# 10. Rotational Motion ??

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```					         10. Rotational Motion        轉動

1.   Angular Velocity & Acceleration       角速度和加速度
2.   Torque                                力距
3.   Rotational Inertia & the Analog of Newton’s Law
轉動慣量和牛頓定律的比照
4.   Rotational Energy                     轉動能
5.   Rolling Motion                        滾動
Examples of rotating objects:

• Planet Earth.             地球
• Wheels of your bike.      你腳踏車的輪子
• DVD disc in the player.   放映機內的DVD碟
• Circular saw.             圓鋸
• Pirouetting dancer.       腳尖旋轉中的舞者
• Spinning satellite.       自轉中的衛星

How should you engineer the blades
so it’s easiest for the wind to get the
turbine rotating?
你應該怎樣設計旋槳才能讓渦輪最容
易被風轉動？

Ans. blade mass toward axis
答：旋槳的質量朝軸心累積
10.1. Angular Velocity & Acceleration
Polar coord ( r,  )
ˆ
角速度和加速度
θ              ˆ
r
Average angular velocity      平均角速度


ω       ω  ω
ˆ    ˆ
t
 = angular displacement 角位移
( positive if CCW 逆時針為正)                          逆時針

 in radians                              ˆ
ω // rotational axis   轉軸
ˆ
ω
1 rad = 360 / 2  = 57.3
(Instantaneous) angular velocity   (瞬時)角速度
ˆ
ω
     d
ω  lim      ω 
ˆ      ω  ω
ˆ    ˆ
t 0 t     dt

d
Angular speed:                     in radians
角速率               dt
1ds v
Circular motion:
圓周運動
s r                    
r dt   r
ds
Linear speed:     v      r
直線速率               dt
Example 10.1. Wind Turbine 風動渦輪機

A wind turbine’s blades are 28 m long & rotate at 21 rpm.            rpm =

Find the angular speed of the blades in rad / s,                     per minute
圈 / 分鐘

& determine the linear speed at the tip of a blade.

  21 rpm 
 21 rev / min   2 rad / rev 
 2.2 rad / s
60 s / min

v  r    28 m 2.2 rad / s   62 m / s
Angular Acceleration                         角加速度
We shall restrict ourselves to rotations about a fixed axis.     以下限於繞固定轉軸的旋轉

 d     d2 
(Instantaneous) angular acceleration           lim                                  α  ω
ˆ
(瞬時)角加速度                    t 0 t   dt   d t2
at          Trajectory of point on rotating rigid body is a circle ( r = const)
旋轉剛體上一點的軌跡是個圓圈 ( r 為常數)。
ˆ
v//θ
a
Its velocity v is always tangential:         ˆ
vrθ
其速度 v必在切線方向：
ar
Its acceleration is in the plane of rotation (   ) :
它的加速度在轉動平面上 (   ) :
ˆ
ω                                                         ˆ
dv     d ˆ      dθ
a       r     θ  r        at  a r
dt     dt        dt
Tangential component: a  r d  θ  r  θ
ˆ         ˆ
t
切線分量：             dt
ˆ
Radial component: a  r  d θ   v r   r  2 r
2

r
ˆ          ˆ
徑分量：             dt       r
ˆ
dθ    d
    ˆ
r           r
ˆ
dt    dt
Angular vs Linear                   角性比線性
Table 10.1 Angular & Linear Position, Velocity, & Acceleration
表 10.1 角和線位置，速度和加速度
Linear Quantity 線性量                        Angular Quantity 角性量
Position 位置 x                            Angular position 角位置 
d x                                           d
Velocity 速度 v                            Angular velocity 角速度              
dt                                            dt
d v d2 x                                        d  d 
Acceleration 加速度 a                   Angular Acceleration 角加速度      
dt   d t2                                       dt    d t2

Eqs for Constant Linear Acceleration       Eqs for Constant Angular Acceleration
等直線加速的方程式                                   等角加速的方程式
v
1
2
 v0  v                                
1
2
 0   
v  v0  a t                                     0   t
1                                                1
x  x0  v0 t  a t 2                             0   0 t   t 2
2                                                2
v2  v0  2a  x  x0 
2
 2   0 2   0 
2
Example 10.2. Spin Down                                   轉停

When wind dies, the wind turbine of Example 10.1 spins down with
constant acceleration of magnitude 0.12 rad / s2.

How many revolutions does the turbine make before coming to a stop?

 2  02  2 

0   2.2 rad / s 
2
# of rev.       2    2
         0
                                         3.2
圈數        2   4                4  3.14  0.12 rad / s      2

10.2. Torque                 力距
離 O 最近， 最小。

Rotational analog of force
力在轉動的比照
Torque力距:      τ  τ
ˆ

ˆ
τ    plane of r & F
垂直於 r 和 F 所成面

[  ] = N-m ( not J )
= 牛頓米 (不是焦耳)

  sin 
  r
Example 10.3. Changing a Tire                                 換車胎
You’re tightening the wheel nuts after changing a flat tire of your car.
你替你的車子換胎後要把輪子的螺絲上緊。
The manual specify a tightening torque of 95 N-m.
使用手冊規定上緊力距為 95 N-m。
If your 45-cm-long wrench makes a 67 angle with the horizontal,
如果你長 45-cm 的扳手與水平線成 67，
with what force must you pull horizontally to do the job?
你需用多少力才能完工？
  r F sin 

95 Nm   0.45 m F sin 180  67

F  230 N

Note:     sin     sin 
10.3. Rotational Inertia & the Analog of Newton’s Law

轉動慣量和牛頓定律的比照

Linear acceleration:
F ma
直線加速
遠一點就      Rotating baton         旋轉中的指揮棒
比較難       (massless rod of length R + ball of mass m at 1 end):
( 無質量的棒長 R + 在一端質量為 m 的球 )
轉軸

  R Ft

Tangential force on ball :
施於球在切線方向的力：
Ft  m at  m  R

  m  R2  I 
= moment of inertia
I mR    2       轉動慣量                 of the baton
= rotational inertia    (指揮棒的)
轉動慣量
Calculating the Rotational Inertia                                 計算轉動慣量
Rotational inertia of discrete masses        I   m i ri 2
離散質點的轉動慣量                        i

ri = perpendicular distance of mass i to the rotational axis.
質點 i 離轉軸的垂直距離

Rotational inertia of continuous matter
連續物質的轉動慣量
I   r2 d m         r 2   r  dV

r = perpendicular distance of point r to the rotational axis.
點 r 離轉軸的垂直距離                                                    質量單元 dm 提供轉動慣量 r2 dm。
( r) = density at point r. 點 r 處的密度
Example 10.4. Dumbbell                                 啞鈴
A dumbbell consists of 2 equal masses m = 0.64 kg
on the ends of a massless rod of length L = 85 cm.

Calculate its rotational inertia about an axis ¼ of the way from one end & perpendicular to it.

GOT IT懂嗎 ? 10.2
Would I                  I會
(a)increase              比較大
 L  2  3L  2  5
I  m        m L2                              (b)decrease              比較小
 4   4  
                  8                            (c)stay the same         保持一樣
if the rotational axis were   若轉軸是
5
  0.64 kg   0.85 m   0.29 kg m2
2
(b) (1)at the center of the rod 在桿的中點
8
(a) (2) at one end?               在桿的端點？
Example 10.5. Rod                                   桿
Find the rotational inertia of a uniform, narrow rod of mass M and length L about an axis
through its center & perpendicular to it.

質量單元的質量為
dm，長度為 dx。

L /2
I   r d m   r  dV                      x 2  dx                    M
L /2

2               2
 L /2
x2     dx
 L /2          L
L /2
M 1 3                  M 2
     x                    L
L 3       L /2        12
Example 10.6. Ring                                      環
Find the rotational inertia of a thin ring of radius R and mass M about the ring’s axis.

2
I   R2 d m  0 R  R d 
2

M R2                 2                       M

2              0
d            
2 R
 M R2                       R2  d m

Pipe of radius R & length L :
半徑為 R ，長度為 L 的管子：
2                                   M

L
I                R2  R d d z
0     0                                      2 R L
M
 R3            2 L  M R 2  R 2 d m
2 R L                   
I = MR2 for any thin ring / pipe
任何幼環或管子
Example 10.7. Disk                             盤
Find the rotational inertia of a uniform disk of radius R & mass M about an axis
through its center & perpendicular to it.

M
I   r2 d m                          
 R2
2M
dm    2 r dr             2
r dr
R

2M         R              1
I 2
R      
0
r3 d r 
2
M R2
Table 10.2. Rotational Inertia 轉動慣量
實心球對其直徑      平板對垂直於它的軸

空心球對其直徑     平板對通過它中心的軸

Parallel - Axis Theorem 平行軸定理
Parallel - Axis Theorem:
I  I cm  M d 2
平行軸定理

軸移了 d = R 遠

Ex. Prove the theorem
for a set of particles.
習題：証明此定理適用
於一組粒子
GOT IT 懂嗎? 10.3.

Explain why the rotational inertia for a solid sphere is less than that of
a spherical shell of the same M & R.

2                            2
I sphere      M R2           I shell      M R2
5                            3

Mass of shell is further away from the axis.
球殼的質量離軸較遠。
Example 10.8. De-Spinning a Satellite
停止人做衛星的自轉
A cylindrical satellite is 1.4 m in diameter, with its 940-kg mass distributed uniformly.

The satellite is spinning at 10 rpm but must be stopped for repair.

Two small gas jets, each with 20-N thrust, are mounted on opposite sides of it & fire tangent
to its rim. 兩部裝在它兩邊的小型噴氣引擎，每部推力 20-N，可朝它邊緣的切線方向噴射。
How long must the jets be fired in order to stop the satellite’s rotation?

To stop the spin:
  0
要自轉停止：

Time required for a const ang accel            t 
等角加速下所需時間                               
  r F sin   I                        1     
2R F   M R2  
2     

10 rpm   2 rad / rev               
1
                                           min / s 
 60         
t       MR                                                     940 kg  0.7 m     8.6 s
4F                          4  20 N 
Example 10.9. Into the Well                             到井裏
A solid cylinder of mass M & radius R is mounted on a frictionless horizontal axle over a well.

A rope of negligible mass is wrapped around the cylinder & supports a bucket of mass m.

Find the bucket’s acceleration as it falls into the well.

Let downward direction be positive.
設朝下的方向為正

Bucket         Fnet  mg  T  m a
桶:
a                         a
Cylinder 圓        T R  I   I                   T I
柱：                                R                         R2

                  a
mg  I       2
ma
R

g             g
a                
I             M
1              1
mR 2          2m
GOT IT 懂嗎? 10.4.
Two masses m is connected by a string that passes over a frictionless pulley of mass M.

One mass rests on a frictionless table; the other vertically.

Is the magnitude of the tension force in the vertical section of the string

(a) greater than,   (b) equal to,          or (c) less than
(a)大於，              (b)等於，                 或(c) 小於
that in the horizontal 在水平部份的 ?            Explain 請解釋.

(a):
There must be a net torque to increase the
pulley’s clockwise angular velocity.
必需有淨力距才能提高滑輪的順時針角速度。
10.4. Rotational Energy 轉動能
Rotational kinetic energy = sum of kinetic energies of all mass elements,
taken w.r.t the rotational axis.
轉動能 =所有質量單元對轉軸的動能的和。

1            1
Set of particles:
一組粒子：
K     
2 i
mi v i2   mi ri 2  2
2 i                  
1
I 2
2

dm   v 2    dm  r 
1                 1
dK      
2

2                 2

1             1
K rot   dK        r  d m   2     
2
r2 d m
2             2

1
K rot      I 2
2
Example 10.10. Flywheel Storage                              飛輪儲能
A flywheel has a 135-kg solid cylindrical rotor
with radius 30 cm and spins at 31,000 rpm.
一飛輪以一個 135-kg ，半徑為 30 cm 的實心圓
柱為轉子，其轉速為 31,000 rpm 。
How much energy does it store?
它可存多少能量？

I 2   M R2  2
1       1 1
K rot                                         Flywheel for hybrid bus (30% fuel saving).
2       22     
2
1                   2                                 1       
 135 kg  0.30 m   31, 000 rpm  2 rad / rev   min / s  
4                                                     60      
Modern flywheels 10s of kW of power for up to a
 32 MJ                           min.
現代的飛輪可提供數十 kW 的功率達一分鐘之久。
~ energy in 1 liter of gasoline        Carbon composite to withstand strain of 30,000 rpm.
一公升汽油的能量                             碳基複合材料可承受 30,000 rpm 的應變。
Magnetic bearings to reduce friction.
磁性軸承可減低磨擦。
Supercondutor to reduce electrical losses.
超導體以減低電的損耗。
Energy & Work in Rotational Motion
旋轉運動的能和功

Work-energy theorem for rotations:
轉動的功-能定理：

2
W    d  K rot  1 I  2  1 I i2
1                     f
2         2
Example 10.11. Balancing a Tire                              平衡一個輪胎

An automobile wheel with tire has rotational inertia 2.7 kg m2.

What constant torque does a tire-balancing machine need to apply in order to spin
this tire up from rest to 700 rpm in 25 revolutions?

1
W             I 2
f
2
2
                              1      
I 2
 2.7 kg m 2   700 rpm  2 rad / rev   min / s  
 60     
     f
                                                                 46 N m
2                       2  25 rev  2 rad / rev 
10.5. Rolling Motion                                 滾動
Composite object:                                V = velocity of CM                    質心速度

 mi  V 2  ui2  2V  ui 
1                   1                             1
Ktotal       mi vi2            mi  V  ui          
2

2 i                 2 i                           2 i

1        1                                                    d
      M V 2   mi ui2                                 mi ui       mi  ri  R cm   0
2        2 i                                       i          dt i

K total  K cm  K internal

Moving wheel:

1        1
Ktotal      M V 2   u2 dm             
1        1
M V 2   2  r 2 dm
2        2                        2        2

1        1                                   is w.r.t. axis thru CM
Ktotal      M V 2  I cm  2
2        2                                   是對通過質心的軸來算
V = velocity of CM 質心速度
Moving wheel:               1        1
運動中的輪子：          Ktotal      M V 2  I cm  2    is w.r.t. axis thru CM
2        2
 是對通過質心的軸來算

Rolling wheel:   X  R               V  R               磨擦防止輪子滑動
滾動中的輪子：

輪子底部的速度是零！

這兩個速度在底部的和為零
Example 10.12. Rolling Downhill                               滾到山下
A solid ball of mass M and radius R starts from rest & rolls down a hill.
一個質量為 M ，半徑為 R 的實心球從靜止滾往山下。
Its center of mass drops a total distance h.        它的質心總共落下 h 遠。
Find the ball’s speed at the bottom of the hill.    求球在山底的速率。

Initially:
開始時：         E0  K trans 0  K rot 0  U 0  M g h

Finally:     E  K trans  K rot  U
最後時：
1       1
     M v2  I  2
2       2
2
1      12      v 
 M v2   M R2     7 M v 2
2      25      R  10

10
E  E0       v           gh              2g h
Note: v is independent of M & R
7
sliding ball
注： v 與 M 和 R 無關                                                                  滑動的球
GOT IT 懂嗎? 10.5.

A solid ball & a hollow ball roll without slipping down a ramp.

Which reaches the bottom first? Explain.

Solid ball.          實心球
Smaller I  smaller Krot      larger v.
I 較小  Krot 較小                v 較大

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