Chapter 12Rotational Motion
1) 1 radian = angle subtended by an arc (l)
whose length is equal to the radius (r)
2)
q=l
r
l
r
q
3) 3600 = 2p radians
4) Radians are dimensionless
A bird can only see objects that subtend an
angle of 3 X 10-4 rad. How many degrees is
that?
3 X 10-4 rad
360o = 0.017o
2p rad
How small an object can the bird
distinguish flying at a height of
100 m?
q=l
r
q
r
l=qr
l = (3 X 10-4 rad)(100m)
l = 0.03 m = 3 cm
l (approx.)
How at what height would the bird
be able to just distinguish a rabbit
that is 30 cm long (and tasty)?
q
r
(ANS: 1000 m)
l (approx.)
The Mighty Thor
swings his hammer at
400 rev/min. Express
this in radians/s.
400 rev 1 min 2p rad
1 min 60 s 1 rev
= 13.3p rad/s or
41.9 rad/s
Formula Review
v = rw
at = ra
ar = v2
r
or
ar = w2r
Converting between Angular and
Linear Quantities
atan
Linear = Radius X Angular
v = rw
atan = ra
Note the use of atan to differentiate from
centripetal acceleration, ac or ar:
ar
Angular Kinematics
v=vo + at
x = vot + ½ at2
v2 = vo2 + 2ax
w=wo + at
q = wot + ½ at2
w2 = wo2 + 2aq
A DVD (Encino Man, Director’s Cut), rotates from
rest to 31.4 rad/s in 0.892 s.
a. Calculate the angular acceleration. (35.2 rad/s2)
b. How many revolutions did it make? (2.23
revolutions)
A car engine idles at 500 rpm. When the light
turns green, it accelerates to 2500 rpm in 3.0 s.
a. Convert the angular velocities to rad/s
b. Calculate the angular acceleration
c. Calculate the number of revolutions the wheel
undergoes.
A bicycle slows from vo = 8.4 m/s to rest over a
distance of 115 m. The diameter of each wheel
is 68.0 cm.
a. Calculate the angular velocity of the wheels
before braking starts. (24.7 rad/s)
b. How many revolutions did each wheel
undergo?( 53.8 rev)
c. What was the angular acceleration? (-0.903
rad/s2)
Center of mass
• one point on a particle that follows the same
path.
• Point at which the force of gravity can be
considered to act (uniform gravity field, Center
of gravity)
General Motion
1. Translational Motion
1. all points of an object follow the same path
2. Sliding a book across a table
2. Rotational Motion
3. General Motion – combination of
translational and rotation motion
Translational
Translational and Rotational
xcm = 1 ∫ mixi = m1x1 + m2x2 + ….
M
m1 + m2
xcm = 1 ∫ x dm
M
A 500 g ball and a 2.0 kg ball are connected by
a massless 50 cm rod.
a. Calculate the center of mass (0.10 m)
b. Calculate the linear speed of each ball if they
rotate at 40 rpm. (0.42 m/s, 1.68 m/s)
Where is the center of mass for the Earth-Moon system.
Assume the center of the Earth is the origin. Some
values you need are:
mEarth = 5.97 X 1024 kg
mMoon = 7.35 X 1022 kg
Earth-Moon distance = 3.84 X 108 m
(4.45 X 106 m)
Using calculus, find the center of mass of a thin
uniform rod of mass M and length L. Use this result
to rind the tangential acceleration of a 1.60 m long
rod rotates about its center with an angular
acceleration of 6.0 rad/s2.
xcm = 1 ∫ x dm
M
dm = M dx
L
(substitute and integrate from 0 to L)
Translational and Rotational Speed
• Does a rotating helicopter blade have kinetic
energy before the helicopter takes off?
• How about afterwards?
• Does all of the energy of the fuel go into moving
the helicopter?
Translational and Rotational Speed
Translational Speed (v)
– speed of the center of a wheel with respect to the
ground
– Can also be called linear speed
– Use regular KE = ½ mv2
Rotational speed (w)
– angular speed of the wheel
– Use KE = ½ Iw2
Deriving the Rotational KE
KE = S½ mv2
v = rw
KE = S ½ m(rw)2
KE = S ½ mr2w2
I = S mr2
KE = ½ Iw2
(I is moment of inertia)
Law of Conservation of Mechanical
Energy
Emech= KEt + KEr + PEt
a. Calculate the moment of inertia of the following
“widget.” Assume the connecting rods are
massless. (2.14 X 10-3 kg m2)
b. Calculate the angular velocity in rpm if the
system has 100 mJ of rotational kinetic energy
(92 rpm)
Rotational KE: Example 1
What will be the translational speed of a log (100
kg, radius = 0.25 m, I= ½ mr2) as it rolls down a
4 m ramp from rest?
4m
(KEt + KEr + PEt)i = (KEt + KEr + PEt)f
(0 + 0 + mgy)i = ( ½ mv2 + ½ Iw2 + 0)f
mgy = ½ mv2 + ½ Iw2
2mgy = mv2 + Iw2 (multiplied both sides by 2)
v = rw so w = v/r
2mgy = mv2 + Iv2
r2
I = 1/2 mr2
2mgy = mv2 + 1mr2v2
2r2
2mgy = mv2 + 1mr2v2
2r2
2gy = v2 + 1v2
2
2gy = 2v2 + 1v2
2
2
2gy = 3v2
2
v = 4gy = 4(9.8m/s2)(4.00 m) = 7.23 m/s
3
3
Now we can calculate the angular speed
v = rw so w = v/r
w = 7.23 m/s = 2.9 rad/s
0.25 m
2.9 rad
1 rev
=
s
2p rad
0.46 rev/s
Moment of Inertia (I)
• Measure of Rotational Inertia
• An objects resistance to a change in angular
velocity
• Would it be harder to push a child on a
playground merry-go-round or a carousel?
• I = moment of inertia
• I = mr2
• More properly
I = Smr2 = m1r12 + m2r22 +….
St= Ia
Would it be harder (require more torque) to twirl a
barbell in the middle (pt. M) or the end (Pt. E)
E
M
Moment of Inertia: Example 1
Calculate the moment of inertia (I) for the barbell
when rotated about point M. We will assume the
barbell is 1.0 m long, and that each weight is a
point mass of 45.4 kg.
I = Smr2 = (45.4 kg)(0.50 m)2 + (45.4 kg)(0.50 m)2
I = 22.7 kg-m2
M
Moment of Inertia: Example 2
Now calculate I assuming Mr. Fredericks twirls the
barbells from point E.
I = Smr2 = (45.4 kg)(0 m)2 + (45.4 kg)(1 m)2
I = 45.4 kg-m2
E
Calculate I for Bouncing Boy (75 kg, radius = 1.2 m).
Use the formulas from the book.
I = 2/5 MR2
I = (2)(75 kg)(1.2 m)2
5
I = 43.2 kg-m2
What will be the translational speed of Bouncing
Boy (75 kg, radius = 1.2 m) as he rolls down a 3
m ramp from rest? (6.50 m/s)
3m
What would his speed be if he just slid down the
ramp?
(KEt + KEr + PEt)i = (KEt + KEr + PEt)f
(0 + 0 + mgy)i = ( ½ mv2 + 0 + 0)f
mgy = ½ mv2
gy = ½ v2
v2 = 2gy
v = \/2gy = (2 X 9.8 m/s X 3.00 m)1/2
v = 7.7 m/s
Why is this larger than if he rolls?
Calculate whether a 5.0 kg sphere, a 5.0 kg hoop,
or a 5.0 kg cylinder will reach the bottom of a 1.0
m tall ramp first.
A 1.0 m long, 200 g rod is hinged at one end and
allowed to fall. (I = 1/3ML2) . Note: use the
height of the center of mass for potential
energy height.
a. Derive the formula for the linear speed of the tip
of the bar at the bottom on the fall. (v=(3gL)1/2)
b. Calculate the speed of the tip of the rod as it hits
the wall. (5.4 m/s)
Calculus and Moment of Inertia.
I = ∫ r2 dm
Always need to substitute for dm in terms of r
Calculate the moment of inertia of a thin rod of
length L and mass M that pivots at one end.
I = ∫ x2 dm
dm = M dx
L
I = ∫ x2 M dx
L
Calculate the moment of inertia of a thin disk of
radius R and Mass M.
I = ∫ r2 dm
dm = M dA
A
dA = 2prdr
Calculate the moment of inertia of a thin rod of
length L and mass Mthat pivots through the
middle. (Hint: integrate from -1/2L to 1/2L)
Parallel Axis Theorem
I = Icm + Md2
Determine the moment of
inertia of a thin rod of mass
M and length L one-third
of the length from one end.
Determine the formula for the moment of inertia of
a solid sphere about an axis on the edge of the
sphere. (7/5Mr2)
Calculate the moment of inertia if the mass is 12.0
kg and the radius is 0.75 m. (9.45 kg m2)
Torque – tendency of a force to rotate a body
about some axis (the force is always
perpendicular to the lever arm)
t = Frsinq
t = Ia
r
pivot
F
Torque Sign Conventions
Counter-clockwise
Torque is positive
Clockwise
Torque is negative
A wrench is 20.0 cm long and a 200.0 N force is
applied perpendicularly to the end. Calculate the
torque.
t = Fr
t = (200.0 N)(0.20 m)
t = 40.0 m-N
20.0 cm
200.0 N
Suppose that same 200.0 N force is now applied at
a 60o angle as shown. Calculate the Torque.
(34.6 m-N)
20.0 cm
200.0 N
60o
The biceps muscle exerts a 700
N vertical force. Calculate
the torque about the elbow.
t = Fr = (700 N)(0.050 m)
t = 35 m-N
Two wheels, of radii r1 = 30 cm and r2 =50 cm
are connected as shown.
a. Calculate the net torque on this compound
wheel when two 50 N force act as shown. (-6.5
m-N)
50 N
30o
r2
r1
50 N
Note that
Fx will
pull the
wheel
Out in space, two rockets are docked, connected
by a 90 m tube. One has a mass of 100,000 kg,
the other 200,000 kg. They both fire their
rockets with 50,000 N of thrust in opposite
directions.
a. Calculate the center of mass of the system (60
m)
b. Calculate the moment of Inertia about the center
of mass (use mr2 for the rockets) (5.4 x 108)
c. Calculate the net torque. (4.5 X 106 Nm)
d. Calculate the angular acceleration. (8.33 X 10-3
rad/s2)
50,000 N
100,000 kg
200,000 kg
90 m
50,000 N
An airplane engine provides a torque of 60 Nm
to a 2.0 m long, 40 kg propeller. It starts from
rest.
a. Calculate the moment of inertia. (13.33 kg m2)
b. Calculate the angular acceleration (4.50 rad/s2)
c. Calculate how long it will take to get up to 200
rpm. (4.6 s)
A 5.0 kg disk rotates on a vertical
axle that is off center. A
vertical cable exerts a 100 N
force as shown.
a. Calculate the torque due to
gravity at the center of mass.
(1.23 kg m2)
b. Calculate the (3.73 kg m2)
c. Calculate the moment of inertia
using the parallel axis theorem.
(9.38 X 10-3 kg m2)
d. Calculate the angular
acceleration. (400 rad/s2)
A 15.0 N force is applied to a
cord wrapped around a pulley
of radius 33.0 cm. The pulley
reaches an angular speed (w)
of 30.0 rad/s in 3.00 s.
a. Calculate the torque (4.95 m-N)
b. Calculate the angular
acceleration. (10.0 rad/s2)
c. Calculate the moment of inertia
of the pulley. (0.495 kg-m2)
33.0 cm
15.0 N
A 15.0 N force is applied to a cord
wrapped around a pulley of
radius 33.0 cm. The pulley
reaches an angular speed (w) of
30.0 rad/s in 3.00 s. Since this is
a real pulley, there is a frictional
torque (tfr= 1.10 m-N) opposing
rotation.
a. Calculate the net torque. (3.85
Nm)
b. Calculate the angular acceleration
(10.0 rad/s2)
c. Calculate the moment of inertia
of the pulley. (0.385 kg-m2 )
33.0 cm
15.0 N
Using the same pulley as
the previous problem
(I=0.385 m-N), hang a
15.0 N bucket (1.53 kg)
from the cord. There is a
frictional torque of 1.10
m-N opposing rotation.
Calculate the angular and
linear acceleration of the
bucket.
33.0 cm
FT
We will break this problem into two parts: pulley
and bucket. Let’s first look at the pully:
St = FTr – tfr
St = Ia
Ia = FTr – tfr
Now we will deal with the bucket
SF = mg – FT
ma = mg – FT
a= ra
mra = mg – FT
33.0 cm
FT
FT
mg
tfr
Two equations, two unknowns
Ia = FTr – tfr
mra = mg – FT
FT = mg - mra
Ia = (mg - mra)r – tfr
Ia = mgr - mr2a - tfr
Ia + mr2a = mgr – tfr
a(I + mr2) = mgr – tfr
Solve for FT
Substitute
a = mgr – tfr = (15.0 N)(0.330m)-1.10m-N
I + mr2
(0.385 m-N) + (1.53 kg)(0.330m)2
a = 6.98 rad/s2
a = ra = (0.330 m)(6.98 rad/s) = 2.30 m/s2
Now calculate the rotational speed of the pully (w) and the
linear speed of the bucket after 3.00 s.
w=wo + at
w= 0 + at = (6.98 rad/s2)(3.00 s) = 20.9 rad/s
v=rw
v = (0.330 m)(20.9 rad/s) = 6.90 m/s
Two 50 lb. children sit on a see-saw. Will they
balance in all three cases?
Review of Torque
Mr. Fredericks can’t quite
budge this rock. How can
he increase his torque so
he can move it?
t
= Fr
1. Increase the lever arm
length
2. Decrease “r”
Conditions for Static Equilibrium
St = 0
SF = 0
St = 0: Example 1
Two children sit on a see-saw as shown. The board of the
see-saw has a mass of 2.00 kg centered at the pivot.
Where should the 25.0 kg child sit so that they are in
perfect balance?
?
2.50 m
30.0 kg
25.0 kg
Let’s pick the pivot as the origin
St = 0
0=(245 N)(x) – (294 N)(2.5 m) – (FN)(0) + (mg)(0)
0 = (245 N)(x) – (294 N)(2.5 m)
(245 N)(x) = 735 m-N
x = 3.0 m
2.5 m
FN
x
mg
294 N
245 N
SF = 0 and St = 0: Example 1
A heavy printing press is
placed on a large beam
as shown. The beam
masses 1500 kg and
the press 15,000 kg.
Calculate the forces on
each end of the beam.
SFy = 0 and St = 0 (we’ll ignore Fx)
SFy = 0 = F1 + F2 - (1500 kg)(g) – (15,000 kg)(g)
0 = F1 + F2 – 1.617 X 105 N
F1 + F2 = 1.617 X 105 N
We’ll choose F1 as the pivot
St = 0
0 = (F1)(0 m) – (1500kg)(g)(10m) - (15,000 kg)(g)(15 m) +
(F2)(20 m)
0 = -2.352 X 106 m-N + (F2)(20 m)
(F2)(20 m) = 2.352 X 106 m-N
F1 + F2 = 1.617 X 105 N
(F2)(20 m) = 2.352 X 106 m-N
(F2) = 2.352 X 106 m-N/(20m) = 1.176 X 105 N
F1 + 1.176 X 105 N = 1.617 X 105 N
F1 = 4.41 X 104 N
SF = 0 and St = 0: Example 3
The beam below has a mass of 1200 kg. Calculate
F1 and F2 for the cantilever as shown. Assume
the center of gravity is at 25 m.
F1
F2
20.0 m
30.0 m
(1200 kg)(g)
SFy = 0 and St = 0
SFy = 0 = F1 + F2 - (1200 kg)(g)
0 = F1 + F2 – 11,760 N
F1 + F2 = 11,760 N
We’ll choose F1 as the pivot
St = 0
0 = (F1)(0 m) + (F2)(20m) - (11,760 N)(25 m)
0 = (F2)(20m) – 294,000 N
(F2)(20m) = 294,000 N
F1 + F2 = 11,760 N
(F2)(20m) = 294,000 N
F2 = 294,000 N/20 m = 14,700 N
F2 = 14,700 N
F1 = 11,760 N - F2
F1 = 11,760 N - 14,700 N
F1 = -2940 N
(we picked the wrong direction for Force 1)
SF = 0 and St = 0: Example 4
A sign of mass M = 280
kg is suspended from a
25.0 kg beam that is
2.20 m long. Angle q is
30o, Calculate FH and
FT, the forces at the
hinge and the tension in
the wire.
SFy = 0
0 = FHY + FTY –mg - Mg
0 = FHY + FTY – (25 kg)(g) – (280 kg)(g)
0 = FHY + FTY – 2989N
FHY + FTY = 2989N
St = 0
(choose sign point as origin)
0 = -(FHY)(2.20m) + (mg)(1.10 m) + (FTY)0 + (Mg)0
0 = -(FHY)(2.20m) + (270 m-N)
(FHY)(2.20m) = (270 m-N)
FHY + FTY = 2989N
(FHY)(2.20m) = (270 m-N)
FHY = 270 m-N/2.20 m = 123 N
FHY + FTY = 2989N
123 N + FTY = 2989N
FTY = 2870 N
Also, SFX = 0
0 = FHX - FTX
FHX = FTX
What we know
FHX = FTX
FHY = 123 N
FTY = 2870 N
qT= 30o
FT
FTY
qT
FTX
sin q = FTY/F
F = FTY/sin q = 2870 N/sin 30o = 5740 N
FTX = Fcos q = (5740 N)(cos 30o) = 4970 N
Hinge Side
FHY = 123 N
FHX = FTX
FTX = 4970 N
FHX = 4970 N
FHY
FH
FHX
FH2 = FHX2 + FHY2
FH = \/ FHX2 + FHY2 = \/ (4970 N)2 + (123 N)2
FH = 4972 N (the y-component doesn’t add much)
SF = 0 and St = 0: Example 5
A crane is designed to
hold a maximum of
10,000 kg (M). The
top crossbeam has a
mass of 450 kg and
the angled beam
makes and angle q =
35o.Calculate the
forces exerted by the
post (Fp) and the
angled beam (FA).
5.0 m
15.0 m
q
FA
Fp
M
FAx
Fpy
Fp
FAy
q
FA
Fpx
SFx =0
0 = Fax – Fpx
Fax = Fpx
SFy =0
0 = FAy – Fpy – mg – Mg
0 = FAy – Fpy – (450 kg)(g) – (10,000kg)(g)
0 = FAy – Fpy – 1.02 X 105 N
St = 0 (choose top of post as origin)
0 = (Fpy)(0) – (450 kg)(g)(10m) – (10,000 kg)(g)(20m) + (Fay)(15m)
0 = (Fay)(15m)– 2.004 X 106 m-N
(FAy)(15m) = 2.004 X 106 m-N
FAy = 2.004 X 106 m-N/15 m
FAy = 1.34 X 105 N
0 = FAy – Fpy – 1.02 X 105 N
Fpy = FAy - 1.02 X 105 N
Fpy = 3.20 X 104 N
sin 35o = FAy
FA
FA = Fay/sin 35o
FA = 1.34 X 105/sin 35o = 2.34 X 105 N
FAx = Fcos 35o = 1.92 X 105 N
FAx
FAy
q
FA
Fax = Fpx = 1.92 X 105 N
Fp2 = Fpx2 + Fpy2
Fp2 = Fpx2 + Fpy2
Fp = \/ Fpx2 + Fpy2 =\/(1.92 X 105 N)2 + (3.20 X 104)2
Fp = 1.95 X 105 N
The Ladder: Example 1
A 5-m long ladder leans
against a wall at a point
4 m above the ground.
The ladder has a mass of
12.0 kg and is uniform.
Assume the wall is
frictionless, but the
ground is not. Calculate
the force from the wall
(Fw) and the force from
the ground (FG).
SFx = 0
0 = FGx – Fw
FGx = Fw
SFy = 0
0 = FGy – mg
FGy = mg = (12.0 kg)(g) = 118 N
Working with the Triangle
52 = 42 + x 2
x2 = 52 – 42
x=3
sin q = 4/5
q = 53o
5m
4m
q
x
37o
5m
53o
3m
4m
Fw
Back to the ladder
53o
37o
FG
37o
53o
Tilting the Ladder
mg
Fw
FG
53o
37o
mg
Calculating the Torque forces
Fw
FG
53o
37o
Fwsin53o
mgsin37o
mg
Choose the point at ground as the pivot
St = 0
0 = (5m)(Fw)(sin53o) – (2.5m)(mg)(sin 37o) – 0
(Fw)(3.99m) = 177 m-N
Fw = 44.3 N
Fw = 44.3 N
FGx = Fw
FGx = 44.3 N
FGy = 118 N
F2 = FGx2 + FGy2
F2 = (44.3 N)2 + (118 N)2
F = 126 N
The Ladder: Example 2
A man who weighs 800 N
climbs to the top of a 6
meter ladder that is leaning
against a smooth (ie,
frictionless) wall at an
angle of 60°. The nonuniform ladder weighs 400
N and its center of gravity
is 2 meters from the foot of
the ladder. What must be
the minimum coefficient of
static friction between the
ground and the foot of the
ladder if it is not to slip?
FG
SFx = 0
0 = Fw – Ffr
Fw = Ffr
SFy = 0
0 = FN – 800 N – 400 N
FN = 1200 N
Working with the Triangle
Fw
60o
30o
800 N
o
30
400 N
FG
60o
Tilt the Ladder
Fw
FG
60o
30o
30o
800 N
400 N
Calculating the Torque forces
Fw
Fwsin60o
800sin30o
FG
60o
30o
400sin30o
800 N
30o
400 N
Choose the ground as the pivot
St = 0
0 =(2m)(400sin30o) + (6m)(800sin30o) (6m)(Fwsin60o)
0 = 2800 m-N – (Fw)(5.20)
Fw = 538 N
Fw = Ffr
Ff = 538 N
Ffr = mFN
m = Ffr/FN = 538 N/1200 N
m = 0.45
Stable vs. Unstable Equilibrium
A body will fall if its center of gravity is no longer
above the base
Conservation of Angular Momentum
• The total angular momentum of a rotating body
remains constant if the net torque acting on it is
zero
• Angular Momentum (L) is conserved
• p = mv (linear momentum)
• L = Iw (angular momentum)
• Iiwi = Ifwf
The skater in more detail
For now, let’s just consider the skater’s arms:
Arms out (initial)
Arms in (final)
Iiwi = Ifwf
I = mr2
Her mass will not change
when she moves her arms
mri2wi = mrf2wf
ri2wi = rf2wf
Since rf will be smaller (arms are in) wf must
increase to compensate
“I wanna go fast!”
A child sits on a merry-go-round at point A. If the
child wants to go faster, should he walk towards
the center or the outside?
A
A skater holds her arms at a length of 56 cm. She
spins at 9.43 rad/s. What will be her new speed
if she pulls her arms tight to her body, 20 cm?
Iiwi = Ifwf
I = mr2
mri2wi = mrf2wf
ri2wi = rf2wf
wf = ri2wi /rf2
wf = [(0.56 m)2(9.43 rad/s)]/(0.20 m)2 = 74 rad/s
A mass m is attached to the end of a string
which passes through a whole in a table.
Initially the r1 is 0.80 m and the block moves
at 2.4 m/s. What will be the new speed as the
radius of the string is reduced to 0.48 m?
Iiwi = Ifwf
I = mr2
mri2wi = mrf2wf
ri2wi = rf2wf
v = rw so w = v/r
ri2vi = rf2vf
ri
rf
rivi = rfvf
vf = rivi
=
(0.80m)(2.4 m/s) = 4.0 m/s
rf
(0.48 m)
Vectors and Angular
Momentum
• The direction of the the
vector w is defined by the
right-hand rule.
• L = Iw
• L is defined as going in the
same direction.
Merry-Go-Round
What will happen if you start to walk on a still
merry-go-round? Can you walk around it
without it moving?
• Assume this merry-goround is initially at rest
(Ltotal = 0)
• As the man walks
counter-clockwise, the
platform spins
counterclockwise
• Total Angular
Momentum is still zero.
What will happen if the
student flips the
bicycle wheel while
standing on the merrygo-round?
A 20.0 cm, 2.00 kg solid disk rotates at 200 rpm.
A 20.0 cm, 1.00 kg circular loop drops down
onto the disk.
a. Calculate the final angular velocity of the
combined system.
A 50 kg ball of clay is thrown at 10 m/s tangent to the
edge of a 30 cm diameter, 2.0 kg disk that can spin.
a. Calculate the initial angular momentum of the ball of
clay just before it sticks to the edge of the disk. (0.075
kg m2)
b. Calculate the final angular velocity of the disk and clay
(3.17 rad/s or 30 rpm)