General ProblemsGeneral Problems

( ( ))

rad/s rad/s 3 3 ..

3 3 rad/s rad/s 27 27 ..

3 3 ss

60 60

min min 1 1 rev rev rad rad 2 2^π^π min min rev 700 rev 700 m

m 0.064 0.064

m m 0.050 0.050 m/s

m/s 9.81 9.81 2 2

2 2

2 2 p

== ==

⎠ ⎟⎟⎟⎟

⎠ ⎞

⎜⎜⎜⎜ ⎞

⎝

⎛

⎛ ×× ××

==

ω ω

((bb) Express the speed of the center) Express the speed of the center of mass in terms of its angular speed of mass in terms of its angular speed of precession:

of precession:

(

( ))( ( ))

cm/s cm/s 16 16

rad/s rad/s 3.27 3.27 m

m 0.050 0.050

p p cm

==

D D vv

((cc) Relate the acceleration of the) Relate the acceleration of the center of mass to its angular speed of center of mass to its angular speed of precession:

precession:

(

( ))( ( ))

2 2 2

2 p p cm

cm/s cm/s 54 54 m/s

m/s 0.535 0.535

rad/s rad/s 3.27 3.27 m

m 0.050 0.050

==

D Dω ω a

((d d ) Use Newton’s second law to) Use Newton’s second law to relate the

relate the vertical component vertical component of theof the force exerted by the pivot to the force exerted by the pivot to the weight of the disk:

weight of the disk:

( ( )) ( ( ))

N N 25 25

m/s m/s 9.81 9.81 kg

kg 2.5

2.5 ²²

v v

==

Mg Mg F

Relate the horizontal component of Relate the horizontal component of the force exerted by the pivot on the the force exerted by the pivot on the axle to the acceleration of the center axle to the acceleration of the center of mass:

of mass:

( ( )) ( ( ))

N N 3 3 ..

1 1

m/s m/s 535 535 ..

0 0 kg kg 2.5

2.5 ²²

cm cm H

==

Ma Ma F

General Problems General Problems

71 • • [SSM][SSM] A partA particle icle whose whose mass mass is is 3.0 3.0 kg mkg moves ioves in tn thehe xy xy plane with plane with velocity

velocity_r_r vv^r^r

== ( (

33..0 0 mm // ss

))

ii^ˆˆ along the linealong the line y y = 5.3 m. ( = 5.3 m. (aa) Find the angular momentum) Find the angular momentum L

L_r_r about the origin when the particle is at (12 m,about the origin when the particle is at (12 m, 5.3 m). (5.3 m). (bb) A force) A force F

F = (–3.9 N)= (–3.9 N) ˆˆii is applied to the particle. Find the torque about the origin due tois applied to the particle. Find the torque about the origin due to this force as the particle passes through the

this force as the particle passes through the point (12 m, 5.3 m).point (12 m, 5.3 m).

Picture the Problem

Picture the Problem While the 3-kg particle is moving in a straight line, it hasWhile the 3-kg particle is moving in a straight line, it has angular momentum given by

angular momentum given by L L r r ^r^r p p^r^r

== ××

wherewhere r r ^r^ris its position vector andis its position vector and p p^r^r is itsis its linear

linear momentum. momentum. The The torque torque due due to to the the applied applied force force is is given given by by ^τ^τ r r F F ..

r r r r r

== ××

((aa) The angular momentum of the) The angular momentum of the particle is given by:

particle is given by:

p p r r L

L ^r^r ^r^r

== ××

Express the vectors

Substitute for

Substitute for r r ^r^randand p p^r^r:and simplify:and simplify to find

to find L L

((bb) Using its definition, express the) Using its definition, express the torque due to the force:

torque due to the force:

Substitute for

Substitute for r r ^r^randand F F

r r

and simplify to and simplify to find angular momentum and net torque, about the origin, acting on the particle.

angular momentum and net torque, about the origin, acting on the particle.

Picture the Problem

Picture the Problem The angular momentum of the particle is given byThe angular momentum of the particle is given by

== ××

wherewhere r r ^r^ris its position vector andis its position vector and p p^r^r is its linear momentum. The torque is its linear momentum. The torque aaccttiinng g oon n tthhe e ppaarrttiicclle e iis s ggiivveen n bbyy^τ^τ d d L L dt dt ..

r r r r

==

The angular momentum of the The angular momentum of the particle is given by:

particle is given by:

Substitute for

Substitute for mm andr r ^r^rand dt dt d d r r ^r^r

and simplify to find and simplify to find L L

Find the net torque due to the force:

( ( ))

[ [ ]]

( ( ))

^k^k

k L k

τ L

m ˆˆ m N N 72 72

ss ˆˆ JJ 72

net 72

net

⋅⋅

==

⋅⋅

==

t t

dt dt d d dt dt d d

r r

73 ••

73 •• Two Two ice ice skaters, skaters, whose whose masses masses are are 55 55 kg kg and and 85 85 kg, kg, hold hold hands hands andand rotate about a vertical axis that passes between them, making one revolution in rotate about a vertical axis that passes between them, making one revolution in 2.5 s. Their centers of mass are separated by 1.7 m and their center of mass is 2.5 s. Their centers of mass are separated by 1.7 m and their center of mass is stationary. Model each skater as a point particle and find (

stationary. Model each skater as a point particle and find (aa) the angular) the angular momentum of the system about their ce

momentum of the system about their center of mass and (nter of mass and (bb) the total kinetic) the total kinetic energy of the system.

energy of the system.

Picture the Problem

Picture the Problem The ice skaters rotate about their center of mass; a point weThe ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by

angular momentum of the system is then given by L L

= =

I I _cm_cm and its kinetic energy and its kinetic energy can

can be be found found fromfrom K K

= =

L L²²

( (

22 I I _cm_cm

))

((aa) Express the angular momentum) Express the angular momentum of the system about the center of of the system about the center of mass of the skaters:

mass of the skaters:

I cm

I L L

= =

Using its definition, locate the center Using its definition, locate the center of mass, relative to the 85-kg skater, of mass, relative to the 85-kg skater, of the system:

of the system:

(

( ))( ( ) ) ( ( ))( ( ))

m m 0.668 0.668

kg kg 85 85 kg kg 55 55

0 0 kg kg 85 85 m

m 1.7 1.7 kg kg 55 55

cm cm

==

++

== ++

x x

Calculate

Calculate I I ::_cm_cm

( ( ))( ( ))

(

( ))( ( ))

2 2

2 2 cm

m m kg kg 5 5 ..

96 96

m m 0.668 0.668 kg

kg 85 85

m m 0.668 0.668 m

m 1.7 1.7 kg kg 55 55

⋅⋅

==

++

−−

==

I I

Substitute to determine Substitute to determine L L::

( ( ))

ss kJ kJ 24 24 ..

0 0 ss JJ 243 243

rev rev rad rad 2 2^π^π ss 2.5 2.5 rev rev 1 m 1 m kg kg 96.5

96.5 ²²

⋅⋅

==

⋅⋅

==

⎠ ⎟⎟⎟⎟

⎠ ⎞

⎜⎜⎜⎜ ⎞

⎝

⎛

⎛ ××

⋅⋅

==

L L

((bb) Relate the total kinetic energy of) Relate the total kinetic energy of the system to its angular momentum the system to its angular momentum and evaluate

and evaluate K K ::

cm cm 2 2

2 2 I I

L K L K

==

Substitute numerical values and Substitute numerical values and evaluate

evaluate K K ::

( ( ))

( (

^96.5^96.5^kg^kg ^m^m

))

⁰⁰^..³¹³¹^kJ^kJ

2 2

ss JJ 243 243

2 2 2 2

⋅⋅ ==

== ⋅⋅

K K

74 ••

74 •• A A 2.0-kg 2.0-kg ball ball attached attached to to a a string string whose whose length length is is 1.5 1.5 m m movesmoves counterclockwise (as viewed from above)

counterclockwise (as viewed from above) in a horizontal circle (Figure 10-56).in a horizontal circle (Figure 10-56).

The string makes an angle

The string makes an angle θ θ = 30° with the vertical. ( = 30° with the vertical. (aa) Determine both the) Determine both the horizontal and vertical components of the angular momentum

horizontal and vertical components of the angular momentum L L

r r

of the ball about of the ball about the point of support

the point of support P P . (. (bb) Find the magnitude of) Find the magnitude of d d L L dt dt

r r

and verify that it equals and verify that it equals the magnitude of the torque exerted by gravity about the point of support.

the magnitude of the torque exerted by gravity about the point of support.

Picture the Problem

Picture the Problem Let the origin ofLet the origin of the coordinate system be at the pivot.

the coordinate system be at the pivot.

The diagram shows the forces acting on The diagram shows the forces acting on the ball. We’ll apply Newton’s second the ball. We’ll apply Newton’s second law to the ball to determine its speed.

law to the ball to determine its speed.

We’ll then use the derivative of its We’ll then use the derivative of its position

position vector vector to to express express its its velocityvelocity and the definition of angular and the definition of angular momentum to show that

momentum to show that L L

r r

has both has both horizontal and vertical components. We horizontal and vertical components. We can use the derivative of

can use the derivative of L L

r r

with with respect to time to show that the rate at respect to time to show that the rate at which the angular momentum of the which the angular momentum of the ball

ball changes changes is is equal equal to to the the torque,torque, relative to the pivot point, acting on it.

relative to the pivot point, acting on it.

θ θ

x x y

z z

m m

r r

P P

T T

g m g m

((aa) Express the angular momentum) Express the angular momentum of the ball about the point of support:

of the ball about the point of support:

vv r r p

p r r L

L ^r^r ^r^r ^r^r ^r^r

== ×× ==

××

(1)(1)

Apply Newton’s second law to the Apply Newton’s second law to the ball:

ball:

∑ ∑

^F^F^x^x

⁼⁼

^T^T^sin^sin^θ^θ

⁼⁼

^m^m_r_r_sin_sin^vv²²_θ_θ

and

∑ ∑

^F^F^z^z

⁼⁼

^T^T^cos^cos^θ^θ

⁻⁻

^mg^mg

⁼⁼

⁰⁰

Eliminate

Eliminate T T between these equations between these equations and solve for

and solve for vv to obtain: to obtain:

θ θ θ θ tantan sin sin rg rg vv

= =

Substitute numerical values and Substitute numerical values and evaluate

evaluate vv::

( ( )) ( ( ))

m/s m/s 2.06 2.06

tan30 tan30 sin30

sin30 m/s

m/s 9.81 9.81 m

m 1.5

1.5 ²²

==

°°

==

Express the position vector of the Express the position vector of the ball:

ball:

( ( )) ( ( ))

( ( ))

k k

j j i

i r

30 ˆˆ 30 cos cos m m 5 5 ..

1 1

sin ˆˆ ˆˆ sin cos

cos 30 30 sin sin m m 5 5 ..

1 1

°°

−−

++

°°

==

ω ω t t ω ω t t

r r

The velocity of the ball is:

( ( )) ( (

ⁱⁱ ^j^j

))

Evaluating ω ω yields: yields:

( (

11..55mm

))

sinsin3030 ²²^..⁷⁵⁷⁵^rad/s^rad/s

Substitute for

Substitute for ω ω to obtain: to obtain: ^vv^r^r

== ( (

22..0606m/sm/s

)) −−

sinsinω ω t t ⁱⁱˆˆ

++

coscosω ω t t ^j^j^ˆˆ Substitute in equation (1) and evaluate

Substitute in equation (1) and evaluate L L

The horizontal component of The horizontal component of L L

r r

is the component in the

is the component in the xy xy plane: plane:

(

( ) )

ⁱⁱ

( ( ))

^j^j

L^r^r_hor_hor

==

55..44JJ

⋅⋅

ss coscosω ω t t ˆˆ

++

55..44JJ

⋅⋅

ss sinsinω ω t t ˆˆ

The vertical component of The vertical component of L L

((bb) Evaluate) Evaluate dt

Evaluate the magnitude of Evaluate the magnitude of

Express the magnitude of the torque Express the magnitude of the torque exerted by gravity about the point of exerted by gravity about the point of support:

Substitute numerical values and Substitute numerical values and evaluate

75 ••

75 •• A A compact compact object object whose whose mass mass isis mm resting on a horizontal, frictionless resting on a horizontal, frictionless surface is attached to a string that wraps around a vertical cylindrical post

surface is attached to a string that wraps around a vertical cylindrical post attached to the surface. Thus,

attached to the surface. Thus, when the object is set into motion, when the object is set into motion, it follows a pathit follows a path that spirals inward. (

that spirals inward. (aa) Is the angular momentum of the o) Is the angular momentum of the object about the axis ofbject about the axis of the post conserved? Explain your answer. (

the post conserved? Explain your answer. (bb) Is the energy of the object) Is the energy of the object conserved? Explain your answer. (

conserved? Explain your answer. (cc) If the speed of the ob) If the speed of the object isject is vv00 when the when the unwrapped length of the string is

unwrapped length of the string is r r , what is its speed when the unwrapped length, what is its speed when the unwrapped length has shortened to

has shortened to r r /2?/2?

Picture the Problem

Picture the Problem The pictorialThe pictorial representation depicts the object representation depicts the object rotating counterclockwise around the rotating counterclockwise around the cylindrical post. Let the system be the cylindrical post. Let the system be the object. In Part (

object. In Part (aa) we need to decide) we need to decide whether a net torque acts on the object whether a net torque acts on the object and in Part (

and in Part (bb) the issue is whether any) the issue is whether any external forces act on the object. In external forces act on the object. In Part (

Part (cc) we can apply the definition of) we can apply the definition of kinetic energy to find the speed of the kinetic energy to find the speed of the object when the unwrapped length has object when the unwrapped length has shortened to

shortened to r r /2./2.

R R

m m

r r

T T

((aa) The net torque acting on the) The net torque acting on the object is given by:

object is given by: RT RT

dt dt dL dL

==

net

==

τ net

Because

Because τ τ netnet

≠≠

0, angular momentum is not conserved. 0, angular momentum is not conserved.

((bb) Because, in this frictionless environment, the net external force acting on the) Because, in this frictionless environment, the net external force acting on the object is the tension force and it acts at right angles to the object’s velocity, the object is the tension force and it acts at right angles to the object’s velocity, the energy of the object is conserved.

energy of the object is conserved.

((cc) Apply conservation of mechanical) Apply conservation of mechanical energy to the object to obtain:

energy to the object to obtain:

0 0

Δ Δ Δ

Δ Δ

Δ E E

==

K K

++

U U

==

or, because

Δ Δ

U U = 0, = 0, 0

Δ K K rotrot

==

Substituting for the kinetic energies Substituting for the kinetic energies yields:

yields:

2 0

2 0 2 0 2 1 2 1 2 2 2 1

1 I' I' ω ω ' '

−−

I I ω ω

==

or or

2 0

2 0 0 2

−−

ω ω

==

ω ω ' ' I I I'

Substitute for

Solving for

Solving for vv

′′

yields: yields:

length L L, and a moment of inertia, and a moment of inertia ML ML²²/10. Inside the cylinder are two disks each/10. Inside the cylinder are two disks each of mass

of mass mm, separated by a distance, separated by a distance^l^l, and tied to a central post by a thin string., and tied to a central post by a thin string.

The system can rotate about a vertical axis through the center of the cylinder. You The system can rotate about a vertical axis through the center of the cylinder. You are designing this cylinder-disk apparatus to shut down

are designing this cylinder-disk apparatus to shut down the rotations when thethe rotations when the strings break by triggering an electronic

strings break by triggering an electronic

″″

shutoff shutoff

″″

signal (sent to the rotating signal (sent to the rotating motor) when the disks hit the ends of the cylinder. During development, you motor) when the disks hit the ends of the cylinder. During development, you notice that with the system rotating at some critical angular speed

notice that with the system rotating at some critical angular speed ω ω , the string, the string suddenly breaks. When the disks reach the ends of the cylinder, they stick. Obtain suddenly breaks. When the disks reach the ends of the cylinder, they stick. Obtain expressions for the final angular speed and the initial and final kinetic energies of expressions for the final angular speed and the initial and final kinetic energies of the system. Assume that the inside walls of the cylinder are frictionless.

the system. Assume that the inside walls of the cylinder are frictionless.

Picture the Problem

Picture the Problem Because the net torque acting on the system is zero; we canBecause the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular use conservation of angular momentum to relate the initial and final angular velocities of the system. See Table

velocities of the system. See Table 9-1 for the moment of inertia of a 9-1 for the moment of inertia of a disk.disk.

Using conservation of angular Using conservation of angular momentum, relate the initial and momentum, relate the initial and final angular speeds to the initial and final angular speeds to the initial and final moments of inertia:

final moments of inertia:

Solving for

Solving for _f_fyields:yields:

Use the parallel-axis theorem to Use the parallel-axis theorem to express the moment of inertia of express the moment of inertia of each of the disks with respect to the each of the disks with respect to the axis of rotation:

axis of rotation:

( ( ))

Express the initial moment of inertia Express the initial moment of inertia I

I ii of the cylindrical tube plus of the cylindrical tube plus disksdisks system:

When the disks have moved out to When the disks have moved out to the end of the cylindrical tube:

the end of the cylindrical tube:

( (

²² ²²

))

Substitute for

Substitute for I I ii and and I I f f in equation (1) in equation (1) and simplify to obtain:

and simplify to obtain:

( ( ))

The initial kinetic energy of the The initial kinetic energy of the system is:

system is:

Substituting for

Substituting for I I ii and simplifying and simplifying yields:

The final kinetic energy of the The final kinetic energy of the system is:

system is:

Substitute for

Substitute for I I f f and and ω ω f f and simplify to obtain:and simplify to obtain:

( ( ))

77 •• [SSM][SSM] Repeat Repeat Problem Problem 76, 76, this this time time friction friction between tbetween the he disks disks andand the walls of the cylinder is not negligible. However, the coefficient of friction is the walls of the cylinder is not negligible. However, the coefficient of friction is not great enough to prevent the disks from reaching the ends of the cylinder. Can not great enough to prevent the disks from reaching the ends of the cylinder. Can the final kinetic energy of the

the final kinetic energy of the system be determined without knowing thesystem be determined without knowing the coefficient of kinetic friction?

coefficient of kinetic friction?

Determine the Concept

Determine the Concept Yes.Yes. The solution depends only upon conservation ofThe solution depends only upon conservation of angular momentum of the system, so it depends only upon the initial and final angular momentum of the system, so it depends only upon the initial and final moments of inertia.

moments of inertia.

78 ••

78 •• Suppose Suppose that that in in Figure Figure 10-5710-57^l^l = 0.60 m, = 0.60 m, L L = 2.0 m, = 2.0 m, M M = 0.80 kg, and= 0.80 kg, and m

m = 0.40 kg. The string breaks when the system’s angular speed approaches the = 0.40 kg. The string breaks when the system’s angular speed approaches the critical angular speed

critical angular speed ω ω ii, at which time the tension in the string is 108 N. The, at which time the tension in the string is 108 N. The masses then move radially outward until they

masses then move radially outward until they undergo perfectly inelasticundergo perfectly inelastic collisions with the ends of the cylinder. De

collisions with the ends of the cylinder. Determine the critical angular speed andtermine the critical angular speed and the angular speed of the system after the inelastic collisions. Find the total kinetic the angular speed of the system after the inelastic collisions. Find the total kinetic

energy of the system at the critical angular speed, and again after the inelastic energy of the system at the critical angular speed, and again after the inelastic collisions. Assume that the inside walls of the cylinder are frictionless.

collisions. Assume that the inside walls of the cylinder are frictionless.

Picture the Problem

Picture the ProblemBecause the net torque acting on the system is zero; we canBecause the net torque acting on the system is zero; we can use conservation of angular

use conservation of angular momentum to relate the initial and final angularmomentum to relate the initial and final angular speeds of the system.

speeds of the system.

final moments of inertia:

Express the tension in the string Express the tension in the string as a function of the

as a function of the criticalcritical angular speed of the system:

angular speed of the system:

Substitute numerical values and Substitute numerical values and evaluate

evaluate _ii::

( ( ))

Substitute numerical values and Substitute numerical values and evaluate

evaluate I I ii::

(

Substitute numerical values and Substitute numerical values and evaluate

Substitute numerical values in Substitute numerical values in equation (1) and evaluate

equation (1) and evaluate ω ω f f ::

( ( ))

The total kinetic energy of the The total kinetic energy of the system at the critical angular speed

In document Cap10 (Page 62-82)