Physic 231 Lecture 20
•
•
Main points of today’s lecture:
I = mi ri 2
Moment of inertia:
∑
i
2
I = I CM + MRCM
•
Parallel axis theorem:
•
•
Analogies between rotational and translational motion
Solutions involving: I ; τ = Iα
•
Rolling motion:
KE rot =
1 2
Iω ;
2
KEtotal =
1
1
Mv 2 + Iω 2
2
2
L = Iω
Moment of inertia for collections of masses
•
Let’s consider the tangential acceleration of a
single mass m1 on attached to a mass rod of
length r1 and fixed to rotate about axis O:
F1t = m1a1t = m1 (r1α ) ⇒ τ 1 = F1t r1 = m1r12α = I1α
•
If we attach a second mass m2 on attached to a
mass rod of length r2 which is rigidly attached
to the pivot so its angular acceleration is the
same:
F2 t = m2a2 t = m2 (r2α ) ⇒ τ 2 = F2 t r2 = m2 r22α
τ net = τ 1 + τ 2 = m1r12α + m2 r22α = (m1r12 + m2 r22 )α = Iα
•
Similarly for three particles:
τ net = τ 1 + τ 2 + τ 3 = (m1r12 + m2 r22 + m3r32 )α = Iα
•
In general:
I = m1r12 + m2 r22 + m3r32 + m4 r42 + .... ≡ ∑ mi ri 2
i
F1t
m2
r2
rr1
1
m3
m1
Fig 8.14, p.229
Slide 25
Moments of inertia of extended objects
•
To get the moments of inertia, one
calculates I=Σmiri2 as well. Some
values are given in the table to the
right.
•
Last row is an illustration of the
“parallel axis theorem”:
2
I = I CM + MRCM
•
Here ICM is the moment of inertial
about an axis going through the
center of mass. I is the moment of
inertia about another axis which is
parellel to and displaced a distance
RCM from the center of mass.
Analogies between straight line and rotational motion
•
We already introduced the following analogous equations and quantities:
∆θ = ω t
ω = ω0 + αt
∆x = v t
v = v0 + at
1
(ω + ω0 )
2
1
∆θ = (ω + ω 0 )t
2
1
∆θ = ω 0t + αt 2
2
ω 2 = ω02 + 2α∆θ
1
(v + vo )
2
1
∆x = (v + vo )t
2
1
∆x = v0t + at 2
2
v 2 − v02 = 2a∆x
ω=
v=
•
New analogies:
m
I
F = ma
KEtrans =
p = mv
τ = Iα
1 2
mv
2
KErot
1 2 rotational
= Iω kinetic energy
2
L = Iω
angular momentum
Example
•
A turntable platter has a radius of 0.150 m and is rotating at 3.49 rad/s.
When the power is shut off, the platter slows down and comes to rest
in 15.0 s, due to a net retarding torque of 6.2x10-3 N⋅m. Assume the
platter to be a uniform solid disk. Determine the mass of the platter.
(Hint: calculate α, plug into Newton’s second law to find M.)
1
– a) 1.37 kg
I = MR 2
2
– b) 2.37 kg
ω − ω0 − 3.49rad / s
– c) 3.37 kg
ω = ω0 + αt ⇒ α =
=
= −.233rad / s 2
15s
t
– d) 4.37 kg
1
τ = Iα = MR 2α
2
τ
− 6.2 x10−3 Nm
= 2.37kg
⇒M =
=
1 2
1
(0.15m )2 (− 0.233rad / s 2 )
Rα
2
2
Example
•
vi
ri
A car is designed to get its energy from a rotating flywheel with a
radius of 2.00 m and a mass of 500 kg. Before a trip, the flywheel is
attached to an electric motor, which brings the flywheel’s rotational
speed up to 5000 rev/min. (a) Find the kinetic energy stored in the
flywheel. (b) If the flywheel is to supply energy to the car as would a
10.0-hp motor, find the length of time the car could run before the
flywheel would have to be brought back up to speed.
1
1
1
1
2
KEi = mi vi2 = mi (riω ) = mi ri 2ω 2 = I iω 2
2
2
2
2
mi
1
1
1
KEtotal = ∑ mi vi2 = ∑ I iω 2 = I totalω 2
2
i 2
i 2
a ) ω = 5000 / 60 ⋅ 2πrad / s = 5.23x102 rad / s
1
1
2
MR 2 = ⋅ 500kg ⋅ (2m ) = 1000kg ⋅ m 2
2
2
2
1
1
KE = Iω 2 = (1000kg ⋅ m 2 )(5.23x102 rad / s )
2
2
KE = 1.37 x108 J
I=
KE
1.37 x108 J
b) KE = P∆t ⇒ ∆t =
=
= 1.8 x104 s ≈ 5h
(10)(746W / hp )
P
Which one wins?
•
Which object will be fastest?
– a) hoop
– b) cylinder
– c) sphere
Rolling down the hill
•
Starting from rest, an object with moment of mass M, moment of
inertia I, and radius R rolls down an incline of height h. What is the
velocity at the bottom? (Hint use conservation of energy.)
KEtotal = KEtrans + KErot
h
Conservation of energy :
KEtotal
2
Mgh =
v=
What
foraslides
acylinder?
hoop?
••• What
If theisisobject
vvfor
without
sphere?
2) 2 2
(I=MR
(I=1/2MR
rolling?
(I=0)
))
(I=2/5MR
1
1 2
2
= Mv + Iω = Mgh
2
2
Rolling motion : v = ωR ⇒ ω =
v
R
1
1  v  1
1 I  2
Mv 2 + I   =  M +
v
2
2
2  R  2
2R 
Mgh
1 I 
1
M
+
 2
2 R 2 
1
1 2
2
= Mv + Iω
2
2
a ) v = gh
4
b) v =
gh
3
10
c) v =
gh
7
d) v = 2 gh
Conceptual question
•
A force F is applied to a dumbbell for a time interval ∆t, first as in (a) and
then as in (b). In which case does the dumbbell acquire the greater centerof-mass speed?
–
–
–
–
a) (a)
b) (b)
c) no difference
d) The answer depends on the rotational inertia of the dumbbell.
Angular momentum
•
v
v
In analogy to the linear momentum
p = mv we have
v
v
the angular momentum L = Iω. For a single particle
moving a circle, the magnitude of L is:
vt
rr1
m1
L = Iω = mr 2ω = mrvt
Fig 8.14, p.229
Slide 25
•
•
Recall that if there are no external forces, the linear momentum will be
conserved. Similarly, if there no external torques, the angular momentum will
be conserved.
For example, this means that an isolated object will keep the same value of the
angular momentum throughout a problem.
v
v
– In linear motion, conservation of p = mv for an isolated object meant that
v
the velocity v must remain constant.
v
– In angular motion, conservation of L = Iωv for an isolated object does not
v
mean ω must remain constant, because I can change!
L f = L0
⇒ I f ω f = I 0ω0
I0
⇒ ω f = ω0
If
Example
•
A woman stands at the center of platform. The woman and the
platform rotate with an angular speed of 5.00 rad/s. Friction is
negligible. Her arms are outstretched and she is holding a dumbbell in
each hand. In this position, the total moment of inertia of the rotating
system (platform, woman and dumbbells) is 5.4 kg⋅m2. By pulling in
her arms, the moment of inertia is reduced to 3.8 kg⋅m2. Find her new
angular speed.
ωf =
I0
ω0
If
ωf =
5 .4
5rad / s = 7.1rad / s
3.8