Transcript Dynamics_RotationI - University of Manchester

Dynamics
Dynamics
READ the
Textbook!
Rotational Motion
Part III – “What we
Angular velocity
know is a drop, what we
don't know is an ocean.”
Angular “suvat”
Kinetic energy – rotation
Moments of Inertia
KE – linear, rotation
http://www.hep.manchester.ac.uk/u/parkes/Chris_Parkes/Teaching.html
October 2012
Chris Parkes
Rotation of rigid body
about an axis
• θ in radians
s = rq – 2πr is circumference
• Angular velocity
dq – same for all points in
w=
dt body
ds
r dq
s = rq , = v =
= wr
dt
dt
• Angular acceleration
dw d 2q
a= = 2
dt dt
v
s = rθ
r
θ
r is perpendicular distance to axis
r
dθ
ds
Rotational Motion
This week – deal with components, next week with vectors
Linear
Rotational
Position, x
Velocity, v
Acceleration
Angle, θ
angular velocity, ω
angular acceleration, α
Rotational “suvat”
w
derive:
t
ò dw = ò a dt
w0
t
Þ w - w0 = at
0
2
1
w
dt
Þ
q
q
=
w
t
+
a
t
ò
0
0
2
0
w = w0 + at
q = q 0 + w 0t + 12 a t 2
(w + w 0 )
q = q0 +
t
2
w 2 = w 02 + 2a (q - q 0 )
constant angular acceleration
Rotational Kinetic Energy
X-section of a rigid body rotating about an axis through O which is perpendicular
to the screen
ω
m1
m2
r2
r1
O
r3
m3
r1 is perpendicular dist of
m1 from axis of rotation
r2 is perpendicular dist of
m2 from axis of rotation
r3 is perpendicular dist of
m3 from axis of rotation
Kinetic Energyi = 12 mi vi2 = 12 mi (w ri )2
Total KE = 12 w 2 Smi ri 2
Moment of Inertia Smi ri2
Moment of Inertia, I
• corresponding angular quantities for linear
2
masses m
quantities
I   mi ri  r 2 dm distance
from

i
rotation axis r
– x; v; pL
– Mass also has an equivalent: moment of Inertia, I
2
1
K
.
E
.

mv
2
– Linear K.E.:
– Rotating body v, mI: K.E.  12 I 2
– Or p=mv becomes: L  I 
Conservation of ang. mom.: I11  I 2  2
e.g.
frisbee
solid sphere
hula-hoop
2
2
I  1 M (R  R )
pc hard disk
I  MR
1
2
2
neutron star
I  52 MR2

R

space station

1
2
R1
R2
R
2
Moment of Inertia Calculations
Systems of discrete particles
Four equal point masses , each of mass 2 kg are arranged in the xy plane as
shown. They are connected by light sticks to form a rigid body. What is the
moment of inertia of the system about the y-axis ?
y
I   mi ri 2
2 kg
i
x
2 kg
2 kg
I = 2 x (2 x12) + (2 x 22)
= 12 kg m2
2 kg
1m
2m
Parallel-Axis Theorem
• Moment of Inertia around an axis
axis parallel to first at distance d
– Co-ordinate system origin at centre-of-mass
I = ICM + Md 2
yA
xA
• Proof
dI = mi d 2 = mi ((xi - x A )2 + (yi - yA )2 )
I = Smi ((xi - x A )2 + (yi - yA )2 )
I = Smi (xi2 + yi2 ) - 2x A Smi xi - 2yA Smi yi + (x A2 + yA2 )Smi
Smi xi = xCM = 0, Smi yi = yCM = 0
I = I CM + Md 2
Since centre-of-mass is origin of co-ordinate system
Table 9.2 Y& F p 291 (14th Edition)
Translation & Rotation
• Combining this lecture and previous ones
• Break problem into
– Velocity of centre of mass vCM
– Rotation about axis w
ω
2
Kinetic Energyi = 12 mi (vCM + vi )2 = 12 mi (vCM
+ vi2 + 2vCM vi )
2
Total KE = 12 MvCM
+ 12 Smi r 2w 2 + 0
vCM
(since å m i vi = 0)
CM
2
Total KE = 12 MvCM
+ 12 ICM w 2
Rolling without Slipping
•
•
•
•
Bicycle wheel along road
Centre – pure translation
Rim –more comlex path known as cycloid
If no slipping vcm = rw

R
R
M
M
m
h
h
m
v
A light flexible cable is wound around a flywheel of mass M and radius R. The
flywheel rotates with negligible friction about a stationary horizontal axis. An
object of mass m is tied to the free end of the cable. The object is released from rest
at a distance h above the floor. As the object falls the cable unwinds without
slipping. Find the speed of the falling object and the angular speed of the flywheel
just as the object strikes the floor.
KEY POINTS

• Assume flywheel is a solid cylinder
R
• Note – cable is light – hence ignore its mass
M
• Note – cable unwinds without slipping - speed
of point on rim of flywheel is same as that of the
cable
• If no slipping occurs, there must be friction
between the cable and the flywheel
• But friction does no work because there is no
movement between the cable and the flywheel
h
m
v
A light flexible cable is wound around a flywheel of mass M and radius R. The
flywheel rotates with negligible friction about a stationary horizontal axis. An
object of mass m is tied to the free end of the cable. The object is released from rest
at a distance h above the floor. As the object falls the cable unwinds without
slipping. Find the speed of the falling object and the angular speed of the flywheel
just as the object strikes the floor.
Example
A bowling ball of radius R and mass M is
rolling without slipping on a horizontal surface
at a velocity v. It then rolls without slipping up
a slope to a height h before momentarily
stopping and then rolling back down. Find h.
v=0
ω=0
h
v
No slipping so no work done against
friction – Mechanical energy
conserved