實驗3：轉動-剛體的轉動運動
Lab. 3 : Rotation
-Rotational Motion of Rigid Body
實驗目的：測量作用在圓盤上之力矩所產生的角加速
度和角速度的大小
證明剛體繞固定軸所做的轉動運動，亦遵守牛頓第二
運動定律。
並探討轉動物體的能量守恆關係。
Object: To measure angular acceleration  and angular
velocity  of a rotating rigid disk applied by a torque .
To prove the rotational motion of a rigid body obey the
Newtonian 2nd motional law.
The
rotational motion also obeys the conservation
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ofMFTai-戴明鳳
energy.
HyperPhysics
Most fundamental concepts are subtracted from the web site:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
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MFTai-戴明鳳
Rotational Dynamics
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
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Circular / Rotational Motion
For circular motion at a constant speed v, the
centripetal acceleration of the motion can be derived.
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http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#rotcon
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Basic Rotational Quantities

The angular displancement is defined by:

For a circular path it follows that the
angular velocity is

The angular acceleration:
i.e., Tangential acceleration:

Centripetal acceleration:
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Radian and Conversions

q is a pure number, but commonly is given
the artificial unit, radian.
 One radian: is the angle subtended by an arc
length equal to the radius of the arc (s = r).
 Conversion between degrees and radians
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Angular Displacement
 The angular displacement is
defined as the angle the object
rotates through during some
time interval.
 This is the angle that the
reference line of length r sweeps
out.
Fig 10.2
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Average Angular Speed &
Instantaneous Angular Speed

The average angular speed, , of a rotating rigid object
is the ratio of the angular displacement to the time
interval.

The instantaneous angular speed is defined as the limit of
the average speed as the time interval approaches zero.



Units: radians/sec, rad/s or s-1 (radians have no dimensions)
 > 0: for q increasing (counterclockwise)
 < 0: for q decreasing (clockwise)
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Average and Instantaneous
Angular Acceleration
 The average angular acceleration, ,
of an object is defined as the ratio of the change in the angular
speed to the time it takes for the object to undergo the change.
 The instantaneous angular acceleration is defined as the limit
of the average angular acceleration as the time goes to 0
 Units of angular acceleration are rad/s2 or s-2 since radians
have no dimensions.
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Angular Motion, General Notes
 When a rigid object rotates about a fixed axis in a given time
interval,
 every portion on the object rotates through the same angle in a
given time interval and has the same angular speed and the
same angular acceleration.
– So q, ,  all characterize the motion of the entire rigid
object as well as the individual particles in the object.
Directions of 
 Strictly speaking, the speed and
acceleration (, ) are the
magnitudes of the velocity and
acceleration vectors.
 The directions are actually given by
the right-hand rule.
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Fig 10.3
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Rotational Kinematics

Under constant angular acceleration, the motion of the
rigid object can be described using a set of kinematic
equations.
– These are similar to the kinematic equations for linear
motion.
– The rotational equations have the same mathematical form
as the linear equations.
Rotational Kinematic Equations
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Comparison Between
Rotational and Linear Equations
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Relationship Between Angular
and Linear Quantities

Displacements

Speeds

Accelerations

Every point on the rotating
object has the same
angular motion.

Every point on the rotating
object does not have the
same linear motion.
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MFTai-戴明鳳
Speed Comparison

The linear velocity is always
tangent to the circular path.
– called the tangential velocity

The magnitude is defined by the
tangential speed.
Fig 10.4
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Tangential Acceleration
Centripetal Acceleration

Fig 10.5
An rotating rigid object traveling in a circle, even
though it moves with a constant speed, will have an
centripetal acceleration:
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Resultant Acceleration

Tangential component of the acceleration
 change speed

Centripetal component of the acceleration
 change direction

Total acceleration:
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MFTai-戴明鳳
Notes for
Speed and Acceleration

All points on the rigid object will have the same
angular speed, but not the same tangential speed.

All points on the rigid object will have the same
angular acceleration, but not the same tangential
acceleration.

The tangential quantities depend on r, and r is not
the same for all points on the object.
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Torque vs. Force
= I vs. F = ma

Forces can cause a change in linear motion.
– Described by the Newton’s Second Law

Torque can cause a change in rotational
motion.
– The effectiveness of this change depends on the
force and the moment arm.
– The change in rotational motion depends on the
torque.
– Also obey the Newton’s 2nd Motional Law
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Torque

Be the vector product or
cross product of two vectors
r and F.

SI units of torque: N.m

Torque is a force multiplied by
a distance, it is very different
from work and energy (unit:
Joules).
Fig 10.12
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Definition of Vector Product

Any two vectors,

The vector product
and
=
Fig 10.13

The direction of C is given by the right-hand rule.
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Conditions for Equilibrium

An object at equilibrium has no net influences to cause it
to move, either in translation (linear motion) or rotation.
The basic conditions for equilibrium are:
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Rotational-Linear Parallels
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The Rigid Object In
Equilibrium
Fig 10.15 (a) & (b)
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Moment of Inertia, Rotational inertia
 Moment
of inertia (rotational inertia) ~ mass for linear motion.
 It appears in the relationships for the dynamics of rotational motion.
 The moment of inertia must be specified with respect to a chosen axis of
rotation.
 For a point mass the moment of inertia is just the mass times the square of
perpendicular distance to the rotation axis, I = mr2.
 That point mass relationship becomes the basis for all other moments of
inertia since any object can be built up from a collection of point masses.
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Moment of Inertia
 Dimensions: ML2, SI units: kg.m2
 It is more easily to calculate the moment of inertia of
an object by assuming it is divided into many small
volume elements, each of mass Dmi.
 For continuous body,
If r is constant, the integral can be evaluated with
known geometry, otherwise its variation with PhysicsNTHU
position
must be known.
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Fig 10.7
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Parallel Axis Theorem



The I (rotational inertia) of any object about an
axis through its center of master (C.M.), Icm, is
the minimum moment of inertia for an axis in
that direction in space.
The moment of inertia about any axis parallel
to that axis through the C.M. is given by the
parallel axis theorem.
The expression added to the Icm will be recognized as the
moment of inertia of a point mass - the moment of inertia about
a parallel axis is the center of mass moment plus the moment
of inertia of the entire object treated as a point mass PhysicsNTHU
at the
center of mass.
MFTai-戴明鳳
Analogy between Rotational & Linear
Kinetic Energies, KErotational & KELinear
For a given fixed axis of rotation, the rotational
kinetic energy can be expressed in the form
a rotating object is analogous to KELinear and can be expressed in terms
of the moment of inertia and angular velocity.
 The total kinetic energy of an extended object can be expressed as the sum
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of the translational kinetic energy of the center of mass and the rotational
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kinetic energy about the center of mass.

KErotational
Work-Energy Principle
 The
expressions for KErotational & KELinear kinetic energy
can be developed in a parallel manner from the workenergy principle.
 Consider the following parallel between a constant torque
exerted on a flywheel with moment of inertia I and a
constant force exerted on a mass m, both starting from rest.
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Rotational Kinetic Energy
of the rigid object


The total rotational kinetic energy is
the sum of the energies of all its particles:
[units: Joules (J)]:


where I is called the moment of inertia
Be similar to the linear kinetic energies associated with
motion (K = 1/2mv2)
Be not a new type of energy, the form is different because it
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is applied to a rotating object.
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
For the linear case, starting from rest, the acceleration
from Newton's second law is equal to the final velocity
divided by the time and the average velocity is half the
final velocity,
showing that the work done on the block gives it a
kinetic energy equal to the work done.
 For the rotational case, also starting from rest, the
rotational work is τθ and
 the angular acceleration α given to the flywheel is
obtained from Newton's second law for rotation.
 The angular acceleration is equal to the final angular
velocity divided by the time and the average angular
velocity is equal to half the final angular velocity.
 It follows that the rotational kinetic energy given to the
flywheel is equal to the work done by the torque.
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MFTai-戴明鳳
Work-Energy Principle

This is a general principle which can be applied
specifically to rotating objects.
 For pure rotation, the net work is equal to the change
in rotational kinetic energy and for a constant torque,
the work can be expressed as
For a net torque, Newton's 2nd law for rotation gives
Combining this expression gives a useful relationship
for describing rotational motion.
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Rolling Objects without Slipping

The K.E. (kinetic energy) is divided between linear
kinetic energy and rotational kinetic energy.
 Another key is that for rolling without slipping, the
linear velocity of the center of mass is equal to the
angular velocity times the radius.
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Rotation Vector Examples
This is an active graphic. Click on any example.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
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MFTai-戴明鳳
原理(Principle)
物體對旋轉軸轉動(Body rotates about a fixed rotational axis) :
角位移(angular displacement): q [rad] [360o = 2p radians]
角速度(angular velocity):
 = dq/dt [rad/s]
角加速度(angular acceleration):  = d/dt = d2q/dt2 [rad/s2]
外力矩(torque)  = r x F (F: 作用力, r : 外力與原點距離) [m.N]
then
   or  = I = dL/dt
比率常數: 對轉動軸之轉動慣量(rotational inertia): I [kg.m2]
I = Simiri2 = rr2dV
mi: 質量, ri: 質量與原點距離, r: 均勻質量密度,
r: 與原點距離, dV: 體積單元
力矩(torque)  = constant,  = constant  等角加速度運動
力矩  = 0,  = 0, L = constant,  = constant  等角速度運動
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MFTai-戴明鳳
(例) 均勻剛體(rigid body)圓盤(disc)或圓柱(質量M, 半徑R)
I = MR2/2
(旋轉軸為通過圓心原點之垂直軸)
I = MR2/2 + Mh2 (旋轉軸為距圓心距離h之垂直軸)
實驗: 基本設計
利用水平氣墊桌上無摩擦之圓盤(含輪軸及輪軸架)轉動
(非均勻圓盤/旋轉軸通過圓心, I  MR2/2)
輪軸(小輪軸或大輪軸)上纏繞細繩, 經滑輪懸吊砝碼質量m
圓盤(含輪軸及輪軸架)受力矩[旋轉軸z軸/右手定則/逆時針轉動]:
 = r x T = I (r = r1 = 1 cm or r2 = 2 cm, T: 水平繩張力)
T = I/r [r x T = rTsinq z= rTz, if q = 90o]
砝碼受力(氣墊水平, 無摩擦時)[方向-z軸]
Fnet = mg - T = ma = mr (a = r)
角加速度[旋轉軸z軸]  = mgr/(I + mr2) [討論滑輪等摩擦力f]
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MFTai-戴明鳳
1.小輪軸(small rotational axis)
2.次小輪軸(other rotational axis)
3.外側插栓(outer insert pin)
4.內側插栓(Inner insert pin)
5.送風口(Wind port)
6.滑輪(frictionless pulley)
T = mg
7.輪軸架(Pulley support)
8.圓盤(Turntable Disc)
圖2 氣墊桌裝置
9.砝碼(weights)
(Fig.2 Air Bearing Turntable)
10.摩擦橡皮圈(Resist elastic band)
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11.氣墊桌(Air Table)
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頂視圖 (Top view)
1.小輪軸(small rotational axis)
2.次小輪軸(other rotational axis)
3.外側插栓(outer insert pin)
4.內側插栓(Inner insert pin)
7.輪軸架(Support of rotation axis )
8.圓盤(Disc)
9.砝碼(weights)
mg – T = ma
T = mg – ma
a=r
T = mg - m  r
Tension, T = mg
T is the force applied
by the weight , mg.
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MFTai-戴明鳳
mg – T = ma
T = mg – ma
a=r
T = mg - m  r
If friction force, f = 0
 = Tr = (mg  mr )r = I
If friction force, f  0
(mg  f  mr ) = I
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MFTai-戴明鳳
實驗流程: 利用氣墊桌系統 (保護儀器:送風後才將圓盤放上)
A. 角加速度測量:
固定輪軸半徑(r1 or r2)及砝碼質量(m1, m2 or m3)
角位移 θ (圈數): 1/4, 1/2, 1, 3/2, … (1 圈 = 2p rad)
測量時間 t (s): t1, t2, t3, t4,…
角速度  (rad/s):  1,  2,  3,  4, … (平均值)
角加速度  : 等角加速度運動?
 畫圖: θ(t2) : 通過原點之直線(θ = t2/2 )
 (t ) : 通過原點之直線(  = t )
 求不同輪軸半徑 (r1, r2) 及砝碼質量 (m1, m2, m3) 時，
所獲得的角加速度 (1, 2, 3, 4, 5, 6)
 估計力矩值  = r x T = r (mg – ma) ~ rmg (if a << g)
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(t1, t2, t3, t4, t5, t6)
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計算慣性矩 I ?
畫  () 圖: 理論上是通過原點、斜率= I 的直線 = I
 故求斜率即可得慣性矩 I
 與其它方法求得之 I = 13.3 g.m2 比較之
B. 驗證  = I 
輪軸半徑 r 及砝碼質量 m 固定
固定力矩：  = r x F ~ rmg ~ constant
改變轉動慣量(慣性矩) I 值, 求角加速度變化。
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MFTai-戴明鳳
 利用輪軸架兩組插栓(內,外)，使之安置在離旋轉軸不
同的距離的位置上 h1, h2
可改變轉動慣量(慣性矩) I 值
 圓柱體(質量mc, 內徑ri, 外徑ro)放進插栓，一次放兩個,
保持平衡，增加之慣性矩
Ic = Icm + mch2 = mc(ro2 - ri2)/2 + mch2 (x2, 一次放兩個)
總轉動慣量＝ Itotal = I + 2Ic
求不同插栓距離(h1, h2)及圓柱體質量(mc1, mc2)之角加
速度
Itotal =  = rmg? [畫圖或列表]
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MFTai-戴明鳳
C. 由角速度求角加速度
調整細繩長度, 使砝碼m由高度h加速落下著地時, 細繩
恰好脫離輪軸(半徑r)
此時力矩  = 0 ( = 0)
圓盤轉動不在外力和力矩作用，變成等角速度()運動
2 – 02 = 2qt
因 0 = 0, 總角位移 qt = h/r (rad)
T
故  = 2r/2h
m
測量等角速度運動之頻率 f
或週期 T
  = 2pf = 2p/T
 代入求角加速度  ，與A的結果比較。
h
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MFTai-戴明鳳
D. 能量守恆
無摩擦時: 重力位能 = 轉動動能 + 平移動能
mgh = I2/2 + mv2/2
末速v (Final veclocity) = r
T
m
h
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