Example 10 – 9: The Ballistic Pendulum Revisited - A thin rod of mass M and length d hangs vertically from a pivot attached to one end. A piece of clay of mass m and moving horizontally at speed hits the rod a distance x from the pivot and sticks to it (Figure 10-34). Find the ratio of the clay–rod system‘s kinetic energy just after the collision to its kinetic energy just before the collision.

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xi. Example 10 – 10: Tipping the Wheel Conceptual - A student sitting on a stool that rests on a turntable with frictionless bearings (Figure 10-35a) is holding a rapidly spinning bicycle wheel. The rotation axis of the wheel is initially horizontal, and the magnitude of the spin-angular-momentum vector of the spinning wheel is . What will happen if the student suddenly tips the axle of the wheel (Figure 10-35b) so that after the rotation the spin axis of the wheel is vertical and the wheel is spinning counterclockwise (when viewed from above)?

V. Proofs (If time).

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As with conservation of energy and conservation of … · Web viewTitle As with conservation of energy and conservation of linear momentum, conservation of angular momentum is one