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Physics 111
Tuesday,
November 2, 2002
• Ch 11: Rotational Dynamics
Torque
Angular Momentum
Rotational Kinetic Energy
Tues
Phys
Nov
Announcements 111
.02.
• Wednesday, 8 - 9 pm in NSC 118/119
• Sunday, 6:30 - 8 pm in CCLIR 468
Tues
Phys
Nov
Announcements 111
.02.
This week’s lab will be on oscillations.
There will be a short quiz.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Let’s look at a balance scale...
L X?
50 g 250 g
If I hang a 50 g mass at …where should I hang
the end of the left side the 250 g mass to get
of the balance... the scale to balance?
Worksheet #1
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L/5
50 g 250 g
How did we arrive at such an answer?
W1 L2 Perhaps we used one of these
= formulae
W2 L1 m1 L1 = m2 L2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L/5
50 g 250 g
Let’s restate our balance condition so that we
get all the subscript 1’s on the same side and
all the subscript 2’s on the same side...
W1 L2
= L1W1 = L2W2
W2 L1
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L/5
50 g 250 g
L1W1 = L2W2
So in the case of this balance, we have...
L(50 g)(9.8 m/s ) = ( L / 5)(250 g)(9.8 m/s )
2 2
Good! The balance
490 gm/s2 L = 490 gm/s2 L is in balance!
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L/5
What happens to
250 g our balance if I
25 g
change the mass
on the left side
from 50 g to 25 g?
L L/5
25 g
250 g The scale begins
to rotate around
the pivot point.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Let’s look at one more case with our balance...
L L
X?
100 g 100 g 50 g X = L/2
How did we arrive at
Worksheet #2 this answer?
Now where should I hang the 100 g
mass to get the scale to balance?
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L
X?
100 g 100 g 50 g
First, we examined the left side of the balance:
gm
L1W1 = L(100 g)(9.8 m/s ) = 980L
2
s2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L
X?
100 g 100 g 50 g
Then, we examined the right side:
L2W2 + L3W3 = X (100 g)(9.8 m/s 2 ) + L(50 g)(9.8 m/s2 )
gm gm
980 X s 2 + 490L s2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L L
X?
100 g 100 g 50 g
If the scale is balanced, the two sides should
equal one another:
gm gm gm
980L s 2 = 980 X s 2 + 490L s2
X = L/2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
• In our scale experiments, we’ve been playing
around with a physical quantity that we’ve
not yet encountered formally.
• Our experiments show us that this quantity is
related to a force applied at a distance.
• The farther away from the pivot point the force
is applied, the greater our new quantity is.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
• When present and unbalanced, our new physical
quantity causes the scale to rotate in the
direction of the unbalanced force.
• We’ve also seen that the quantity is additive.
That is, if we have two forces on one side of the
pivot at two different distances, the resulting
physical quantity is simply the sum of the two
force * distance products.
• Before we name our new physical quantity,
let’s examine one more thought experiment.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L
50 g
Let’s put a block on
one side of our
balance and let go.
What will eventually happen to this system?
(Assume there is SOME friction in the pivot,
but that the pivot is free to rotate 360o.)
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
The block eventually
comes to rest
with the balance
aligned vertically.
We still have a mass of 50 g
at a distance L from the pivot L
point, so why has the motion
stopped? What has
happened to the value of our
new physical quantity? 50 g
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
For this case, we notice that
the gravitational force on the
mass is acting along the
length of the balance.
In other words, the force is
parallel to the radius of length
L around the pivot point.
L
In this case, the
magnitude of our
new physical
quantity must be 50 g
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Let’s finally give a name to
this new physical quantity. How about...
| τ |= F⊥ d
Where d is the distance from the pivot point (the
point of rotation) to the point at which the force
is applied and…
is the component of the applied
F force that is perpendicular to the
⊥ line joining the pivot point to the
point at which the force is applied.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
| τ |= F⊥ d
[| τ |] = [F⊥ ][d]
[| τ |] = N m
Although torque has the same units as energy
(J), we generally denote torques as forces
acting at a distance, and therefore leave the
units in the form N m.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
0.5 m
pivot Fg
What is the 50 g
magnitude of the
torque experienced Here, the force of
by this unbalanced gravity acts
balance? perpendicularly
to the radius joining
| τ |= F⊥ d the two yellow
points above.
| τ |= mgd = (0.05 kg)(9.8 m/s )(0.5 m) = 0.245 Nm
2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #3
≥
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #3
You are using a wrench and
trying to loosen a rusty nut.
Which of the arrangements
shown is most effective in
loosening the nut? List in order
of descending efficiency the
following arrangements:
1) 1, 2, 3, 4 5) 1, 3, 4, 2
2) 4, 3, 2, 1 6) 4, 2, 3, 1
3) 2, 1, 4, 3 7) 2, 3, 4, 1
4) 2, 4, 1, 3 8) 2, 3, 1, 4
PI, Mazur (1997)
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
m
Ft
r
pinned to
table
The tangential force results
in a tangential acceleration. Ft = mat
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
m
Ft
r
pinned to
table
It also creates a torque about the pinned point.
| τ |=| Ft || r |= m | at || r |
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
m
Ft
r
pinned to
table
Recalling that tangential acceleration is related
to angular acceleration, we get:
2
| τ |= m | at || r |= m(| r || α |) | r |= mr | α |
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
m
Ft
r
pinned to
table
This expression is valid so 2
long as our connecting τ = mr α
rod/string is massless.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
If we examine a more general system and plot
the relationship, we find
The slope of this
τ line is known as
the moment of
inertia, I.
α
For our previous
example,
Note: r is the distance to
the axis of rotation. I = mr 2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
2Imr= [I ] = [m][r ] 2
[I ] = kg m 2
If we have an extended object, we can compute
its moment of inertia about some axis from the
sum of the moments of each of the individual
pieces of that object:
N
I = m1r12 + m2r22 + m3r32 + ... + mN rN = ∑ mi ri2
2
i=1
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
So we can rewrite our
expression for torque
in terms of our new
quantity, the moment of
τ = Iα
inertia...
Notice the similarity
to our linear motion
expression of F = ma
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #4
≥
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #4
Two wheels with fixed hubs, each having a mass of 1 kg,
start from rest, and forces are applied as shown. Assume the
hubs and spokes are massless, so that the rotational inertia
is I = mR2. In order to impart identical angular
accelerations, how large must F2 be?
1. 0.25 N
2. 0.5 N
3. 1 N
4. 2 N
5. 4 N
PI, Mazur (1997)
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
For the ball to fall into the
basket, the linear acceleration
of the end of the stick must
exceed that of gravity.
Is that possible? 1) Yes 2) No
The moment of inertia of a stick 1
rotated about its end is given by I = ML2
3
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
τ net = F⊥ R = Iα
The force creating the torque in this case is
that of gravity, acting at the center of the stick.
L 1 2
τ net = Mg(cosθ ) = ML α
2 3
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
To get the linear
3g acceleration of the end of
α= (cosθ ) the stick, we simply multiply
2L
by the length of the stick.
atan = α L = 1.5g(cosθ )
So, if the cos θ > 2/3, the end of the stick
θ < 480
will move with a linear acceleration > g.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Recall in our Δv Δp
discussion of force, F = ma = m =
we discovered Δt Δt
where p was the
momentum p = mv
in the system
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
We can construct a completely analogous
argument to define angular momentum
Δω ΔL
τ = Iα = I =
Δt Δt
where L is the
angular momentum L = Iω
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
L = Iω
[ L] = [I ][ω ]
kg m 2
[ L] = (kg m )(rad/s) =
2
s
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
When there is no NET external torque
on a system, the angular momentum
of the system will be conserved.
Li = Iω i = Iω f = L f
So when examining isolated systems,
total energy, linear momentum, and angular
momentum are all conserved!
Conservation of Angular
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
We’ve studied this quantity
already in looking at masses
at some fixed distance, r, K = mv
1
2
2
from an axis of rotation.
When looking at an object rotating at a constant
angular velocity ω at a distance r from the axis,
K = m(rω ) = mr ω = Iω
1
2
2 1
2
2 2 1
2
2
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
K = Iω
1
2
2
[K ] = [ ][I][ω ]
1 2
2
[K ] = (kg m )(rad/s)
2 2
2
kg m
[K ] = 2
= Nm = J
s
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
So we’ve defined another form of kinetic energy.
Before, we studied translational kinetic energy.
Now we have rotational kinetic energy, too.
So the total kinetic energy of a system now
includes two terms (if the system is rotating
and translating):
Ktot = Kr + Kt
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #5
≥
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #5
A figure skater stands on one spot on the ice (assumed
frictionless) and spins around with her arms extended.
When she pulls in her arms, she reduces her rotational
inertia and her angular speed increases so that her
angular momentum is conserved. Compared to her initial
rotational kinetic energy, her rotational kinetic energy
after she has pulled in her arms must be
1. the same.
2. larger because she’s rotating faster.
3. smaller because her rotational inertia is smaller.
PI, Mazur (1997)
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
So, when we examine the total energy of
a system, we must take care to include the
rotational kinetic energy term as well.
The total energy (including rotational kinetic
energy) will be conserved so long as NO non-
conservative forces are acting upon the system.
Etot = Kr + Kt + U g ( +U spring )
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
We have studied objects in “equilibrium” before.
What is the defining characteristic
of an object in equilibrium?
And from our earlier studies, the fact that it
did not accelerate indicated that:
Fnet = 0
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
But is this a sufficient condition to ensure
that in fact an object does not accelerate
in some way?
This “zero net force” condition
only guarantees that the center of mass
of the object on which the forces are
being exerted does not accelerate.
Let’s look at the meter stick again...
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Worksheet #6
Exert a constant force
tangential to the circle
of motion. Top view
Holding pencil through the
pivot point.
Draw a free-body
Assume the system is
diagram for this Oriented in a horizontal plane
system.
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
Is the net force in this system zero?
So why does the ruler rotate with
increasing speed?
Because the net torque is NOT zero!
We have a new equilibrium condition:
τ net = 0
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
For an object to be in STATIC equilibrium
(i.e., at rest, not moving, not rotating), the
following TWO conditions must be met.
Fnet = 0 τ net = 0
(We’re free to pick ANY origin about which to
computer our torques in this case.)
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
A student sits on a rotating stool holding two
weights, each of mass 3.00 kg. When his arms
are extended horizontally, the weights are each
1.00 m from the axis of rotation, and he rotates
at an angular speed of 0.750 rad/s. The
moment of inertia of the student + stool is 3.00
kg m2 and is assumed to be constant. The
student now pulls the weights horizontally in to
a distance of 0.300 m from the rotation axis. (a)
What’s the new angular speed? (b) What are the
initial and final kinetic energies of this system?
Problem #1
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
(a) What’s the new angular speed? (b) What are the
initial and final kinetic energies of this system?
Pictorial Representation: Knowns:
ωi Istudent = 3.00 kg m2
Initially:
m = 3.00 kg
ri m ri = 1.00 m
rf = 0.30 m
ωi = 0.75 s-1
Finally: ωf
rf m Unknowns:
ωf
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
(a) What’s the new angular speed? (b) What are the
initial and final kinetic energies of this system?
Itotal = Iweights + Istudent = 2 (mr2) + (3 kg m2)
Initially:
Itotal = 2 (3 kg)(1 m)2 + (3 kg m2) = 9.00 kg m2
Finally:
Itotal = 2 (3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2
Ii ωi = If ωf
Ii (9kg m 2 )
ω f = ωi = 2
(0.75rad/s) = 1.91rad/s
If (3.54kg m )
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
(a) What’s the new angular speed? (b) What are the
initial and final kinetic energies of this system?
Initially:
Ki = 0.5 Ii ωi2 = 0.5 (9 kg m2)(0.75 rad/s)2 = 2.53 J
Finally:
Kf = 0.5 If ωf2 = 0.5 (3.54 kg m2)(1.91 rad/s)2 = 6.45 J
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
A uniform horizontal beam weighs 300 N, is 5.00 m
long, and is attached to a wall by a pin connection
that allows the beam to rotate. Its far end is
supported by a cable that makes an angle of 53o
with the horizontal. If a 600-N person stands
1.50 m from the wall, find the tension in the cable
and the force exerted by the wall on the beam.
1.5 m
Problem #2
53o
5.0 m
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
A uniform horizontal beam weighs 300 N, is 5.00 m
long, and is attached to a wall by a pin connection
that allows the beam to rotate. Its far end is
supported by a cable that makes an angle of 53o
with the horizontal. If a 600-N person stands
1.50 m from the wall, find the tension in the cable
and the force exerted by the wall on the beam.
Free-body T
diagram f 5.0 m 53o
for the
N Fc W
plank
1.5 m
2.5 m
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
y
From the free-body T
f x
diagram we can write 5.0 m 53o
down the torque equation
N Fc W
and Newton’s 2nd Law in
1.5 m
the x and y directions.
2.5 m
Compute the torques about the left end of the
plank, where the plank meets the wall:
τ net = f (0m) − Fc (1.5m) − W (2.5m) + (T sin53 )(5.0m) = 0
0
0 = 0 − (600N)(1.5m) − (300N)(2.5m) + (4.0m)T
T = 412.5 N
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
y
From the free-body T
f x
diagram we can write 5.0 m 53o
down the torque equation
N Fc W
and Newton’s 2nd Law in
1.5 m
the x and y directions.
2.5 m
Newton’s 2nd Law in the x-direction:
Fnet,x = N − T cos53 = 0
0
0 = N − (412.5N)cos53 0
N = 248 N
Tues
Phys
Nov
Ch 11: Rotational Dynamics 111
.02.
y
From the free-body T
f x
diagram we can write 5.0 m 53o
down the torque equation
N Fc W
and Newton’s 2nd Law in
1.5 m
the x and y directions.
2.5 m
Newton’s 2nd Law in the y-direction:
Fnet, y = f + T sin53 − Fc − W = 0
0
0 = f + (412.5N)sin53 − 600N − 300N
0
f = 571 N
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