AP Physics
Lyzinski, CRHS-South

Day
#1
Sections 10.1 & 10.2

Polar coordinates (r,

)
CCW is positive, CW is negative s = arc length = r

Where does it come from?

360 o
 s circumfere nce


2

( rad )
 s
2

R
C r

P s
 s

R


= angular position (measured in radians, not degrees)
 
= Angular displacement
  
2

1
Degree to radian conversion

( rad )


180

(deg)

w
= Angular velocity  avg angular velocity = w 
 
 t
Instantaneous angular velocity = w  lim
 t
 
 
 t
 d
 dt
Units of radians per second (rad/s) a
= Angular acceleration  avg angular acceleration = a 
 w
 t
Instantaneous angular acceleration = a  lim
 t
 
 w
 t
 d w dt
 d
2
 dt 2
Units of radians per second squared (rad/s 2 )

v
 dx dt a
 dv
 dt d
2 x dt
2 v
2
 v
1
 at v
2
2  v
1
2 
2 a
 x
 x

1
2 at
2  v
1 t
 x

1
2 t ( v
1
 v
2
) a w
2
 w  d w dt d

 dt d
2
 dt
2
 w
1
 a t w
2
2  w
1
2 
2 a  
  
1
2 a t
2  w
1 t
  
1
2 t ( w
1
 w
2
)
CAUTION!!! These equations can only be used if a or a are constant!!!!!

#6
A rotating wheel requires 3.00 s to rotate through 37.0 revolutions.
Its angular speed at the end of the 3.00-s interval is 98 rad/s.
What is the constant angular acceleration of the wheel?
t

3 sec
  w
2

37 rev


98 rad s
37 rev
1


2
 rad
1 rev a 
???



232 .
478 rad
  
1
2 t ( w
1
 w
2
)

232 .
478

1
2
( 3 )( w
1

98 )
 w
1

56 .
985 rad s w
2
 w
1
 a t

98

56 .
985

3 a  a 
13 .
67 rad s
2

• A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is
98 rad/s. What is the constant angular acceleration of the wheel?

#9
The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 8.00 s, at which time it is turning at 5.00 rev/s. At this point, the person doing the laundry opens the lid and a safety switch turns off the machine. The tub smoothly slows to rest in 12.0 s. Through how many revolutions does the tub turn while its in motion?
t

 
8 sec

1
2 t ( w
1 w
1

 w
2
0
)

1
2 w
2

5
( 8 )( 0 rev s

5 )

20 rev w
1
 


5
1
2 rev s t ( w
1 w
2
 w
2
)


0
1
2 t

12 sec
( 12 )( 5

0 )

30 rev
50 revs total

A Mr. L original 
A certain wheel begins to rotate. Its position varies with time according to the equation  

3 t
2 
6 t

5
 rad
Find the wheel’s average angular acceleration between 5 and 9 sec.
a 
 w
 t d
 dt w
( 5 )
 w



6 ( 5 )
6 t


6
 rad s
6

36 rad s w
( 9 ) a 

 w
 t
6 ( 9 )


6 w
( 9 )


60 rad s w
( 5 )
9

5 sec

60 rad
9 s


36
5 sec rad s
Duh!!! The acceleration is constant 

24 rad s
4 sec

6 rad s
2

pp. 299-300
Do problems 1-7 all (skip # 6 and #7b)

Day
#2
Section 10.3

v w
Relating Rotational Motion (
, w, and a
) to Translational Motion (x, v, and a) r
 s
Point of rotation
Note: As r increases, v and a get larger, while w and a stay the same.
v
 ds dt
 d ( r

) dt
 r d
 dt
 r w a t
 dv
 dt d ( r w
) dt
 r d w dt
 r a a r
 v
2 r

( r w
)
2 r
 r w
2 a
 a t
2  a r
2 
( r a
)
2 
( r w
2
)
2  r a
2  w
4

#16
A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. The tires have diameter 58.0 cm and do not slip on the pavement. (a) Find the number of revolutions each tire makes during this motion. (b) What is the final angular speed of a tire in revolutions per second?
v
1

0 v
2

22 m s t

9 sec
 v
2
 v
1
 at
 a

2 .
4 m s
2 a
 a
R
 a  a
R

2 .
4 m s
.
29 m

8 .
429 rad s
2
  
1
2 a t 2  w
1 t

1
2
( 8 .
429 rad s
2
)( 9 ) 2 
0

341 .
37 rad

54 .
33 rev w
2
 w
1
 a t

0

( 8 .
429 rad s
2
)( 9 )

75 .
9 rad s

12 .
1 rev s

#12
The drive train of a bicycle is shown. The wheels have a diameter of 67.3 cm and the pedal cranks are 17.5 cm long. The cyclist pedals at a steady cadence of 76.0 rev/min. The chain engages with a front sprocket 15.2 cm in diameter and a rear sprocket 7.00 cm in diameter. (a) Calculate the speed of a link in the chain relative to the bicycle frame. (b) Calculate the angular speed of the bicycle wheels. (c) Calculate the speed of the bicycle relative to the road.

w pedals r wheel

76 rev min

.
3365 m

7 .
959 r front sprocket

.
152 m rad s r rear sprocket v chain v chain

.
07 m
 w pedals r front sprocket
 w wheel r rear sprocket v wheels
 w wheel r wheel


7 .
959
 rad s

(.
076 m )
.
604

.
604
 w w
(.
035 )

( 17 .
257 )(.
3365 )

5 .
8 m s m s
 w w

17 .
257 rad s

pp. 300-301
Do problems 10, 13, 14, 15, 16, 19

Day
#3
Section 10.4

A solid object is a collection of particles. When the object rotates, each of these particles moves, thus possessing kinetic energy. If we add up all these individual energies, we can find the energy associated with the rotating object.
However, as we have learned previously, the velocity of each particle depends on how far the particle is from the axis of rotation. Particles close the axis move slower than particles far from the axis (according to v = r w
). Therefore, it might be useful to express each individual kinetic energy in terms of w
(which is the same for each particle) instead of v (which changes based on distance from the axis).

K
R

Rotational Kinetic Energy
  i
K i
  i
1
2 m i v i
2   i
1
2 m i
( r i w i
)
2 
1
2


 i m i r i
2

 w
2
KE of each individual particle
Notice that w was taken out of the summation because it is the same for every particle, no matter how far the particle is from the axis of rotation.
 i m i r
2
The term has been given the name “the Moment of Inertia”, or “ I ” .
Therefore,
K
R

1
2
I w
2
“ I” has units of kg
 m
2

Remember, in translational motion, an object’s inertia is its
“tendency” to want to either remain at rest or moving at a constant velocity. An object’s mass is a direct measure of its inertia.
In rotational motion, the individual particles have masses at different distances from the axis of rotation. The Moment of inertia is the rotational analog of mass. It is a measure of how difficult it is to change an object’s motion about its axis of rotation. The closer the mass is to the axis, the easier it is to change its rotational motion. Thus, objects with more of their mass far from the axis rotation have a higher moment of inertia.

Translational vs. Rotational Motion (revised) v
 dx dt a
 dv
 dt d
2 x dt
2 v
2
 v
1
 at v
2
2  v
1
2 
2 a
 x
 x

1
2 at
2  v
1 t
 x

1
2 t ( v
1
 v
2
)
K

1
2 mv
2 w  d
 dt a  d w dt
 d
2
 dt
2 w
2
 w
1
 a t w
2
2  w
1
2 
2 a  
  
1
2 a t
2  w
1 t
  
1
2 t ( w
1
 w
2
)
K
R

1
2
I w
2

A Mr. L original 
A Penn State Baton Twirler is spinning her 2 ft long baton, which has identical end masses of 300 grams. Assuming the rod itself to be mass-less, find the moment of inertia of the baton if she rotates it about (a) line “a” (which is through the center of the rod) or (b) line
“b” (which is 4 inches off-center).
1
16 ft in

.
3048

.
4064 m m
8
I a in


.
2032
 m i r i
2 m

(.
300 kg )(.
3048 m )
2

(.
300 kg )(.
3048 m )
2

.
0557 kg
 m
2
I a
  m i r i
2 
(.
300 kg )(.
4064 m )
2

(.
300 kg )(.
2032 m )
2 
.
0619 kg
 m
2 a b

#25
2.86 m
0.12 kg 60.0 kg
Find the speed that the small mass leaves the Trebuchet. Assume the rod to be mass-less.
Before
14.0 cm v
I
E b



E a m i r i
2 
(.
12 )( 2 .
86 )
2 
( 60 )(.
14 )
2 mgh

Mgh

(.
12 )( 9 .
8 )(.
14 )
1
2

I w
2  mgH
( 60 )( 9 .
8 )(.
14 )

2 .
158 kg
 m
2

1
2
( 2 .
158 ) w
2 
(.
12 )( 9 .
8 )( 3 )
After w v


8 .
w r
55 rad s

( 8 .
55 )( 2 .
86 )

24 .
5 m s

pp. 301-302
Do problems 21, 22, 23

Day
#4
Section 10.5

Calculating
Moments of Inertia
I
  i m i r i
2  lim
 m i

0
 i r i
2  m i
  r
2 dm
I

1
2
MR
2
Notice that the object with more of its mass further away from the axis of rotation has a larger moment of inertia
(and thus it will be harder to change it rotational motion)
I

MR
2
I

MR
2
I

2
5
MR
2
I

1
2
MR
2
I

2
3
MR
2
I

1
12
MR
2
I

1
3
MR
2

Calculating
Moments of Inertia
I
  i m i r i
2  lim
 m i

0
 i r i
2  m i
  r
2 dm
I

1
2
MR
2
I

MR
2
I

MR
2
I

2
5
MR
2
I

1
12
MR
2 Again, notice that the object with more of its mass further away from the axis of rotation has a larger moment of inertia
I

1
2
MR
2
I

2
3
MR
2
I

1
3
MR
2

Using Calculus to find Moments of Inertia
• First , make sure to figure out what your “tiny pieces” look like.
• Second , choose the appropriate density function for your “tiny piece” ( l for linear, s for area, or r for volume).
• Third , use the appropriate density function and solve for dm.
• Fourth , make sure that dm only has one variable in it, and then plug it into
I
  r
2 dm

L y dm with thickness dL x
Example: Find the moment of inertia of a thin rod that rotates about its end.
I l  m
L
 l
L
 m

 
L
0 r
2 dm
 
L
0 r
2
( l dx )
 
L
0 x
2
( l dx ) l dL
 dm
 l 
L
0 x
2 dx
 l
1
3
 
0
L  l
L
3
3

 m
L
3
L
3
 mL
3
3 L

1
3 mL
2 l dx
 dm

R y dm with thickness dr
Example: Find the moment of inertia of a solid cylinder that rotates about its central axis.
r dr r
Top view of the “thin” cylindrical slices r r dV
 m
V
 r
V
 m
 r dV
 dm



( r
 dr ) 2   r 2
 
 height area of ring


2
 rdr zero, b/c tiny times tiny
 
( dr ) equals super tiny
2

 dm
 r
( 2

Lrdr )
I
 
L
0 r
2 dm
 
L
0 r
2
2
r
Lrdr

2
r
L

R
0 r
3 dr

2
r
L
 
4 0
R

2
r
L
 
4
R

2
 m
V
L
 
4
R

2
 m

R 2 L
L
 
4

1
2
MR
2

R dm with thickness dr
Example: Find the moment of inertia of a sphere about its central axis.
r dV
 m

V

3
4

( r

 dr )
3 r
V

4
3
 m
 r
3

 r dV
 dm
( r
 dr )
Volume of outer
3  sphere r
3 
Volume of r
2 dr inner sphere
 r ( dr )
2  r
2 dr
 r ( dr )
2 
( dr )
3

I dm


4
3
r
( r
2 dr
 r ( dr )
2
4
3
r
( 2 r
2 dr )
 r
2 dr
 r ( dr )
2

( dr )
3
) zero, b/c tiny times tiny equals super tiny
 
R
0 r
2 dm
 
R
0 r
2
4
3
r
( 2 r
2 dr )


8
3
r
 
5
R

8
3
 m
V
 
5
R

8
3
r 
R
0 r
4 dr

8
3
r
 
5 0
R
8
3



4
3 m

R
3


 
5

2
5 mR
2

If you know the moment of inertia of an object about a given axis, you can use the equation
I

I
CM

MD
2 to find the moment of inertia of this object about any axis parallel to the given axis. The distance between these axis is “D”.
I

1
12
ML
D
I

I

1
12

ML
2
MD
2

M
 
2
2

1
12
ML
2 
1
4
ML
2 
1
3
ML
2

R
I

1
2
MR
D = R/2
I


I
1
2

MR
2
MD
2

M
 
2
2

1
2
MR
2 
1
4
MR
2 
3
4
MR
2

Sections 10.6 & 10.7

Torque ( t
): The tendency of a force to rotate an object about a given axis.
F

F cos

F sin
 r d t  rF sin
 
Fd
“d” is known as the “moment arm” or “lever arm”
***Notice that only forces perpendicular to the lever arm cause a torque

• The units of torque ( t
) are N-m
• The direction of a torque is found using the Right-Hand-Rule
– Place your fingers in the direction of the lever arm.
– “Slap” in the direction of the force.
– Your thumb points in the direction of the torque.
– The direction of a torque is found using the Right-Hand-Rule
• Positive torques are CCW and negative torques are CW.
• Torque is NOT a force!!!
• Torque is NOT the same as work. They have the same units, but are VERY different.
• The net torque on an object is the vector sum of the individual torques.
Therefore, t   t i

• Forces can cause a change in motion in translational motion.
• Forces can cause a change motion in rotational motion. HOWEVER, the further the force is from the axis of rotation, the more “effective” it will be in changing motion. Thus, the force as well as the length of the “lever arm” are important in rotational motion. Therefore, instead of speaking only of a “force”, we speak of a
“torque”.

nd




 t t
F t
F t
F t

 r


 ma t m ( r a mr
I a m ( r a
2 a
)
) r
The net force on a particle is proportional to its TANGENTIAL acceleration.
The net torque on a particle is proportional to its ANGULAR acceleration.

nd
F t
F t

 ma t m ( r a
)
F t r
 m ( r a
) r t
 d t
 d t mr


2 a
 

2 dm
  dm
Every “tiny little” mass (dm) in the rigid body is located at a different distance from the axis of rotation, and this needs to be taken into account.
Also, each of these masses is subjected to its own individual “tiny little” torque (d t
). To get the total torque, we need to sum up ALL of the “tiny little” ones (by integrating).
  t  a  r
2 dm
  t 
I a

M
1
M=0
R
M
2
Massless pulleys don’t really exist (but make calculations easy  )
R
T
1
T
2 t net

I a
T
1
R

T
2
R
T
1
R

T
2
R
T
1
R
T
1


T
2
R
T
2

1
2
MR
2 a

1
2

0
( 0 ) R
2 a
All a mass-less pulley does is change the direction of a force.

M
R
M
1
M
2
The pulley in this example is modeled as a solid disk (and thus
I = ½ MR 2 )
R
T
1
T
2 t net

I a
T
1
R

T
2
R
T
1
R

1
2

1
2
MR
2 a
MR
2 a

T
2
R

T
1

T
2
The difference in the tensions causes the net torque which forces the pulley to rotate

A Mr. L original 
The system below is at rest when the 10 kg mass is released. The pulley is not mass-less, but rather has a mass of 6 kg and a radius of 20 cm. If the surface has a coefficient of friction of 0.2, find the acceleration of the system.
20 kg
10 kg

F
N1
F f
20 kg m
1 g
T
1
T
1


 m
1 g m
1 a

 m
1 a
 m
1 g a
T
1
T
1
T
2 a
T
2 a 10 kg m
2 g
T
2


T
2 m
2 g
 m
2 a
 m
2 a m
2 g
 t 
T
2
R

T
1
R

I a
T
2
R

T
1
R


1
2
MR
2
 a
R
T
2
 m
2
 g
T
1

 m
2
1
2 a
Ma
10 ( 9 .
8 )

10 a
 m
1 a
20 a
  m
1 g


1
2
Ma

(.
2 )( 20 )( 9 .
8 )

(.
5 )( 6 ) a
58 .
8

33 a
 a

1 .
78 m s
2

Translational vs. Rotational Motion (revised) v
 dx dt a
 dv
 dt d
2 x dt
2 v
2
 v
1
 at v
2
2  v
1
2 
2 a
 x
 x

1
2 at
2  v
1 t
 x

1
2 t ( v
1
 v
2
)
K

1
2 mv
2
F net
 ma w  d
 dt a  d w dt
 d
2
 dt
2 w
2
 w
1
 a t w
2
2  w
1
2 
2 a  
  
1
2 a t
2  w
1 t
  
1
2 t ( w
1
 w
2
)
K
R

1
2
I w
2 t
I a net


Section 10.8

W
 t  
(similar t o W

F
 x)
  dW dt
 d dt
 t d
 dt
 tw
(similar t o
 
F v )
K
R

1
2
I w
2
(similar t o K
T

1
2 mv
2
)
W
 
K
 
U (where K now equals K
T

K
R
)
If no external torques or forces are present, then E b
= E a
.

T
W
 
K
F
 x

1
2 mv
2
2 
1
2 mv
1
2
R
Rotating Wheel
(where the axis of rotation is fixed)
W
 
K t   
1
2
I w
2
2 
1
2
I w
1
2

T
R
A Rolling Object
(it is rotating and translating at the same time)
K

1
2
I
CM w
2
K due to rotation

1
2
M
CM v
2
K due to translation

T
R
M
1
M
2
R
Solid Disk h
M
3
H
E b
U


K
E a

U

K
M
1 gh

M
3 gh

0

M
3 gH

1
2
M
1 v
2 
1
2
M
1 v
2 
1
2
I w
2 since w  v
R
, H

2h, and I

1
2
MR
2
( M
1

M
3
) gh

M
3 g ( 2 h )

1
2
M
1 v
2 
1
2
M
1 v
2 
1
2
I
 
R
2
( M
1

M
3
) gh

1
2
M
1 v
2 
1
2
M
1 v
2 
1
2

1
2
MR
2
  
R
2
Zero level

T
R
R
E b
U


K
E a

U

K
Mgh

0

Mg ( R )

1
2
Mv 2 
1
2
I w
2 h since w  v
R
, and I

MR
2
(for a thin hoop)
Mgh

0

MgR

1
2
Mv
2 
Mgh

0

MgR

1
2
Mv
2 
1
2
1
2
MR
2
MR
2
 
R
2
 
R
2
Mgh

MgR

Mv
2  v
 g ( h

R )
Find the velocity of the thin hoop
(with radius “R”) at the bottom.

Review Day 

Section 10.9

• In order to roll, an object needs to encounter friction, which applies a torque to the object and causes a rotation about its center of mass.
• If an object does not slip at all while rolling, it is said to undergo
PURE rolling motion.
• For pure rolling motion, v
CM
 ds dt
 d (
 r ) dt
 r d
 dt
 r w
These conditions must hold for non-slip rolling.
a
CM
 dv
CM dt
 d ( r w
) dt
 r d w dt
 r a

A closer look at an object that rolls but doesn’t slip
(part 1)
When an object undergoes PURE rotation, every point on the object has the same angular velocity, Therefore, all points that are equidistant from the axis of rotation have the same tangential velocity.
v = 0 v t
= R w v t
= R w

A closer look at an object that rolls but doesn’t slip
(part 2)
When an object undergoes PURE translation (which is the equivalent of
ALL slip and NO roll), every point on the object has the same velocity, namely the velocity of the center of mass. v
CM v
CM v
CM

A closer look at an object that rolls but doesn’t slip
(part 3)
When an object undergoes PURE Rolling, this is a combination of both
ROTATION and TRANSLATION. While every point on the object has the same angular velocity, the contact point with the floor acts as a pivot point . Thus, points on the rotating object that are furthest from the floor have the largest tangential velocity.
. v t
= v
CM
+ R w  2 v
CM v
CM v = 0

Review for Test

Day
12