Transcript F - Cloudfront.net

Physics: Lecture 11
(Ch. 10-11 Halliday)
• Direction and the right hand rule
• Rotational dynamics and torque
• Work and energy with example
More about rolling
Rotational v.s. Linear Kinematics
Angular
Linear
  constant
a  cons tan t
  0  t
v  v0  at
1 2
   0   0t  t
2
1 2
x  x0  v0 t  at
2
And for a point at a distance R from the rotation axis:
x = R
v = R
a = R
Rotational Dynamics:
What makes it spin?
•
•

Suppose a force acts on a mass
constrained to move in a circle.
Consider its acceleration in the
(tangential) direction at some
instant:
– at = r
Now use Newton’s 2nd Law in
the 
direction:
– Ft = mat = mr
^

^

F
^
F
a
r
Multiply by r :
rFt = mr2
^
r

m
rFt = mr2
Rotational Dynamics:
What makes it spin?
use I  mr 2
 I
• Define torque: t = rFt.
– t is the tangential force Ft
times the lever arm r.
^

t  I
• Torque has a direction:
– + z if it tries to make the
system
spin CCW.
– - z if it tries to make the
system
spin CW.
^
r
F
F
a
r

m
Rotational Dynamics:
What makes it spin?
r
 rF   m
2
• So for a collection of many particles
arranged in a rigid configuration:

i
i
t
ti
i i
i
i
I
Since the particles are connected rigidly,
they all have the same .
t
i
 I
i
t NET  I
m4
F4
F1
m3
F3
r1 m1

r4
r3
r2
m2
F2
Rotational Dynamics:
What makes it spin?
tNET = 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.
Torque

Recall the definition of torque:
t = rF
= r F sin 
= r sin  F
t = r pF

rp = “distance of closest approach”
or torque arm.
Equivalent definitions!
Fr  F

F

r
rp
Torque
t = r Fsin 
• So if  =
0o,
F
then t = 0
r
F
• And if  = 90o, then t = maximum r
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.
L
F
(a) case 1
(b) case 2
F
L
axis
(c) same
case 1
case 2
QuestionSolution
•
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
L
F
L
case 1
case 2
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
F
of the torque.
y
r
x
t
z
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
• The length of C is given by:
C = AB sin 
• 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.
B

A
The Cross Product
• Cartesian components of the cross product:
C=AXB
B
C X = AY B Z - B Y AZ
A
C Y = AZ B X - B Z AX
C Z = AX B Y - B X AY
Note: B X A = - A X B
C
Torque Vector
Fig. 11-10 (a) A force F, lying in an x-y plane, acts on a particle at point A. (b) This force
produces a torque t = r x F on the particle with respect to the origin O. By the right-hand rule for
vector (cross) products, the torque vector points in the positive direction of z. Its magnitude is
given by in (b) and by
in (c).
Sample problem
Calculations: Because we want the torques
with respect to the origin O, the vector required
for each cross product is the given position
vector r. To determine the angle  between
the direction of r and the direction of each
force, we shift the force vectors of Fig.a, each
in turn, so that their tails are at the origin.
Figures b, c, and d, which are direct
views of the xz plane, show the shifted force
vectors F1, F2, and F3. respectively.
In Fig. d, the angle between the directions of
and is 90°.
Now, we find the magnitudes of the torques to
be
Torque & the Cross Product:
• So we can define torque as:
t =rXF
= rF sin 
t X = r Y FZ - FY r Z = y FZ - FY z
tY = rZ FX - FZ rX = z FX - FZ x
F
t
r
y
t Z = r X FY - FX r Y = x FY - FX y
z
x
Comment on t = I
• When we write t = I we are really talking about
the z component of a more general vector
equation. (Recall that we normally choose the zaxis to be the the rotation axis.)
tz = Izz
tz
Iz
z
• We usually omit the
z subscript for simplicity.
z
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
45o
has a mass of 3 kg, and its shape
is that of a thin rod).
F = 200 N
L = 0.5 m
Example
Wrench w/ bolts

Torque t = LFsin  = (0.5 m)(200 N)(sin 45)

If the nut turns freely, t = I
 We know t and we want , so we need to figure out I.
1 2 1
2
I  ML  3 kg 0.5 m   0.25 kgm2
3
3
= 70.7 Nm
45o
F = 200 N
L = 0.5m
So = t / I = (70.7 Nm) / (0.25 kgm2)
 = 283 rad/s2

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:
= F.dr = FR d cos()
= FR d cos(90-)
= FR d sin()
= FR sin() d
dW = t d
– dW

F

R
d
dr = R d
axis
• We can integrate this to find: W = t
• Analogue of W = F •r
• W will be negative if t and  have opposite signs!
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
I  2f   i2  W NET
2
Connection with CM motion...
•
So for a solid object which rotates about its center or mass and whose
CM is moving:
K NET
1
1
2
2
 I CM   MVCM
2
2
VCM

We will use this formula more in the future.
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?
R
F
M
Disk & String...
• The work done is W = F x d
– The displacement is
2 x 0.1m x 10 rev = 2  m
R

So W = (10 N)(2 ) = 62.8 J
 


F
d
F
M
Disk & String...
Flywheel, pulley,
WNET = W = 62.8 J = K  1 I  2
& mass
2
Recall that I for a disk about
its central axis is given by:
1
I  MR 2
2
So

R

11

K   MR 2  2  W
22

4W

2
MR
462.8 J 
.04kg .12
M
 = 792.5 rad/s
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 ?
(a) disk 1
2
1
(b) disk 2
(c) same
F
F
d
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
2
1
So since I1 = I2
1 = 2
F
F
d
Spinning Disk Demo:
• We can test this with our big flywheel.
1 2 1 2
W  K  I   mv
2
2
I
negligible
in this case
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
Applications of Newton’s Second Law for Rotation
Examples
t  Ftr
t net ext  t ext  I
Radius sprocket rs= 7cm.
Find ω after t = 5 s.
1. ω = αt
3.
t ext  Frs
2.
t ext  I

t
Fr
ext
s
4  

I
MR 2
Frs
18 N * 0.07m
t
* 5s  21.4rad / s
5.   t 
2
MR
2.4kg * 0.35m
Known L, M. Find:
(a) α immediately
following the release
(b) The force FA
a.
1
b.
t ext  I
2 t grav
1 Fexty  Macmy
L
 Mg
2
3

t grav
2. Mg-FA=Macmy
I

MgL / 2
3g

(1 / 3) ML2 2L
3. at = rα
4. acmy = rcmα = (L/2)α
Mg  FA L 3g
5. From 2, 3 and 4:

M
2 2L
6. FA = (1/4)Mg
Nonslip Conditions
A string wrapped around a rotating wheel
must move with a tangential velocity vt
that is equal to the tangential velocity of
the rim of the wheel, provided that the
string remains taut and does not slip.
By differentiating the previous equation we get:
Problem Solving Guide
Falling weight & pulley...
An object of mass m is tied to a light string wound
around a pulley that has the moment of inertia I
and the radius R. In non slipping conditions, find
the tension in the string and the acceleration of the
object.
1. Draw a free body diagram,
drawing each force in its point
of application.
2. The only force causing
a torque is T:
t ext  I
TR  I
Falling weight & pulley...
3. Draw a free body diagram
for the suspended object and
apply Newton’s second law:
6. Substitute
T into 3) and
solve for at:
mg  T  mat
at  R
4. Relate the linear and
angular accelerations:
5. α from 2) and
at from 3) in 4)
gives:
Fext  ma
mg  T
TR
R
m
I
mg 
mg
 mat
2
mR
1
I
T
so
at 
1
I
1
mR 2
g
mg
mR2
1
I
Falling weight & pulley...
• Using 1-D kinematics
(Lecture 1) we can solve
for the time required for
the weight to fall a
distance L:
1 2
2L
t
L  at
a
2
 mR 
 g
a  
2
 mR  I 
2
where

I
R
T
m
a
mg
L
Power

The work done by a torque t acting through a displacement  is
given by:
W  t

The power provided by a constant torque is therefore given by:
dW
d
P
t
 t
dt
dt
Review: Torque and Angular
Acceleration
tNET = 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.
Two wheels can rotate freely about fixed axles through
their centers. The wheels have the same mass, but one
has twice the radius of the other.
Forces F1 and F2 are applied as shown. What is F2 / F1 if
the angular acceleration of the wheels is the same?
F2
F1
1. A) 1
2. B) 2
3. C) 4
Solution
We know
but
t  I
t  FR
and
FR  mR 2
so
F  mR
Since R2 = 2 R1
I  mR 2
F2
mR2
R2


F1
mR1
R1
F2
2
F1
F1
F2
Rotation around a moving axis.
• A string is wound around a puck (disk) of mass M and
radius R. The puck is initially lying at rest on a
frictionless horizontal surface. The string is pulled
with a force F and does not slip as it unwinds.
– What length of string L has unwound after the
puck has moved a distance D?
M
R
F
Top view
Rotation around a moving axis...
A
• The CM moves according to F = MA



D
The distance moved by the CM is thus

The disk will rotate about
its CM according to t = I
1 2
I  MR
2

I

1 2
F 2
At 
t
2
2M
RF
1
MR 2
2

2F
MR
1
F 2
  t 2 
t
2
MR
So the angular displacement is
M
t
F
M
A
R
F
Rotation around a moving axis...
•
So we know both the distance moved by the CM and the angle of
rotation about the CM as a function of time:
F 2
D
t (a)
2M

Divide (b) by (a):
D

F 2

t (b)
MR
2
R
R  2 D
The length of string
pulled out is L = R:
L  2D

F
F
D
L
Comments on CM acceleration:
• We just used t = I for rotation about an axis through the CM even
though the CM was accelerating!
– The CM is not an inertial reference frame! Is this OK??
(After all, we can only use F = ma in an inertial reference
frame).
• YES! We can always write t = I for an axis through the CM.
– This is true even if the CM is accelerating.
– We will prove this when we discuss angular momentum!

M
A
R
F
Chapter 11
(Halliday)
Rolling, Torque, and
Angular Momentum
Rolling Objects
Rolling Without Slipping
Rolling as Translational and
Rotation Combined
Although the center of the object moves in a straight line parallel to the surface, a point on the rim
certainly does not.
This motion can be studied by treating it as a combination of translation of the center of mass and rotation
of the rest of the object around that center.
The part of the spool that is moving
slowest is blurred the least.
Rolling = Translation + Rotation
Note: “com” – center of mass
The wheel of radius R
is rotating without
slipping on a flat
surface.
Point P on the wheel moves as shown with the linear velocity:
where r is the perpendicular
distance from P to the axis of
rotation.
The CM of the wheel
moves with the speed:
Note: The top of the wheel @ 2R moves @ v = 2vcm
By differentiating each side of the equation 9-26
we obtain the acceleration of the CM
A falling yo-yo unwinding from a string with the top
fixed, follows the same rules as the wheel.
As the wheel rotates through the angle Φ, the
point of contact b/w the wheel and the surface
moves a distance:
If the wheel is rolling on a flat surface, the CM
remains directly over the point of contact, so it also
moves through s.
The Forces of Rolling: Friction and Rolling
A wheel rolls horizontally without sliding while accelerating
with linear acceleration acm. A static frictional force fs acts on
the wheel at P, opposing its tendency to slide.
If the wheel does slide when the net force acts on it, the
frictional force that acts at P is a kinetic frictional force, fk .
The motion then is not smooth rolling, and the above relation
does not apply to the motion.
The Kinetic Energy of Rolling
If we view the rolling as pure rotation about an axis through P,
then
( is the angular speed of the wheel and IP is the rotational
inertia of the wheel about the axis through P).
Using the parallel-axis theorem (I P= Icm +Mh2),
IP = ICM + MR2
(M is the mass of the wheel, Icm is its rotational inertia about an
axis through its center of mass, and R is the wheel’s radius, at a
perpendicular distance h).
Using the relation vcm =R, we get:
A rolling object, therefore has two types of kinetic energy:
1)a rotational kinetic energy due to its rotation about its center of mass (=½ Icm2),
2) and a translational kinetic energy due to translation of its center of mass (=½ Mv2cm)
Example 1: A bowling ball
with R = 11 cm, vi = 2m/s,
and M = 7.2 Kg is rolling
up an incline and stops
momentarily at h, before
rolling back. Find h.
Solution
Apply conservation of energy:
v
Substitute:   i
R
Solve for h:
U i  Ki  U f  K f
2
2
and I  MR
5
Mgh 
1
1
2
Mvi  Ii2
2
2
vi2
1 2 1 2
7
2
Mgh  Mvi  ( MR )( 2 )  Mvi2
2
2 5
R
10
7vi2
h
 0.285m  28.5cm
10 g
A cue stick hits a ball x above CM. Find x
for which the ball rolls w/o slipping. Express
your answer in terms of the radius of the
sphere R.
1. Free body diagram,
negligible friction.
2. Torque about CM:
3. 2nd Law:
F  maCM
4. Nonslip condition:
5. Substitute:
6. Solve for x:
t  Fx
and
t  I CM 
aCM  R
F
Fx
R
m
I CM
I CM (2 / 5)mR 2 2
x

 R
mR
mR
5
Remark: Striking the ball at a point higher or lower than (2/5)R
from CM, will result in the ball rolling and slipping (skidding)
which is often desirable in the game of billiard.
Objects of Different I Rolling Down The
Plane
A sphere, a cylinder, and a hoop with the same mass are released together from
rest at the top of an incline. The sphere reaches the bottom first, followed by
the cylinder and then the hoop.
Rolling Motion
Roll objects
down ramp
• Objects of different I rolling down an inclined plane:
K = - U = Mgh
R
M
v=0
=0
K=0
1
1 2
2
K  I CM   MvCM
2
2
h
vCM = R
Rolling...
• If there is no slipping:
vCM
In the lab reference frame
2v
v

v
Where v = R
In the CM reference frame
Rolling...
1
1
2
2
K  I CM   MvCM
2
2
Use vCM = R and ICM = cMR2 .
1
1 2
1 c
2 2
2
K  c MR   MvCM    1MvCM
2
2
2
So:
1
2
 c  1MvCM
 Mgh
2
vCM
hoop:
c=1
disk:
c = 1/2
sphere: c = 2/5
etc...
1
 2 gh
c 1
The rolling speed is always lower than in the case of simple sliding
since the kinetic energy is shared between CM motion and rotation.
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
y
– CCW rotation is in
the + z direction
x
z
y
– CW rotation is in
the - z direction
x
z
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 your thumb will point in
the direction of the rotation vector!
We normally pick the z-axis to be the rotation axis
as shown.
– = z
– = z
– = z
For simplicity we omit the subscripts unless
explicitly needed.
x
z
y
x
z
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 t when  = 0 :
0
t 

500 rad / s
So in this case t 
 1000 s  16.7 min
2
0.5 rad / s


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
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
Recap of today’s lecture
• More about rolling
• Direction and the right hand rule
• Rotational dynamics and torque
• Work and energy with example