# Physics 211 Lecture 18 Todayâ€™s Agenda by fsm30412

VIEWS: 9 PAGES: 34

• pg 1
```									                 Physics 211: Lecture 18

Today’s Agenda

   Direction and the right hand rule

   Rotational dynamics and torque

   Work and energy with example

Physics 211: Lecture 18, Pg 1
Rotational v.s. Linear Kinematics

Angular                                 Linear

 = constant                            a  constan t

 = 0 + t                              v = v 0 + at

1 2                                       1 2
 = 0 + 0 t + t                       x = x0 + v 0 t + at
2                                         2

And for a point at a distance R from the rotation axis:

x = Rv = Ra = R

Physics 211: Lecture 18, Pg 2
Roll objects
Rolling Motion              down ramp

   Objects of different I rolling down an inclined plane:

v=0             K = - U = Mgh
R       = 0
M      K=0
1 2 1
h                                                     K = I  + Mv 2
2     2

v = R

Physics 211: Lecture 18, Pg 3
Rolling...

   If there is no slipping:

       v
2v         v
v
Where v = R

In the lab reference frame             In the CM reference frame

Physics 211: Lecture 18, Pg 4
Rolling...

1 2 1                                             hoop:       c=1
K = I  + Mv 2       Use v = R and I = cMR2 .
2     2                                           disk:       c = 1/2
1            1      1                     sphere: c = 2/5
K=     c MR 2 2 + Mv 2 = ( c + 1)Mv 2
2            2      2
etc...

So:     1                                               1
( c + 1)Mv 2 = Mgh                v = 2 gh
c +1
2

The rolling speed is always lower than in the case of simple sliding
since the kinetic energy is shared between CM motion and rotation.

We will study rolling more in the next lecture!

Physics 211: Lecture 18, Pg 5
Direction of Rotation:
   In general, the rotation variables are vectors (have direction)
   If the plane of rotation is in the x-y plane, then the convention is

 CCW   rotation is in
the + z direction                 y

x

z

 CW   rotation is in
the - z direction                 y

x

z

Physics 211: Lecture 18, Pg 6
Direction of Rotation:
The Right Hand Rule
y
   To figure out in which direction the rotation
vector points, curl the fingers of your right
hand the same way the object turns, and                                   x
your thumb will point in the direction of the
rotation vector!
z
y
   We normally pick the z-axis to be the
rotation axis as shown.
 = z
x
 = z
 = z                                                 z
   For simplicity we omit the subscripts
unless explicitly needed.

Physics 211: Lecture 18, Pg 7
Example:

   A flywheel spins with an initial angular velocity 0 = 500 rad/s.
At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long
does it take to stop?

   Realize that  = - 0.5 rad/s2.

   Use      0  t     to find when  = 0 :

0
t 

   So in this case t                  2
 1000 s  16.7 min

Physics 211: Lecture 18, Pg 8
Lecture 18, Act 1
Rotations
   A ball rolls across the floor, and then starts up a ramp as
shown below. In what direction does the angular acceleration
vector point when the ball is on the ramp?

(a) down the ramp
(b) into the page
(c) out of the page

Physics 211: Lecture 18, Pg 9
Lecture 18, Act 1
Solution
   When the ball is on the ramp, the linear
acceleration a is always down the ramp (gravity).
   The angular acceleration is therefore counter-clockwise.

   Using your right hand rule,  is out of the page!


a

Physics 211: Lecture 18, Pg 10
Rotational Dynamics:
What makes it spin?

   Suppose a force acts on a mass constrained to move in a
circle. Consider its acceleration in the ^ direction at some

instant:                                                     ^
 a = r                                             ^     r

F
                                   
Now use Newton’s 2nd Law in the ^              F
direction:
a
 F = ma = mr                                            m

   Multiply by r :                                    r
rF = mr2                                

Physics 211: Lecture 18, Pg 11
Rotational Dynamics:
What makes it spin?

rF = mr2   use     I   = mr 2
= I
   Define torque:  = rF.
^
^        r
  is the tangential force F                         
times the lever arm r.
F
F
 = I                                      a
m
   Torque has a direction:
 + z if it tries to make the system         r
spin CCW.
 - z if it tries to make the system
spin CW.


Physics 211: Lecture 18, Pg 12
Rotational Dynamics:
What makes it spin?

 ri Fi ,   mi ri  i
   So for a collection of many particles                             2
arranged in a rigid configuration:           i              
i 
i         I

   Since the particles are connected rigidly,
they all have the same .
 i  I 
i
m4                                        F1
F4                                r1 m1
 NET  I                     r4

m3
r3       r2
m2
F2
F3
Physics 211: Lecture 18, Pg 13
Rotational Dynamics:
What makes it spin?

      NET = I

   This is the rotational analogue
of FNET = ma
   Torque is the rotational analogue of force:
 The amount of “twist” provided by a force.
   Moment of inertia I is the rotational analogue of mass.
 If I is big, more torque is required to achieve a given
angular acceleration.
   Torque has units of kg m2/s2 = (kg m/s2) m = Nm.

Physics 211: Lecture 18, Pg 14
Torque

   Recall the definition of torque:

 = rF

= r F sin                           Fr  F
= r sin  F

F
 = r pF

rp = “distance of closest approach”       r

   Equivalent definitions!
rp

Physics 211: Lecture 18, Pg 15
Torque

 = r Fsin                                    F

   So if  = 0o, then  = 0                  r

   And if  = 90o, then  = maximum
F
r

Physics 211: Lecture 18, Pg 16
Lecture 18, Act 2
Torque
   In which of the cases shown below is the torque provided by the
applied force about the rotation axis biggest? In both cases the
magnitude and direction of the applied force is the same.

(a) case 1
L
(b) case 2
(c) same                            F                              F
L
axis

case 1          case 2

Physics 211: Lecture 18, Pg 17
Lecture 18, Act 2
Solution
   Torque = F x (distance of closest approach)
 The applied force is the same.
 The distance of closest approach is the same.

Torque is the same!

F                                  F
L                       L

case 1              case 2

Physics 211: Lecture 18, Pg 18
Torque and the
Right Hand Rule:
   The right hand rule can tell you the direction of torque:
 Point your hand along the direction from the axis to the
point where the force is applied.
 Curl your fingers in the direction of the force.
 Your thumb will point in the direction
of the torque.
F
y

r
x

z
Physics 211: Lecture 18, Pg 19
The Cross Product

   We can describe the vectorial nature of torque in a
compact form by introducing the “cross product”.
 The cross product of two vectors is a third vector:

AXB=C                                B

   The length of C is given by:                            
C = AB sin                                       A

   The direction of C is perpendicular to         C
the plane defined by A and B, and in
the direction defined by the right hand
rule.

Physics 211: Lecture 18, Pg 20
The Cross Product

   Cartesian components of the cross product:

C=AXB
B
CX = AY BZ - BY AZ

CY = AZ BX - BZ AX                                        A

CZ = AX BY - BX AY                        C

Note: B X A = - A X B

Physics 211: Lecture 18, Pg 21
Torque & the Cross Product:

   So we can define torque as:

 =rXF
= rF sin 

X = rY FZ - FY rZ = y FZ - FY z               F
Y = rZ FX - FZ rX = z FX - FZ x
Z = rX FY - FX rY = x FY - FX y     
r
y

x
z

Physics 211: Lecture 18, Pg 22
Comment on  = I

   When we write  = I we are really talking about the z
component of a more general vector equation. (Recall that
we normally choose the z-axis to be the the rotation axis.)

z = Izz
z

Iz
z

   We usually omit the
z subscript for simplicity.
z

Physics 211: Lecture 18, Pg 23
Example

   To loosen a stuck nut, a (stupid) man pulls at an angle of
45o on the end of a 50 cm wrench with a force of 200 N.
 What is the magnitude of the torque on the nut?
 If the nut suddenly turns freely, what is the angular
acceleration of the wrench? (The wrench
has a mass of 3 kg, and its shape
is that of a thin rod).                             45o
F = 200 N

L = 0.5 m

Physics 211: Lecture 18, Pg 24
Example
Wrench w/ bolts

   Torque  = LFsin  = (0.5 m)(200 N)(sin 45)    = 70.7 Nm

   If the nut turns freely,  = I
 We know  and we want , so we need to figure out I.

1     1                                                  45o
I  ML2  3 kg 0.5 m 2  0.25 kgm2
3     3                                   F = 200 N

L = 0.5m
So = / I = (70.7 Nm) / (0.25 kgm2)



Physics 211: Lecture 18, Pg 25
Work

   Consider the work done by a force F acting on an object
constrained to move around a fixed axis. For an
infinitesimal angular displacement d:

 dW      .
= F dr = FR d cos()                                    F
= FR d cos(90-)                                     
= FR d sin()                          R
d        dr = R d
= FR sin() d
axis
 dW =  d

   We can integrate this to find: W = 
   Analogue of W = F •r
   W will be negative if  and  have opposite signs!

Physics 211: Lecture 18, Pg 26
Work & Kinetic Energy:

   Recall the Work/Kinetic Energy Theorem:       K = WNET

   This is true in general, and hence applies to rotational
motion as well as linear motion.

   So for an object that rotates about a fixed axis:

K      
1 2
2

I f   i2  WNET

Physics 211: Lecture 18, Pg 27
Example: Disk & String

   A massless string is wrapped 10 times around a disk of
mass M = 40 g and radius R = 10 cm. The disk is
constrained to rotate without friction about a fixed axis
though its center. The string is pulled with a force F = 10 N
until it has unwound. (Assume the string does not slip, and
that the disk is initially not spinning).

 How fast is the disk spinning after the string has
unwound?

M
R

F

Physics 211: Lecture 18, Pg 28
Disk & String...

   The work done is W = 
 The torque is = RF (since  = 90o)
 The angular displacement  is
M
R
   So W = (.1 m)(10 N)(20rad) = 62.8 J
      
    
                                        F

Physics 211: Lecture 18, Pg 29
Disk & String...
Flywheel, pulley,
1 2                 & mass
WNET = W = 62.8 J = K        I
2
its central axis is given by:
1
M
R
I  MR 2
2

So K   MR 2   2  W
1 1
     
2 2    

4W         4 62 .8 J 
MR 2
.04 kg .1    2

Physics 211: Lecture 18, Pg 30
Lecture 18, Act 3
Work & Energy
   Strings are wrapped around the circumference of two solid disks
and pulled with identical forces for the same distance.
Disk 1 has a bigger radius, but both have the same moment of
inertia. Both disks rotate freely around axes though their
centers, and start at rest.
 Which disk has the biggest angular velocity after the pull ?

1                    2
(a) disk 1
(b) disk 2
(c) same
F                     F

Physics 211: Lecture 18, Pg 31
Lecture 18, Act 3
Solution
    The work done on both disks is the same!
W = Fd
    The change in kinetic energy of each will
therefore also be the same since W = K.
1
But we know K      I 2
2                1                        2
So since I1 = I2

1 = 2
F                     F

d

Physics 211: Lecture 18, Pg 32
Spinning Disk Demo:
I
   We can test this with our big flywheel.

negligible
1 2 1
W  K      I  mv 2     in this case
2     2                                 m

2W

I

In this case,     I = 1 kg - m2
W = mgh = (2 kg)(9.81 m/s2)(1 m) = 19.6 J

 = 6.26 rad/s ~ 1 rev/s

Physics 211: Lecture 18, Pg 33
Recap of today’s lecture

   Direction and the right hand rule

   Rotational dynamics and torque

   Work and energy with example

Physics 211: Lecture 18, Pg 34

```
To top