Physics 231
Topic 8: Rotational Motion
Alex Brown
October
21-26
2015
MSU Physics
231 Fall 2015
1
MSU Physics 231 Fall 2015
2
MSU Physics 231 Fall 2015
3
Key Concepts: Rotational Motion
Rotational Kinematics
Equations of Motion for Rotation
Tangential Motion
Kinetic Energy and Rotational Inertia
Moment of Inertia
Rotational dynamics
Rolling bodies
Vector quantities
Mechanical Equilibrium
Angular Momentum
Covers chapter 8 in Rex & Wolfson
MSU Physics 231 Fall 2015
4
Review: Radians & Radius
Circumference= 2r
s = r
r
s

 in radians!
360o=2 rad = 6.28… rad
 (rad) = 2/360o  (deg)
MSU Physics 231 Fall 2015
5
Review: Angular speed
 f  i
 Average angular


velocity
t f  ti
t

  lim
t 0 t
Instantaneous
Angular velocity
Angular velocity : rad/s
MSU Physics 231 Fall 2015
6
Review: Angular  Linear Velocities
s  r
s   r
s 

r
t t
v r
linear (tangential) velocity
MSU Physics 231 Fall 2015
7
Angular acceleration
Definition: The change in angular velocity per time unit

 f  i
t f  ti
 Average angular

t acceleration

  lim
t  0  t
Instantaneous angular
acceleration
Unit: rad/s2
MSU Physics 231 Fall 2015
8
Angular  Linear Accelerations
v r
v   r
velocity
change in velocity
a
v 

r
t t
a  r
Change in velocity per time unit
linear or tangential acceleration
MSU Physics 231 Fall 2015
9
Equations of motion
Linear motion
Angular motion
x(t) = xo + vot + ½at2
(t) = o + ot + ½t2
v(t) = vo + at
(t) = o + t
Angular motion = Rotational Motion!
MSU Physics 231 Fall 2015
10
example
=0
A person is rotating a wheel. The
handle is initially at rest at  = 0o.
For 5s the wheel gets a constant angular
acceleration of 1 rad/s2. After that
The angular velocity is constant.
Through what angle will the wheel
have rotated after 10s.
First 5s: (5)= 0
+0t + ½t2
= 0 + (0)(5) + ½1(5)2 = 12.5

(5) = 0 + t = 0 + (1)(5) = 5 rad/s
Next 5s: (5) = 12.5 + (5)(5) + 0 = 37.5
t
37.5 rad = 2150o = 6.0 rev
MSU Physics 231 Fall 2015
11
A Rolling Coin
D
d
0 = 18 rad/s
 = -1.8 rad/s
d = 0.02 m
a) How long does the coin roll before stopping?
b) What is the average angular velocity?
c) How far (D) does the coin roll before coming to rest?
a) (t) = 0 + t
0 = 18 - 1.8t t=10 s
b)  = (0 + (10))/2 = (18 + 0)/2 = 18/2 = 9 rad/s
 = area under  vs t curve = (10)(18)/2 = 90 rad
c) D = s =  r = (90) (0.02/2) = 0.9 m
MSU Physics 231 Fall 2015
12
Torque
d
Top view
O
It is much easier to swing the
door if the force F is applied
as far away as possible (d) from
the rotation axis (O).
F
Torque: The capability of a force to rotate an object about
an axis.
Torque
 =Fd
(Nm)
Torque is positive if the motion is counterclockwise
(angle increases)
Torque is negative if the motion is clockwise
(angle decreases)
MSU Physics 231 Fall 2015
13
Multiple forces causing torque.
F1 = 100 N
Two persons try to go
F2 = 50 N through a rotating door
Top view
at the same time, one on
the l.h.s. of the rotator and
one the r.h.s. of the rotator.
If the forces are applied as
0.3 m
0.6 m
shown in the drawing, what
will happen?
1 = -F1 d1 = -(100)0.3 = -30 Nm
2 = F2 d2 = (50)0.6 = 30 Nm

0 Nm
Nothing will happen! The 2
torques are balanced.
MSU Physics 231 Fall 2015
14
Center of gravity (center of mass)
Fpull
dpull
1 2 3………………………n
Fgravity
dgravity?
 = Fpulldpull + Fgravitydgravity
(side view)
We can assume that
for the calculation
of torque due to gravity,
all mass is concentrated
in one point:
The center of gravity:
the average position of
the mass
dcg=(m1d1+m2d2+…+mndn)
(m1+m2+…+mn)
MSU Physics 231 Fall 2015
15
Center of mass; more general
center of gravity = center if mass
The center of gravity
m x

m
i
xCG
x1,y1
i
m 1g
i
i
xcg,ycg
i
m y

m
i
y CG
O
x2,y2
m2 g
x3,y3
m3g
i
i
i
i
Note 2: freely rotating systems (I.e. not held fixed at some point)
must rotate around their center of gravity.
MSU Physics 231 Fall 2015
16
examples: (more in the book)
Oxygen molecule relative mass O is 16
H is 1
Where is the center of gravity?
x  L cos   0.060
y  L sin   0.080
m x

m
(16)(0)  (1) x  (1) x 0.12


 0.0067
16  1  1
18
i i
xCG
i
i
i
m y

m
i i
yCG
i
i
(16)(0)  (1) y  (1)( y )

0
16  1  1
i
MSU Physics 231 Fall 2015
17
Quiz
Consider a hoop of mass 1 kg.
The center of gravity is located
at:
a) The edge of the hoop
b) The center of the hoop
c) Depends on how the hoop is
positioned relative to the
earth’s surface.
MSU Physics 231 Fall 2015
18
Object in equilibrium?
Top view Fp
d
d
CG
Fp
Newton’s 2nd law: F=ma
Fp+(-Fp)=ma=0
No acceleration of the center of mass
up or down
But the block starts to rotate!
 = Fpd + Fpd = 2Fpd (both clockwise)
There is rotation!
MSU Physics 231 Fall 2015
19
Torque:
Center of
Gravity:
 = Fd
m x

m
i
xCG
i
m y

m
i
i
y CG
i
i
i
i
i
i
Translational equilibrium: F = ma = 0 The center of gravity
does not move
Rotational equilibrium: =0
The object does not rotate
Mechanical equilibrium: F = ma = 0 &
=0 No movement!
MSU Physics 231 Fall 2015
20
question
20N
1m
1m
20N
20N
COG
Is this “freely rotating” object in mechanical equilibrium?
a) Yes b) No
Freely rotating: can only rotate around its COG:
Rotational eq.:  = -20x1 + 20x0 + 20x1 =0
Translational eq.: F= 20 –20 +20 =+20
No translational equilibrium: no mechanical equilibrium!
MSU Physics 231 Fall 2015
21
Torque: =Fd
quiz
40N
20N
-2
40N
0
1
2
m
A wooden bar is initially balanced. Suddenly, 3 forces are
applied, as shown in the figure. Assuming that the bar can
only rotate, what will happen (what is the sum of torques)?
a) the bar will remain in balance
b) the bar will rotate counterclockwise
c) the bar will rotate clockwise
MSU Physics 231 Fall 2015
22
Somethings to keep in mind
If an object is in equilibrium, it does not matter where
one chooses the axis of rotation: the net torque must
always be zero.
If it is given that an object is at equilibrium, you can
choose the rotation axis at a convenient location to solve
the problem.
MSU Physics 231 Fall 2015
23
T1
T2
0.75m
0.25m
Weight of board: w
What is the tension in each of the
wires (in terms of w) given that the
board is in equilibrium?
Translational equilibrium
F = ma = 0
T1+T2-w = 0 so T1=w-T2
Rotational equilibrium
w
 = 0
T10 – (0.5)w + (0.75)T2=0
T2=0.5/(0.75 w) = 2/3w
0
T1=1/3w
T2=2/3w
Note: I chose the rotation point at T1, but could have chosen anywhere
else. By choosing T1 the torque due to T1
Does not contribute to the equation for rotational equilibrium
MSU Physics 231 Fall 2015
24
Torque and angular acceleration
Ft
r
m
Newton 2nd law: F=ma
Ft = mat
Ftr = mrat
Ftr = mr2
Used at = r
 = mr2 = I
(used  = Ftr)
The angular acceleration goes linear with the torque.
I = mr2 = moment of inertia for one point mass
MSU Physics 231 Fall 2015
25
More General for rotation around a common axis
m2
r2
r1
 = I
r3
m3
m1
compare with: F=ma
The moment of inertia
in rotations is similar to
the mass in Newton’s 2nd law.
Moment of inertia I:
I = (miri2)
ri is the distance to the axis of rotation
I = mr2 = when all mass is at the same distance r
MSU Physics 231 Fall 2015
26
Extended objects - hoop
F
I=(miri2)
=(m1+m2+…+mn)R2
=MR2
M
r
mi
MSU Physics 231 Fall 2015
27
A homogeneous stick
m
m
m
m
m
m
m
m
m
m
F
=(m1r12+m2r22+…+mnrn2) 
=(miri2) 
=I
Moment of inertia I:
I=(miri2)
Rotation point
MSU Physics 231 Fall 2015
28
Two inhomogeneous sticks
r
m
m
m
m
5m
5m
m
m
m
m
F
5m
m
m
m
m
m
m
m
m
5m
=(miri2)
118mr2
18m
Easy to rotate!
F
=(miri2)
310mr2
18 m
Difficult to rotate
MSU Physics 231 Fall 2015
29
Demo: fighting sticks
MSU Physics 231 Fall 2015
30
F
A simple example
A
B
F
r
r
A and B have the same total mass. If the same
torque is applied, which one accelerates faster?
Answer: A
=I
Moment of inertia I:
I=(miri2)
MSU Physics 231 Fall 2015
31
0.2 kg
The rotation axis matters!
0.3 kg
0.5 m
I = (miri2)
I = (miri2)
=(0.2) 0.52 + (0.3) 0.52
+ (0.2) 0.52 + (0.3) 0.52
= 0.5 kg m2
= (0.2) 0 + (0.3) 0.52
+ (0.2) 0 + (0.3) 0.52
= 0.3 kg m2
MSU Physics 231 Fall 2015
32
=I
(compare to F=ma)
Moment of inertia I:
I=(miri2)
: angular acceleration
I depends on the choice of rotation axis!!
MSU Physics 231 Fall 2015
33
Some common cases
R
hoop/thin cylindrical
shell: I=MR2
R
disk/solid cylinder
I=1/2MR2
L
solid sphere
I=2/5MR2
thin spherical shell
I=2/3MR2
Long thin rod with
rotation axis through Long thin rod with
rotation axis through
center: I=1/12ML2
end: I=1/3ML2
MSU Physics 231 Fall 2015
L
34
Falling Bars
mass: m
L
Ibar=mL2/3


Fp = Fgcos = mg cos perpendicular component
Fg=mg
Compare the angular acceleration for 2 bars of different
mass, but same length.
 = I = mL2/3 also  = Fd = mg(cos)L/2
So mg(cos)L/2 = mL2/3 and  = 3g(cos)/(2L)
independent of mass!
Compare the angular acceleration for 2 bars of same mass,
but different length
 = 3g(cos)/(2L) so if L goes up,  goes down!
MSU Physics 231 Fall 2015
35
Rotational kinetic energy
v
ri

i
mi
 
Consider a object rotating
with constant velocity. Each point
moves with velocity vi. The total
kinetic energy is:

1 mi vi2  1 mi i2 ri 2  1  2   mi ri 2   1  2 I
2
2
2  i
2
i

KEr = ½I2
Conservation of energy for rotating object must include:
rotational and translational kinetic energy and potential energy
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
MSU Physics 231 Fall 2015
36
Convenient way of writing KE for rotation
I = kmr2
object
I
k
A
Cylindrical
shell
Mr2
1
B
Solid cylinder
(1/2) mr2
0.5
C
Thin spherical
shell
(2/3) mr2
(2/3)
D
Solid sphere
(2/5) mr2
(2/5)
MSU Physics 231 Fall 2015
37
rolling motion (no slipping)
this means that =v/r
MSU Physics 231 Fall 2015
38
KE for rolling motion (no slipping)
KE = KE (translational) + KE (rotational)
r = radius
= [½mv2 + ½I2] = [½mv2 + ½(kmr2)v2/r2]
= [½mv2+½kmv2] = (1+k)½mv2
used: I=kmr2 and =v/r
KE = (1+k)½mv2
MSU Physics 231 Fall 2015
39
Example.
1m
Consider a ball (sphere) and a block going down the same 1m-high slope.
The ball rolls and both objects do not feel friction. If both
have mass 1kg, what are their velocities at the bottom (I.e.
which one arrives first?). The radius of the ball is 0.4 m.
Block: [½mv2+mgy]initial
v=2gh
0 + mgh
= [½mv2 + mgy]final
= ½mv2 + 0
so v = 4.43 m/s
MSU Physics 231 Fall 2015
40
Example.
1m
Consider a ball and a block going down the same 1m-high slope.
The ball rolls and both objects do not feel friction. If both
have mass 1kg, what are their velocities at the bottom (I.e.
which one arrives first?). The radius of the ball is 0.4 m.
Ball: [½m(1+k)v2+mgy]initial= [½m(1+k)v2 + mgy]final
0 + mgh
= ½m(1+k)v2 + 0
v=[2gh/(1+k)] with k = 2/5 (No mass dependence)
so v = 3.74 m/s
slower than the block since part of the
energy goes into rotational
KE
MSU Physics 231 Fall 2015
41
Which one goes fastest (h=1m)
object
k
v
Block sliding
0
4.43
Cylindrical shell
1
3.13
4
Solid cylinder
1/2
3.61
2
Thin spherical shell 2/3
3.42
3
Solid sphere
3.74
1
2/5
fastest
The larger the moment of inertia, the slower it rolls!
The available Ekin is spread over rotational and translational (linear) Ekin
Or: besides moving the object, you spend energy rotating it
MSU Physics 231 Fall 2015
42
Angular momentum
    0  I  I 0
  I  I 

t
 t 
L  I
L  L0 L


t
t
if    0 then L  0
Conservation of angular momentum: If the net torque
equals zero, the angular momentum L does not change
Ii i = If f
MSU Physics 231 Fall 2015
43
Conservation laws:
In a closed system:
•Conservation of energy E
•Conservation of linear momentum p
•Conservation of angular momentum L
MSU Physics 231 Fall 2015
44
Neutron star
Sun: radius: Rs = (7) 105 km
Period of rotation: 25 days
Supernova explosion
Neutron star: radius: Rn = 10 km
What is frequency of rotation ?
Isphere=(2/5)MR2
s = sun n = neutron star
(assume no mass is lost so Ms=Mn)
s = 2/(25 days x 24 x 3600) = (2.9) 10-6 rad/s
Conservation of angular momentum:
Iss = Inn
n = (Rs/Rn)2 s = 1.4E+04 rad/s
fn = (n /2 ) = 2200 Hz !
MSU Physics 231 Fall 2015
45
The spinning lecturer…
A lecturer (60 kg) is rotating on a platform with i=2 rad/s
(1 rev/s). He is holding two 1 kg masses 0.8 m away from his
body. He then puts the masses close to his body (R=0.0 m).
Estimate how fast he will rotate (marm=2.5 kg).
0.8m
0.4m
Ii = 0.5 MlecR2
(body)
+2(MwRw2)
(two weights)
+2(0.33Marm0.82) (two arms)
= 1.2 + 1.3 + 1.0 = 3.5 kg m2
If = 0.5 MlecR2 = 1.2 kg m2
MSU Physics 231 Fall 2015
46
The spinning lecturer…
A lecturer (60 kg) is rotating on a platform with i=2 rad/s
(1 rev/s). He is holding two 1 kg masses 0.8 m away from his
body. He then puts the masses close to his body (R=0.0 m).
Estimate how fast he will rotate (marm=2.5 kg).
0.8m
0.4m
Conservation of angular mom.
Ii I = If f
f = (Ii /If) i
f = (3.5/1.2) 2 = 18.3 rad/s
(approx 3 rev/s)
For details see next slide
KEi = KEf ? A=yes
MSU Physics 231 Fall 2015
B=no
47
Details of spinning lecturer
Iinitial=Ibody+2Imass+2Iarms
Ibody= 0.5MlecRlec2 (solid cylindrical shell) M=60, R=0.2
Imass=MmassRmass2 (point-like mass at radius R) M=1 R=0.8
Iarms=1/3MarmRarm2 (arm is like a long rod with rotation axis through
one of its ends) M=2.5 kg R=0.8
Iinitial=0.5MlecRlec2+2(MmassRmass2)+2(0.33MarmRarm2)
=1.2+1.3+1.0= 3.5 kgm2
Ifinal =0.5MlecR2=1.2 kgm2 (assumed that masses and arms do not
contribute to moment of inertia anymore)
Conservation of angular mom. Iii=Iff
3.5*2=1.2*f i= was given to be 2 rad/s
f=18.3 rad/s (approx 3 rev/s)
KEi = (1/2)Ii (i )2 = 69
KEf = 201
MSU Physics 231 Fall 2015
48
Orbital motion
Ft
r
m
Ft = mat
I = mr2 = moment of inertia one point mass
L = I
For a point-like mass orbiting a central axis:
I=mr2 and =v/r
So L = (mr2 ) (v/r) and thus
MSU Physics 231 Fall 2015
L=mrv
49
MSU Physics 231 Fall 2015
50
MSU Physics 231 Fall 2015
51
MSU Physics 231 Fall 2015
52
MSU Physics 231 Fall 2015
53
MSU Physics 231 Fall 2015
54
question
B
A
dA
dB
Two people are on a see saw. Person A is lighter than
person B. If person A is sitting at a distance dA from the
center of the seesaw, is it possible to find a distance dB so
that the seesaw is in balance?
A) Yes
B) No
If MA=0.5MB and dA=3 m, then to reach balance, dB=…
A) 1 m B) 1.5 m C) 2 m D) 2.5 m E) 3 m
MSU Physics 231 Fall 2015
55
s=0.5 coef of friction between the wall and
the 4.0 meter bar (weight w). What is the
minimum x where you can hang a weight w
for which the bar does not slide?
sn
T T
y
37o
n
Tx
x
w
(x=0,y=0)
w
Translational equilibrium (Hor.)
Fx = ma = 0
n-Tx = n - Tcos37o = 0 so n = Tcos37o
Translational equilibruim (vert.)
Fy = ma = 0
sn – w – w + Ty=0
sn - 2w + Tsin37o = 0 (use n=Tcos37o)
sTcos370 - 2w + Tsin370 = 0
0.4T - 2w + 0.6T=0
T = 2w
Rotational equilibrium:
 = 0
xw + 2w - 4Tsin370 = 0
(4sin37o) = 2.4 T = 2w
w (x + 2 - 4.8)=0
x = 2.8 m
MSU Physics 231 Fall 2015
56
The direction of the rotation axis
L
L
The rotation in the horizontal plane is
reduced to zero: There must have been a large
net torque to accomplish this! (this is why you
can ride a bike safely; a wheel wants to keep turning
in the same direction.)
The conservation of angular momentum not only holds
for the magnitude of the angular momentum, but also
for its direction.
MSU Physics 231 Fall 2015
57
Rotating a bike wheel!
L
L
A person on a platform that can freely rotate is holding a
spinning wheel and then turns the wheel around.
What will happen?
Initial: angular momentum: Li = Iwheelwheel
Closed system, so L must be conserved.
Final:
Lf = - Iwheel wheel + Ipersonperson = Li = Iwheelwheel
person= 2Iwheel wheel/Iperson
MSU Physics 231 Fall 2015
58