rotation. so far we have looked at motion in a straight line or curved line- translational motion....
TRANSCRIPT
Rotation
• So far we have looked at motion in a straight line or curved line- translational motion.
• We will now consider and describe rotational motion – where an object turns about an axis.
• So far we have looked at motion in a straight line or curved line- translational motion.
• We will now consider and describe rotational motion – where an object turns about an axis.
• We will start by mentioning some keywords.
1. Angular position2. Angular displacement3. Angular velocity4. Angular acceleration
• These terms are analogous to their linear equivalents.
Angular position
• The angular position is an angle measured between a reference line and a fixed direction taken as zero.
Reference line
Angular position
• The angular position θ is:
Reference line
r
s (radians)
s: arc length
Angular displacement
• The angular displacement gives the change in angular position of a rotating body.
12
t1
t2
Δθ
θ1
θ2
Counter-clockwise rotation is positive displacement.
Angular velocity
• If an object moves through the angle an angular position of θ1 to θ2 the average angular velocity is:
12
12
ttt
t1
t2
Δθ
θ1
θ2
(average angular velocity)
Angular velocity
• Thus,dt
d
tt
0lim
t1
t2
Δθ
θ1
θ2
(instantaneous angular velocity)
Units: rads/s
Angular acceleration
• Similarly we can definite the angular acceleration as,
dt
d
tt
0lim (instantaneous angular acceleration)
12
12
tttavg
(average angular acceleration)
Units: rads/s2
Example
• The rotation of a wheel about its central axis is given by . Calculate the angular velocity and acceleration.
4273 tt
Example
• The rotation of a wheel about its central axis is given by . Calculate the angular velocity and acceleration.
• Sol:
4273 tt
273 2 tdt
d
tdt
d6
Equations of linear and angular motionMotion Formula Missing variable
Linear
v=v0 + at x-x0
x-x0=v0t + 1/2at2 v
v2=v02 + 2a(x-x0) t
x-x0=1/2(v + v0)t a
x-x0=vt - 1/2at2 v0
Angular
ω=ω0 + αt θ-θ0
θ-θ0=ω0t + 1/2αt2 ω
ω2=ω02 +2a(θ-θ0) t
θ-θ0=1/2(ω + ω0)t α
θ-θ0=ωt - 1/2αt2 ω0
Relating linear and angular
• Recall that linear and angular variables can be related as follows:
• Position:
• Speed:
rs
dt
dr
dt
dsv
rv (linear velocity)
Relating linear and angular
• Time:
• Acceleration:v
rT
2
rdt
d
dt
dva
2
Relating linear and angular
• Time:
• Acceleration:
• Thus:
v
rT
2
rdt
d
dt
dva
(tangential acceleration)
2
rat
rr
var
22
(radial acceleration)
Example
• A grindstone having a constant angular acceleration of 0.35rad/s2 starts from rest with an arbitrary reference line horizontal at angular position . What is the angular displacement of the reference line at t=18s?
00
• Sol: What equation do we use?
• Sol:2
21
00 tt
• Sol:2
21
00 tt
00 00 (starts from rest)
221 1835.000
revrad 932007.560
• What is the wheel’s angular velocity at t=18s?
• Sol:t 0
srad /3.61835.00
The Kinetic Energy of Rotation
• It is clear that a rotating body has kinetic energy.
• It is clear that a rotating body has kinetic energy.
• However it is not clear how to calculate the KE of the body since (1) the particles making the body move at different velocities, (2) the KE of the body as a whole is zero since the com has a velocity of zero.
• The kinetic energy is found by summing the KE of the particles of the body and writing the velocity in terms of the angular velocity (which is the same for all particles).
2212
22212
1121 ... nnvmvmvmK
221
iivmK
221 ii rmK
2221 iirmK
• The kinetic energy is found by summing the KE of the particles of the body and writing the velocity in terms of the angular velocity (which is the same for all particles).
2212
22212
1121 ... nnvmvmvmK
221
iivmK
221 ii rmK
2221 iirmK
Moment of inertia
• The kinetic energy is found by summing the KE of the particles of the body and writing the velocity in terms of the angular velocity (which is the same for all particles).
2212
22212
1121 ... nnvmvmvmK
221
iivmK
221 ii rmK
22122
21 IrmK ii
Moment of inertia
• The momentum of inertia (rotational inertia) about some rotational point is the measure of the resistance to a change in the angular acceleration due to the action of a torque.
2i
iirmI (Moment of Inertia)
• The rotational inertia depends not only on the mass of the object but how it is distributed wrt the rotational axis.
Axis of rotation
(1)
(2)
Consider the two rods which have identical total mass . Both rods balance at the centre. However rod 2 rotates more freely than rod 1. Rod 1 has a larger moment of inertia than rod 2.
Moment of Inertia of some objects
• The moment of inertia for a continuous body is:
dmrI 2
• The moment of inertia for a continuous body is:
• If the rotational inertia of a body is known about any axis which passes through its com then the parallel-axis theorem can be used to find the moment of inertia about any parallel axis.
dmrI 2
2MhII cm (parallel-axis theorem)
Example
Example
• Find the rotational inertia about the com of the rigid body consisting of two particles of mass m connected by rod (with negligible mass)of length L.
2
L
2
L
cm mm
Example
• Find the rotational inertia about the com of the rigid body consisting of two particles of mass m connected by rod (with negligible mass)of length L.
2
L
2
L
cm mm
i
iicm rmI 2 2212
212
22211 LmLmrmrm 2
21 mL
Example
• Calculate the rotational inertia about the end of the body.
2
L
2
L
cm mm
2MhII cm 2mL
2212
21 2 LmmL
2
L
2
L
Example
• Consider a thin, uniform rod of mass M and length L. What is the rotational inertia of the rod through its com?
2
L
2
L
dmx
dx
Example
• Consider a thin, uniform rod of mass M and length L. What is the rotational inertia of the rod through its com?
dmrI 2
Consider an element dm of width dx. The mass of dm is the mass per unit length times the length of dm
2
L
2
L
dmx
dx
Example
• Consider a thin, uniform rod of mass M and length L. What is the rotational inertia of the rod through its com?
dmrI 2
Consider an element dm of width dx. The mass of dm is the mass per unit length times the length of dm
dxL
Mdm
2
L
2
L
dmx
dx
Example
• Consider a thin, uniform rod of mass M and length L. What is the rotational inertia of the rod through its com?
2
2
2
L
L
dxL
MxI
Consider an element dm of width dx. The mass of dm is the mass per unit length times the length of dm
2
L
2
L
dmx
dx
Example
• Consider a thin, uniform rod of mass M and length L. What is the rotational inertia of the rod through its com?
2
2
2
L
L
dxL
MxI
Consider an element dm of width dx. The mass of dm is the mass per unit length times the length of dm
2
2
3
3
L
L
x
L
M
2121 ML
Torque
• A simple definition of torque is an influence which tends to change the rotational motion of an object.
• The Torque = Force applied X perpendicular distance from the axis or point of rotation to the line of action of the force.
sinFrFrAlt:
• Newton’s 2nd law can be rewritten for a rotating body as:
tt maF (at is the tangential acceleration)
• Newton’s 2nd law can be rewritten for a rotating body as:
tt maF (at is the tangential acceleration)
rmarF tt rrm 2rm
• Hence we can write that, I
Example
• A uniform disk of mass M=2.5kg and radius R=20cm is mounted on a fixed horizontal axle. A block of mass m=1.2kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk and the tension in the cord, assuming that the cord does not slip and there is no friction at the axle.
• Sol: Diagram
• Sol: Free body diagrams
TMg
T
• Considering the block:
Mg
T
mamgT
• Considering the block:
• Considering the disk:Mg
T
mamgT
T
IFr
• Considering the block:
• Considering the disk:Mg
T
mamgT
T
IFr
221 MRRT
where the torque is negative because it causes a clockwise rotation
• Considering the block:
• Considering the disk:Mg
T
mamgT
T
IFr
221 MRRT
R
aMRRT 2
21 MaT 2
1
R
a
• Substituting for T we get,
• Hence,
• Finally,
Mg
T
28.42
2
msmM
mga
T2/24 sradR
a
NMaT 0.621