普通物理－Lecture8－Rotation II

(1)

普通物理

Lecture 8

Rotational Motion (II)

轉動(II) 轉動(II)

(2)



Force vs. Torque

力與扭矩力與扭矩



Torque and Equilibrium

扭矩平衡扭矩平衡



Center of Gravity

重心重心



Center of Gravity

重心重心



Mechanical Equilibrium

力學平衡力學平衡



Work

Work Energy in a Rotating System

Energy in a Rotating System

轉動系統的功與能轉動系統的功與能



Torque and Angular Acceleration

扭矩與角加速度扭矩與角加速度



Work

Work--Energy in a Rotating System

Energy in a Rotating System

轉動系統的功與能轉動系統的功與能



(3)

Force vs. Torque

q



Forces

cause

accelerations



Torques

cause

angular accelerations



Force and torque are related



 The door is free to rotate about The door is free to rotate about

an axis through O an axis through O



 There are three factors that There are three factors that

determine the effectiveness of determine the effectiveness of determine the effectiveness of determine the effectiveness of the force in opening the door: the force in opening the door:

–– The The magnitudemagnitude of the forceof the force

–– The The positionposition of the application of of the application of the force

(4)

Torque

q



 Torque, , is the tendency of a force to rotate an Torque, , is the tendency of a force to rotate an object about some axis

object about some axis



object about some axis object about some axis

––

 is the torqueis the torque

F

r







 is the torqueis the torque

 F is the forceF is the force

–– symbol is the Greek tausymbol is the Greek tau



symbol is the Greek tau symbol is the Greek tau

 r is the length of the position vectorr is the length of the position vector



(5)

Direction of Torque

q



Torque is a vector quantity

Th di ti i

Th di ti i di ldi l t tht th ll

–– The direction is The direction is perpendicular to the plane perpendicular to the plane

determined by the position vector and the force determined by the position vector and the force

–– If the turning tendency of the force is If the turning tendency of the force is

counterclockwise

counterclockwise, the torque will be , the torque will be positivepositive

–– If the turning tendency is If the turning tendency is clockwiseclockwise, the torque will , the torque will be

(6)

Multiple Torques

p

q



When two or more torques are acting on an

object, the torques are added

j

,

q

–– As vectorsAs vectors



If th

t t

i

th

bj t’

t

f



If the net torque is zero, the object’s rate of

rotation doesn’t change

(7)

General Definition of Torque

q



The applied force is not always perpendicular



The applied force is not always perpendicular

to the position vector



The component of the force

perpendicular

to

the object will cause it to rotate

jj



 When the force is parallel to the position vector, no rotation When the force is parallel to the position vector, no rotation

occurs occurs

(8)

General Definition of Torque

q



Taking the angle into account leads to a

more general definition of torque:

g

q







rF

sin

 FF is the forceis the force

 rr is the position vectoris the position vector

 r r is the position vectoris the position vector

(9)

Lever Arm



 The The lever armlever arm, d, is the perpendicular distance from the , d, is the perpendicular distance from the , ,, , p pp p

axis of rotation to a line drawn along the direction of the axis of rotation to a line drawn along the direction of the force

force



(10)

Right Hand Rule

gg

P i h fi i h



 Point the fingers in the Point the fingers in the direction of the direction of the position vector position vector position vector position vector 

 Curl the fingers Curl the fingers toward the force toward the force toward the force toward the force vector

vector 

 The thumb points inThe thumb points in 

 The thumb points in The thumb points in the direction of the the direction of the torque

torque torque torque

(11)

Net Torque

q



The net torque is the sum of all the torques

produced by all the forces

p

y

p

y

–– Remember to account for the direction of the Remember to account for the direction of the tendency for rotation

tendency for rotation tendency for rotation tendency for rotation

 CounterclockwiseCounterclockwise torques are torques are positivepositive

 ClockwiseClockwise torques aretorques are negativenegative

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Torque and Equilibrium

q

Th

Th Fi tFi t C ditiC diti f Ef E ilib iilib i 

 TheThe FirstFirst Condition of EquilibriumCondition of Equilibrium The

The net external force must be zeronet external force must be zero

0 or  F

0 0

x and y

F  F 

–– This is a necessary, but not sufficient, condition to This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical ensure that an object is in complete mechanical equilibrium

equilibrium equilibrium equilibrium

(13)

Torque and Equilibrium

q



To ensure mechanical equilibrium, you need

to ensure rotational equilibrium as well as

q

translational



The

Second

Condition of Equilibrium states



The

Second

Condition of Equilibrium states

–– The The net external torque must be zeronet external torque must be zero

0 

 





0

(14)

Torque and Equilibrium

Th it

q



 The woman, mass m, sits on The woman, mass m, sits on

the left end of the seesaw the left end of the seesaw



 The man mass M sits whereThe man mass M sits where 

 The man, mass M, sits where The man, mass M, sits where

the seesaw will be balanced the seesaw will be balanced



 Apply the Second Condition Apply the Second Condition pp ypp y

of Equilibrium and solve for of Equilibrium and solve for the unknown distance, x

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Example 1

A cook holds a 2.00 kg carton of milk at arm’s length. What force must be exerted by the biceps muscle ?( ignore the weight of the forearm) FB

 y p ( g g ) i h

0 

Require that

0 

 

about an axis through the elbow and perpendicular to the page. This gives

_{2.00 kg 9.80 m s}



2



_{25.0 cm+8.00 cm}  _cos75.0 _{8.00 cm} ₀ B F      _ _         

(16)

Axis of Rotation

If h bj i i ilib i i

If h bj i i ilib i i dd



 If the object is in equilibrium, it If the object is in equilibrium, it does not matterdoes not matter where you put the axis of rotation for calculating where you put the axis of rotation for calculating the net torque

the net torque the net torque the net torque

–– The location of the axis of rotation is completely The location of the axis of rotation is completely arbitrary

arbitrary arbitrary arbitrary

–– Often the nature of the problem will suggest a Often the nature of the problem will suggest a convenient location for the axis

convenient location for the axis

–– When solving a problem, you When solving a problem, you mustmust specify an axis of specify an axis of rotation

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Center of Gravity

Center of Gravityyy



The force of gravity acting on an object

must be considered



In finding the torque produced by the force

of gravity all of the weight of the object can

of gravity, all of the weight of the object can

be considered to be concentrated at a single

point

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Center of Gravity

Center of Gravityyy



 The object is divided The object is divided up into a large number up into a large number of very small particles of very small particles of weight (mg)

of weight (mg) 

 Each particle will have Each particle will have a set of coordinates

a set of coordinates indicating its location indicating its location (x,y)

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Center of Gravity



 We assume the object is free We assume the object is free to rotate about its center

to rotate about its center 

 The torque produced by The torque produced by

each particle about the axis each particle about the axis pp of rotation is equal to its

of rotation is equal to its weight times its lever arm weight times its lever arm

–– For example, For example,

1 1

1



m

gx

(20)

Center of Gravity

Center of Gravityyy



We wish to locate the point of application of

the

single force

g f

whose magnitude is equal to

g

q

the weight of the object, and whose

effect on

the rotation is the same

as all the individual

the rotation is the same

as all the individual

particles.



This point is called the

center of gravity

of

the object

jj

(21)

Center of Gravity

Center of Gravityyy



The coordinates of the center of gravity can

be found from the sum of the torques acting

q

g

on the individual particles being set equal to

the torque produced by the weight of the

object

i i i i

m x

m y

d



_i _i



_{i i} cg cg i i

y

x

and y

m





(22)

Center of Gravity

Center of Gravityyy



The center of gravity of a homogenous,

symmetric body must lie on the axis of

y

symmetry.



Often the center of gravity of such an object



Often, the center of gravity of such an object

is the

(23)

Center of Gravity

Center of Gravityyy



 The wrench is hung freely from The wrench is hung freely from two different pivots

two different pivots two different pivots two different pivots 

 The intersection of the lines The intersection of the lines indicates the center of gravity indicates the center of gravity indicates the center of gravity indicates the center of gravity 

 A rigid object can be balanced A rigid object can be balanced by a single force equal in

by a single force equal in by a single force equal in by a single force equal in

magnitude to its weight as long magnitude to its weight as long as

as the force is acting upward the force is acting upward g pg p through the object’s center of through the object’s center of gravity

(24)

Example 2

Consider the following mass distribution, where x- and y- coordinates are given in meters: 5.0 kg at (0.0,0.0) m, 3.0 kg at (0.0, 4.0) m, and 4.0 kg at (3.0,0.0) m. Where should a fourth object of 8.0 kg be placed so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m?g y j g ( )

(25)

Notes About Equilibrium

q



A

zero net torque

does not mean the

absence of rotational motion

–– An object that rotates at An object that rotates at uniform angular uniform angular velocity

velocity can be under the influence of a zero netcan be under the influence of a zero net

velocity

velocity can be under the influence of a zero net can be under the influence of a zero net torque

torque

 This is analogous to theThis is analogous to the translationaltranslational situation wheresituation where

 This is analogous to the This is analogous to the translationaltranslational situation where situation where a zero net force does not mean the object is not in

a zero net force does not mean the object is not in motion

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Solving Equilibrium Problems

g q



 Draw a diagram of the systemDraw a diagram of the system

–– Include coordinates and choose a rotation axisInclude coordinates and choose a rotation axis



 Isolate the object being analyzed and draw a free Isolate the object being analyzed and draw a free

body diagram showing all the external forces

body diagram showing all the external forces yy gg gg

acting on the object acting on the object

–– For systems containing more than one object, draw a For systems containing more than one object, draw a yy gg jj ,, separate free body diagram for each object

(27)

Problem Solving

Problem Solvinggg

A l th S d C diti f E ilib i



 Apply the Second Condition of EquilibriumApply the Second Condition of Equilibrium

–– This will yield a single equation, often with one This will yield a single equation, often with one

k hi h b l d i di t l

unknown which can be solved immediately unknown which can be solved immediately



 Apply the First Condition of EquilibriumApply the First Condition of Equilibrium

–– This will give you two more equationsThis will give you two more equations



 Solve the resulting simultaneous equations for Solve the resulting simultaneous equations for

all of the unknowns all of the unknowns

(28)

Free Body Diagram

y

g



 Isolate the object to be analyzedIsolate the object to be analyzed 

(29)

Free Body Diagram

y

g

Th f b d di



 The free body diagram The free body diagram includes the directions includes the directions of the forces of the forces of the forces of the forces Fig 8.12, p.228 Slide 17 

 The weights act through The weights act through

th t f it f

the centers of gravity of the centers of gravity of their objects

(30)

Free Body Diagram

y

g



 The free body diagram shows the normal force and The free body diagram shows the normal force and yy gg the force of static friction acting on the ladder at the the force of static friction acting on the ladder at the ground

(31)

Example 3

A 20.0 kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole A cable at an angle of 30 0 degrees with negligible mass that is hinged to a pole. A cable at an angle of 30.0 degrees with the beam helps to support the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by the pole.

30.0°

Tur Hur

(a) Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.

196 N d V ur T sin 30.0  d 196 Nd 0        , giving T  392 N 0 (b)  F_x 0 cos30.0 0 H T  

392 N cos30 0 339 N to the right

H 

(b) From

392 N cos30.0 339 N to the right

H    0 y F   From sin 30.0 196 N 0 V T   

(32)

Torque and Angular Acceleration

q

g



When a rigid object is subject to a

net

torque (≠0),

q

(≠ ),

it undergoes an angular

g

acceleration



The

angular acceleration

is directly



The

angular acceleration

is directly

proportional to the

net torque

–– The relationship is analogous to ∑F = maThe relationship is analogous to ∑F = ma

(33)

Rotational Kinetic Energy

gy



An object rotating about some axis with an

angular speed, ω, has rotational kinetic

g

p

, ,

energy

½Iω

22 



Energy concepts can be useful for



Energy concepts can be useful for

simplifying the analysis of rotational motion

(34)

Total Energy of a System

gy

y



Conservation of Mechanical Energy

(

KE KE PE

) (

KE KE PE

)

R b hi i f i f

(

KE KE PE

_t



_r



_{g i}

) (



KE KE PE

_t



_r



_{g f}

)

–– Remember, this is for conservative forces, no Remember, this is for conservative forces, no dissipative forces such as friction can be present dissipative forces such as friction can be present

–– Potential energies of any other conservative Potential energies of any other conservative forces could be added

(35)

Work

Work--Energy in a Rotating System

Energy in a Rotating System

gy

g y



In the case where there are dissipative forces

such as friction, use the

,,

generalized Work

generalized Work--

gg

Energy Theorem

instead of Conservation of

Energy



(36)

General Problem Solving Hints

gg



The same basic techniques that were used in

linear motion can be applied to rotational

pp

motion.

Analogies:

Analogies: FF becomesbecomesττ mm becomesbecomes II andand aa

–– Analogies: Analogies: FF becomesbecomesττ , , mm becomes becomes II and and aa

becomes

(37)

Problem Solving Hints for Energy

Methods



Choose two points of interest

–– One where all the necessary information is One where all the necessary information is yy given

given

–– The other where information is desiredThe other where information is desiredThe other where information is desiredThe other where information is desired



Identify the conservative and

nonconservative forces

(38)

Problem Solving Hints for Energy

Methods



Write the general equation for the Work

Write the general equation for the

Work--Energy theorem if there are

Energy theorem if there are

gy

nonconservative forces

Use Conservation of Energy if there are no Use Conservation of Energy if there are no

–– Use Conservation of Energy if there are no Use Conservation of Energy if there are no nonconservative forces

nonconservative forces

U

t

bi

t



Use

v = rω

to combine terms



(39)

Example 4

Use conservation of energy to determine the angular speed of the spool after the 3 00 kg bucket has fallen 4 00 m starting the rest The light string attached to the 3.00 kg bucket has fallen 4.00 m, starting the rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

(40)

Angular Momentum

Th t t f li t   i t ti l

gg

The counterpart of linear momentum in rotational motion is a new vector known as angular momentum.

p  mv

The new vector is defined as follows: In the example shown in the fi

r p

     

gure both and r p

lie in the -plane. Using the right hand rule we can see that the direction of is along the -axis

xy

z

 

can see that the direction of is along the axis. The magnitude of angular momentum sin ,

z rmv        

(41)

Angular Momentum

A l t d d th h i f th i i O If th i i N t

gg

2

Angular momentum depends on the choice of the origin O. If the origin is shifted in general we get a different value of

Note:

 

2

SI unit for angular momentum: kg.m /s Sometimes the equivalent J.s is used





r



p

m r v



  

 



  

r

p

m r v









mv

r

_





(42)

Angular Momentum

Newton's Second Law in Angul r Forma

gg

Newton's second law for linear motion has the form: F_net dp . Below we

dt

  





will derive the angular form of Newton's second law for a particle.

d d m r v  m r   _{ } _ _ 



v



m r_  dv  dr v _ m r



  a v v



        





m r v m r dt dt     











 







0 v m r v m r a v v dt dt d F   _    _            ₀

_

 

_{ }

 

_







 net net v v m r dp a r ma r F t d d                 p 



(43)

Angular Momentum

gg

z m₁ m ℓ₁ ℓ₃ O m₃ m₂ ℓ₂ x y x y

The Angular Momentum of a System of Particles

We will now explore Newton's second law in l f f t f ti l th t

g

h

y

angular form for a system of n particles that have angular momentum , , ,...,g   _{  }, , , , _

(44)

Angular Momentum

1 2 3

The angular momentum of the system is L L       ... _n 



n _i

gg

1

The time derivative of the angular momentum is =

i n i i d dL dt dt  



   1

The time derivative for the angular momentum of the i-th part

i dt _ dt , icle i net i d dt     ,

Where is the net torque on the particle. This torque has contributions

from external as well as internal forces between the particles of the system. Thus

net i  , net i dL dt      1

Here is the net torque due to all the external forces.

n net net i   



 

(45)

Angular Momentum

Angular Momentum of a Rigid Body Rotating about a Fixed Axis

gg

We take the z-axis to be the fixed rotation axis. We will determine the z-component of the net angular momentum. The body is

di id d l t f  th t h iti t  divided n elements of mass that have a position vector The angula

i i

m r

 

 

r momentum of the i-the element is:

I i d i 90 Th i i  ri pi     _{ }        

Its magnitude sin90 = The z-compoment of is: sin sin

i i i i i i iz i iz i i i i i i i r p r m v r m v r m v   _             

The z-component of the angular momentu

  2 m is the sum:_z n n n n z iz i i i i i i i i L L   r_ m v  r_ m r_ _ m r_ _  







 



1 1 1 1 2

The sum is the rotational inertia of the rigid body

z iz i i i i i i i i i i i i n i i m r I              



(46)

Example 5

A light rigid rod 1.00 m in length rotates about an axis perpendicular to its length and through its center. Two particles of masses 4.00 kg and 3.00 kg are connected g p g g to the ends of the rod. What is the angular momentum of the system if the speed of each particle is 5.00 m/s? ( Neglect the rod’s mass.)

(47)

Angular Momentum

gg



 Similarly to the relationship between Similarly to the relationship between force and force and

momentum

momentum in a linear system, we can show the in a linear system, we can show the relationship between

relationship between torque and angular torque and angular momentum

momentum 

 Angular momentum is defined as Angular momentum is defined as

–– L = I ωL = I ω –– and and

 





L

t







(48)

Angular Momentum

gg



 If the If the net torque is zeronet torque is zero, the angular momentum , the angular momentum remains

remains constantconstant



 Conservation of Angular MomentumConservation of Angular Momentum states: states: The The

angular momentum of a system is conserved when

angular momentum of a system is conserved when gg yy

the net external torque acting on the systems is the net external torque acting on the systems is zero.

zero.

(49)

Conservation Rules



In an

isolated system

, the following quantities

are conserved:

–– Mechanical energyMechanical energy Linear momentum Linear momentum

–– Linear momentumLinear momentum

(50)

Conservation of Angular

Momentum

Th fi h t d t t d

E l The figure shows a student seated

on a stool that can rotate freely about a

i l i Th d h h b i

Example:

i

vertical axis. The student who has been set into rotation at an initial angular speed , holds two dumbbells in his outstretched hands. His angular momentum vector lies along the L

(51)

Conservation of Angular

Momentum

The student then pulls in his hands as shown in fig b This action reduces the The student then pulls in his hands as shown in fig.b. This action reduces the rotational inertia from an initial value to a smaller final value .

No net external torque acts on the student st

i f

I I

ool system Thus the No net external torque acts on the student-stool system. Thus the angular momentum of the system remains unchanged.

Angular momentum at :t L  I Angular momentum at :t L I 

Angular momentum at : Angular momentum at : Since 1 i i i i f f f f i i i f i i f f f i f f f i f t L I t L I I I L L I I I I I I                   _i f f I I

(52)

Conservation of Angular

Momentum



 With hands and feet With hands and feet drawn closer to the drawn closer to the body, the skater’s body, the skater’s angular speed angular speed increases increases –– LL is is conservedconserved, , II decreases

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Example 6

The system of small objects is rotating at an angular speed of 2.0 rev/s. The objects are connected by light flexible spokes that can be lengthened or objects are connected by light, flexible spokes that can be lengthened or

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Analogies between

Translational & Rotational Motions

Rotational Motion

Translational Motion

Translational & Rotational Motions

x v     2 2 a p     2 2 2 2 mv K m I K I      m F ma I   F I       F  

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Rolling

Rolling as Translation移動 and rotation轉動Combined

Consider an object with circular

t₁ = 0 t₂ = t

Consider an object with circular cross section that rolls along a surface without slipping. This surface without slipping. This motion, though common, is

complicated. We can simplify its study by treating it as a

combination of translation of the center of mass and rotation of

center of mass and rotation of the object about the center of mass

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Rolling

Consider the two snapshots of a rolling bicycle wheel shown in the Consider the two snapshots of a rolling bicycle wheel shown in the figure.

An observer stationary with the ground will see the center of mass O

f th h l f d ith d Th i t P t hi h

of the wheel move forward with a speed ν_com. The point P at which the wheel makes contact with the road also moves with the same speed. During the time interval t between the two snapshots both Op g p and P cover a distance s.

(eqs.1)

During t the bicycle rider sees the wheel rotate by an angle θ about O so that

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Rolling

We have seen that rolling is a combination of purely translational We have seen that rolling is a combination of purely translational motion with speed ν_com and a purely rotational motion about the center of mass with angular velocity . The velocity of each g y y point is the vector sum of the velocities of the two motions.

com

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Rolling

For the translational motion the velocity vector is the same for every point ( , see fig.b). the rotational velocity varies from point to point. Its magnitude is equal to ωr where r is the distance

f th i t f O It di ti i t t t th i l bit

of the point from O. Its direction is tangent to the circular orbit (see fig.a). The net velocity is the vector sum of these two terms. For example the velocity of point P is always zero. The velocity of For example the velocity of point P is always zero. The velocity of the center of mass O is (r = 0). Finally the velocity of the top point T is equal to .

v



R



v



R



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Rolling as Pure Rotation

A

v

v_T _{Another way of looking at rolling is}

shown in the figure

B

v_A

v_O We consider rolling as a _{about an axis of rotation that passes}pure rotation

through the contact point P between

v_B through the contact point P between

the wheel and the road. The angular velocity of the rotation is

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Rolling as Pure Rotation

In order to define the velocity vector for each

A

v_A

v_T y

point we must know its magnitude as well as its direction. The direction for each point on

B

A

v_B

v_O the wheel points along the tangent to its

circular orbit.

For example at point A the velocity vector

v_B _{For example at point A the velocity vector}

is perpendicular to the dotted line that

connects point A with point P. The speed of p p p each point is given by: ν=ωr. Here r is the distance between a particular point and the

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Example 7

An automobile travelling at 80.0 km/h has tires of 75 cm diameter. (a) What is the angular speed of the tires about their axles?

(a) What is the angular speed of the tires about their axles?

(b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires

(without skidding), what is the magnitude of the angular acceleration of the wheels? (c) Har far does the car move during the braking?

(c) Har far does the car move during the braking? The initial speed of the car is

80 km/h (1000 m/km)(1 h/3600 s) 22.2 m/s

v  

The tire radius is R = 0 750/2 = 0 375 m

com0 22.2 m/s

59 3 d/

v

The tire radius is R 0.750/2 0.375 m.

(a) The initial speed of the car is the initial speed of the center of mass of the tire, so

0 59.3 rad/s.

0.375 m

R

   

(b) With θ = (30.0)(2 ) = 188 rad and ω = 0,



  2 2 2 2 0 (59.3 rad/s) 2 9.31 rad/s . 2 188 rad           

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The Kinetic Energy of Rolling

Consider the rolling object g j

shown in the figure. It is easier to calculate the kinetic energy of gy the rolling body by considering the motion as pure rolling about p g the contact point P. The rolling

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The Kinetic Energy of Rolling

2

1

Th ki ti K i th i b th ti K I 2 H I i th The kinetic energy is then given by the equation: . Here is the

2

rotational inertia of the rolling body about point P. We can determine using

P P P K K I I I   2 g y p g

the parallel axis theorem.

P P com I  I MR  1



2



2 2 com K  I MR 



2



2 2 2 2 2 1 1 1 2 com 2 com 2 K  I MR   I   MR  2 2 2

The expression for the kinetic energy consists of two terms. The first term corresponds to the rotation about the center of mass O with angular velocity corresponds to the rotation about the center of mass O with angular velocity . The second term is associated with the kinetic energy due to the translational

1

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Friction & Rolling

When an object rolls with constant speed (see top figure) it has no tendency to slide at the contact point P and thus no frictional force acts there. If a net force acts on the rolling body it results in a non-zero acceleration a_com

for the center of mass (see lower figure). If the rolling object accelerates to the right it has the tendency to slide

com

object accelerates to the right it has the tendency to slide at point P to the left. Thus a static frictional force f_s

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Friction & Rolling

The rolling condition results in a connection beteen the magnitude of the acceleration of the center of mass and its angular acceleration

We take time derivetives of both sides

com a v R a      dvcom d R  R  

We take time derivetives of both sides

com com

v  R  a  R R

dt  dt  

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Rolling down a Ramp

Consider a round uniform body of mass M and radius R

a_com

Consider a round uniform body of mass and radius rolling down an inclined plane of angle . We will

calculate the acceleration of the center of mass

M R

a



calculate the acceleration of the center of mass along the x-axis using

com

a

Newton's second law for the translational and rotational motion

translational and rotational motion

Newton's second law for motion along the -axis: sin (eqs.1) Newton's second law for rotation about the center of mass:

s com x f Mg Ma Rf I        Newton s second law for rotation about the center of mass:

We substitute in the second equation and g

s com com Rf I a R         et: com s com a Rf I R   

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Example 7

The figure gives the speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down at a 30 degree ramp The scale on the velocity axis is set

Example 7

that rolls smoothly down at a 30 degree ramp. The scale on the velocity axis is set by v_s=4.0 m/s. What is the rotational inertia of the object?

a = – 3.5 m/s2 θθ = 30º M = 0 50 kg and R = 0 060 m θθ 30 , M 0.50 kg and R 0.060 m sin g  2 sin 1 com com g a I MR     MR I = 7 2  10-4 kg m2 I = 7.2  10-4 kg.m

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Rolling down a Ramp

2 2 Cylinder Hoop MR I I MR a_com 1 2 1 ₂ 2 ₂ 2 sin sin 1 / 1 / I I MR g g a a I MR I MR       1 ₂ 2 ₂ 1 2 1 ₂ 1 / 1 / sin 1 / 2 I MR I MR g a MR      ₂ 2 ₂ ₂ sin 1 / g a MR MR MR    1MR / 2 1 2 1 / sin sin 1 1/ 2 1 1 MR MR MR g g a  a      

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The Yo

The Yo--Yo

Yo

C id f M di R d l di R

Consider a yo-yo of mass , radius , and axle radius rolling down a string . We will calculate the acceleration

of the center of its mass along the axis using Newton's

o

M R R

a of the center of its mass along the -axis using Newton s y

second law

com

a y

for the tralational and rotational motion as we did in the previous problem

a_com

in the previous problem

Newton's second law for motion along the -axis: (eqs.1) com y Mg T  Ma y ( q )

Newton's second law for rotation about the center of mass:

com

g

Angular acceleration acom

R T I

   . Angular acceleration   We substitute in the second equation and get:

o com o R T I R       

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The Yo

The Yo--Yo

Yo

2 (eqs.2) We substitute from equation 2 into equation 1

com com a T I ₂ T  2 ( q ) q q com o c com om com com R a Mg I Ma a R g I     a_com 2 2 1 com com com c o om o R I MR  y

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普通物理－Lecture8－Rotation II

普通物理

Lecture 8

Rotational Motion (II)

Contents





Force vs. Torque

Force vs. Torque





Torque and Equilibrium

Torque and Equilibrium





Center of Gravity

Center of Gravity





Center of Gravity

Center of Gravity





Mechanical Equilibrium

Mechanical Equilibrium





Work

Work Energy in a Rotating System

Energy in a Rotating System





Torque and Angular Acceleration

Torque and Angular Acceleration





Work

Work--Energy in a Rotating System

Energy in a Rotating System



Force vs. Torque

Force vs. Torque

q

q

Forces

Forces

Forces

Forces

cause

cause

cause

cause

accelerations

accelerations

accelerations

accelerations

Torques

Torques

cause

cause

angular accelerations

angular accelerations

Force and torque are related

Force and torque are related

Torque

Torque

q

q



F

r







Direction of Torque

Direction of Torque

q

q

Torque is a vector quantity

Torque is a vector quantity