Physics 1: Mechanics
Đào Ngọc Hạnh TâmOffice: A1.503,
Email: [email protected], Vietnam National University
Acknowledgment: Slides are supported by Prof. Phan Bao Ngoc
Part A: Dynamics of Mass Point
Chapter 1 Bases of Kinematics
Chapter 2 Force and Motion (Newton’s Laws)
Group Assignment
Part B: Laws of Conservation
Chapter 3 Work and Mechanical Energy
Midterm exam Chapter 4 Linear Momentum and Collisions
Part C: Dynamics and Statics of Rigid BodyChapter 5 Rotation of a Rigid Body About a Fixed Axis
Group Assignment
Chapter 6 Equilibrium and Elasticity
Chapter 7 Gravitation
Final exam
Contents of Physics 1
Part C: Dynamics and Statics of Rigid Body
Chapter 5: Rotation of a Rigid Body About a Fixed Axis
5.1. Rotational Variables
5.2. Rotation with Constant Angular Acceleration
5.3. Kinetic Energy of Rotation, Rotational Inertia
5.4. Torque, and Newton’s Second Law for Rotation
5.5. Work and Rotational Kinetic Energy
5.6. Rolling Motion of a Rigid Body
5.7. Angular Momentum of a Rotating Rigid Body
5.8. Conservation of Angular Momentum
Overview
•We have studied the motion of translation, in which objects move along a straight or curved line.
•In this chapter, we will examine the motion of rotation, in which objects turn about an axis.
5.1. Rotational variables:
We study the rotation of a rigid body about a fixed axis.Rigid bodies: Bodies can rotate with all its parts locked together andwithout any change in its shape. Fixed axis: A fixed axis means the rotational axis does not move.
Angular Position:
Reference line: To determine the angular position, we must define a reference line, which is fixed in the body, perpendicular to the rotation axis and rotating with the body.The angular position of this line is the angle of the line relative to a fixed direction.
(rad) radians :
r
s
1 rev = 3600 = 2 radwhere s is the arc length of a circular path of radius r and angle .
Angular Displacement
12 Convention: • > 0 in the counterclockwise direction.• < 0 in the clockwise direction.
Angular Velocity
Average angular velocity: ttt
12
12avg
Instantaneous angular velocity:dt
d
tt
0lim
Angular Acceleration
Average angular acceleration: ttt
12
12avg
Instantaneous angular acceleration:
dt
d
tt
0lim
Unit: rad/s or rev/s or rpm; (rev: revolution)
Unit: rad/s2 or rev/s2
Note: Angular displacement, velocity, and acceleration can be treatedas vectors (see page 246).
5.2. Rotation with Constant Angular Acceleration
For one dimensional motion:dt
dva
dt
dxv ;
Let’s change variable names: avx ,,
)(2
2
1
020
2
200
0
tt
t
Checkpoint 2 (p. 248):In four situations, a rotating body has angular position (t) given by (a) =3t-4, (b) =-5t3+4t2+6, (c) =2/t2-4/t, and (d) =5t2-3.
To which situations do the angular equations above apply?
)(2
2
1
0
2
0
2
2
00
0
xxa
attxx
at
Ans: a) and d)
5.3. Kinetic Energy of Rotation
a. Linear and Angular Variable Relationship
•The position: rs where angle measured in rad; s: distance along a circular arc; r: radius of the circle
•The speed:
rv (rad)
rdt
d
dt
ds
The period of revolution:
22
v
rT
•The Acceleration: rdt
d
dt
dv
Tangential acceleration: rat
Radial acceleration:r
r
var
22
b. Kinetic Energy of Rotation:
The KE of a rotating rigid body is calculated by adding up the kinetic energies of all the particles:
2233
222
211
2
1...
2
1
2
1
2
1iivmvmvmvmK
222
2
1)(
2
1iiii rmrmK
2
iirmI
2
2
1IK
Unit for I: kg m2
Rotation Inertia (or moment of inertia)
(radian)K (J)
c. Calculating the Rotational Inertia:
2
iirmI
• For continuous bodies:
dmrI 2
• If the rigid body consists of a few particles:
• Parallel-Axis Theorem: calculate I of a body of mass M about a given axis if we already know Icom:
2MhIIcomO
h: the perpendicular distance between the given axis at 0 and the axis through the COM of the body.
Some Rotational Inertias Table 10-2
Sample problem:
In 1985,Test Devices, Inc. (www.testdevices.com) was spin testing asample of a solid steel rotor (a disk) of mass M= 272 kg and radiusR = 38.0 cm. When the sample reached an angular speed of14000 rev/min, the test engineers heard a dull thump from thetest system, which was located one floor down and one room overfrom them.Investigating, they found that lead bricks had been thrown out inthe hallway leading to the test room, a door to the room had beenhurled into the adjacent parking lot, one lead brick had shot fromthe test site through the wall of a kitchen, the structural beams ofthe test building had been damaged, the concrete floor beneath thespin chamber had been shoved downward by about 0.5 cm, and the900 kg lid had been blown upward through the ceiling and had thencrashed back onto the test equipment. The exploding pieces had notpenetrated the room of the test engineers only by luck.
How much energy was released in the explosion of the rotor?
2
2
1IK
Energy was released in the explosion of the rotor
Rotational Inertia of the disk
(1)
222 .64.19)38.0)(272(2
1
2
1mkgmkgMRI
Angular speed =14000 rev/min = 14000. 2/60= 1.466×103 rad/s
2322 )/10.466.1)(.64.19(2
1
2
1sradmkgIK
= 2.1 × 107 J
(1)
5.4. Torque, and Newton’s Second Law for Rotation
a. Torque (“to twist”):•A force applied at point P of a body that is free to rotate about an axis through O. The force has no component parallel to the rotation axis.
F
F
F
Rotation axis
f
The ability of to rotate the body is defined by torque F
F
trFFr )sin(
F of armmoment the:
r
Fr
(Unit: N.m)
FrFr )sin(
Resolve into 2 components: Ft (tangential) and Fr (radial).
F
f
F
Direction of torque: Use the right hand rule to determine
Important Note: Torque is a vector quantity, however, because we consider only rotation around a single axis, we therefore do not need vector notation. Instead, a torque is a positive value if it would produce a counterclockwise rotation and a negative value for a clockwise rotation.
f1counterclockwise
clockwise
f2
tnet =t1 -t2 = F1r1sinf1 -F2r2 sinf2
b. Newton’s Second Law for Rotation
•We consider a simple problem, the rotation of a rigid body consisting of a particle of mass m on one end of a massless rod.
tt maF
The torque acting on the particle:
)()( 2mrrrmrmarF tt
(at: tangential acceleration)
: angular accelerationmeasure)(radian I
if more than one force applied to the particle:
Inet
This is valid for any rigid body rotating about a fixed axis.
53/270. The figure below shows a uniform disk that can rotatearound its center like a merry-go-round. The disk has a radiusof 2.0 cm and a mass of 20.0 grams and is initially at rest.Starting at time t=0, two forces are to be applied tangentiallyto the rim as indicated, so that at time t=1.25 s the disk hasan angular velocity of 250 rad/s counterclockwise. Force F1 hasa magnitude of 0.1 N. What is magnitude F2?
Inet IRFRFnet 1221
For a uniform disk: 2
2
1MRI
12 FR
IF
For rotation:t
tt
0
(N)14.01.025.12
)250)(100.2)(100.20(
2
23
12
Ft
MRF
+
5.5. Work and Rotational Kinetic Energy
The work-kinetic theorem applied for rotation of a rigid body:
WIIKKK ifif 22
2
1
2
1
•We consider a torque of a force F accelerates a rigid body in rotation about a fixed axis:
The work done by the torque:
f
i
dK
If is a constant:
)( ifW
The power:
dt
dWP
Problem 61 (p. 271): A 32.0 kg wheel, essentially a thin hoop with radius 1.2 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s. (a) How much work must be done to stop it? (b) What is the required average power?
•The rotational inertia of a wheel (a thin hoop) about central axis:
)m (kg 1.462.10.32 222 MRI
•To stop the wheel, =0:
(rad/s) 3.29(s) 60
(rad) 2 280 rev/min 2800
(a) The work is needed to stop the wheel:
(kJ) 19.8or (J)197883.291.462
1
2
10 22
0 IKKW if
(W<0: energy transferred from the wheel)
(b) The average power: kW)(32.1or (W) 131915
19788
t
WP
5.6. Rolling Motion of a Rigid Body
We consider a rigid body smoothly rolling along a surface. Its motionconsists of two motions: translation of the center of mass androtation of the rest of the body around that center.
Example: A bike wheel is rollingalong a street. During a time intervalt, both O (the center of mass) and P(the contact point between thewheel and the street) move by adistance s:
Rs where R is the radius of the wheel.
Rvcom
•The speed of the center of the wheel:
•The linear acceleration of the center of the wheel:
Racom
The Kinetic Energy of Rolling
22
2
1
2
1comcom MvIK
Rotational KE Translational KE
Examples: 1. Rolling on a horizontal surface: A 2.0 kg wheel, rolling smoothly on a horizontal surface, has a rotational inertia about its axis I = MR2/2, where M is its mass and R is its radius. A horizontal force is applied to the axle so that the center of mass has an acceleration of 4.0 m/s2. What is the magnitude of the frictional force of the surface acting on the wheel?
Applying Newton’s second law for rotational motion:
(1) comsnet IRf
)(clockwise 0
axis) x theofdirection positive (in the 0,
xcoma
(2) , Ra xcom (1), (2):
appliedF
sf
(for )
+
+
(N)0.42
0.42
22
MR
R
,
2
,2
xcomxcom
coms
Ma
R
aIf
(for acom,x)
-
Examples: 2. Rolling Down a Ramp: We consider a rigid body smoothly rolling down a ramp.Force analysis: Fg, FN, and fs (opposing the sliding of the body, so the force is up the ramp)
(1) sin ,xcoms MaMgf
•Applying Newton’s second law for translational motion:
•Applying Newton’s second law for rotational motion: (2) comsnet IRf
ockwise)(countercl 0
axis) x theofdirection negative (in the 0,
xcoma
2
,
1
sin
MR
I
ga
comxcom
(3) , Ra xcom
(1), (2), (3):
Examples: 3. The Yo-Yo:
y
For translation motion:
(1) ,ycomMaMgT
(2) 0 comITR
For rotational motion:
(3) 0, Ra ycom
20
,
1MR
I
ga
comycom
Examples: 4. Pulley: A 10.0 kg block hangs from a cord which is wrapped around the rim of a frictionless pulley. Calculate the acceleration, a, of the block as it moves down? (The rotational inertia of the pulley is 0.50 kg m2 and its radius is 0.10 m).
• For translation motion:(1) maTmg
(2) pulley0 ITR
• For rotational motion:
(3) 0Ra
)m/s(63.1
1.0
5.010
8.910 2
22
pulley
R
Im
mga
a
T
T
mg
+
(Note: a is the acceleration of the block, is the angular acceleration of the pulley)
Example: (The Kinetic Energy of Rolling) A 10 kg cylinder rolls without slipping. When its translational speed is 10 m/s, What is its translational kinetic energy, its rotational kinetic energy, and its total kinetic energy?
•The rotational inertia of a cylinder about central axis:
2
2
1MRI
•Translational kinetic energy:
(J)50010102
1
2
1 22 MvKk
•Rotational kinetic energy:
22
22
4
1
2
1
2
1
2
1Mv
R
vMRIKr
(J) 250rK•Total kinetic energy:
(J) 750 rk KKK
The Kinetic Energy of Rolling
5.7. Angular Momentum of a Rotating Rigid Body
a. Angular Momentum of a Particle:
)( vrmprl
:r
the position vector of the particle with respect to O
:p
the linear momentum of the particle
•The direction of determined by the right-hand rulel
l
sinrmvl
or
rpprl
(Unit: kg m2 s-1)
l
)( vrmprl
dt
pdrFr net
dt
pdFnet
Newton’s Second Law in Angular Form for a Particle:
For translational motions:
dt
ld
dt
vrdm
vdt
rd
dt
vdrmnet
)(
)(
dt
ldnet
So, Newton’s Second Law in Angular Form:
Note: The torques and the angular momentum must be defined with respect to the same origin O.
l
dt
Ldnet
b. Angular Momentum of a System of Particles:
n
iin lllllL
1321 ...
Newton’s Second Law in Angular Form:
c. Angular Momentum of a Rotating Rigid Body:
Method: To calculate the angularmomentum of a body rotating about afixed axis (here the z axis), we evaluatethe angular momentum of a system ofparticles (mass elements) that formthe body.
iiiiii vmrprl 090sin
il
izlil
is the angular momentum of element i of mass mi:
iiiiii vmrprl 090sin
iiiiiiizi vmrvmrll ))(sin(sin,
i
n
iiii
n
iii
n
iziz rrmrvmlL
111
, )(
n
iiiz rmL
1
2
zz IL
We drop the subscript z: IL
I z : rotational inertia of
the body about the z axis
Note: L and I are the angular momentum and the rotational inertia of a body rotating about the same axis.
Example: At the instant of the figure below, a 2.0 kg particle P has aposition vector of magnitude 5.0 m and angle 1=45
0 and a velocityvector of magnitude 4.0 m/s and angle 2=30
0. Force of magnitude2.0 N and angle 3=30
0, acts on P. All three vectors lie in the xyplane. About the origin, what are the (a) magnitude and (b) directionof the angular momentum of P and the (c) magnitude and (d) directionof the torque acting on P?
r
v
F
prl
(a)
)30sin(0.40.20.5sin 2 rmvl
(b) Using the right-hand rule, points out of thepage and it is perpendicular to the figure plane.
l
l
Fr
(c)
m) (N 5)30sin(0.20.5sin 3 rF
(d) points out of the page and it is perpendicular to the figure plane.
)/m (kg 20 2 s
5.8. Conservation of Angular Momentum
If no net external torque acts on the system:
constantL
fi LL
ffii II
dt
Ldnet
tnet = 0
Example: You stand on a frictionless platform that is rotating atan angular speed of 1.5 rev/s. Your arms are outstretched, andyou hold a heavy weight in each hand. The moment of inertia ofyou, the extended weights, and the platform is 6.0 kg.m2. Whenyou pull the weights in toward your body, the moment of inertiadecreases to 1.8 kg.m2. (a) What is the resulting angular speedof the platform? (b) What is the change in kinetic energy of thesystem? (c) Where did this increase in energy come from?
(a) No net torque acting on the rotating system (platform + you + weights):
(rev/s)58.1
5.10.6
I
IIIconstant
2
1122211
I
(b) The change in kinetic energy:
(J)62125.10.6258.12
1
2
1
2
1 22211
222 IIK
(c) Because no external agent does work on the system, so the increase in kinetic energy comes from your internal energy (biochemical energy in your muscles).
More Corresponding Variables and Relations for Translational and Rotational
1, 3, 7, 14, 20, 26, 39, 43, 53, 56, 62 (p. 267- 271)
2, 5, 9, 17, 35, 38, 41, 43, 54, 60 (p. 297-302)
Homework:
Fr
Inet
•Torque:
Newton’s second law for rotation:
Work – KE Theorem: WIIKKK ifif 22
2
1
2
1
•For a point mass:2mrI
•For a rigid body: I depends on the object shape (table 10-2)
(kg.m2)
• Rotational KE:
2
2
1IK rv
rat
rr
var
22
Angular position: (rad)Angular displacement: (rad)Angular velocity: (rad/s)Angular acceleration: (rad/s2)
1 rev = 3600 = 2 rad
Constant
rs Linear variables
I: Rotation Inertia(or moment of inertia)
)(2
2
1
020
2
200
0
tt
tSummary:
•Rolling motion:Rvcom
Racom 22
2
1
2
1comcom MvIK
Rotational KETranslational KE
•Angular Momentum: prl
sinrmvl
dt
Ldnet
•Newton’s Second Law in Angular Form:
•Angular Momentum of a Rigid Body:
IL •Conservation of Angular Momentum:
If no net external torque acts on the system:
constantL
fi LL
ffii II