Transcript hp1f2013_class15_rolling_motion_and_accelerating_frames

Honors Physics 1
Class 15 Fall 2013
Rolling motion
Noninertial reference frames
Fictitious forces
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Rolling motion
We will treat the case where
1) the axis of rotation passes through the
center of mass.
2) the axis of rotation does not change
direction.
Example = Rolling wheel
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11.2 Rolling
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11.3 The Kinetic Energy of Rolling
If we view the rolling as pure rotation about an axis
through P, then
(w 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= Icom +Mh2),
(M is the mass of the wheel, Icom 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 vcom =wR, 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 Icomw2),
2. a translational kinetic energy due to translation of its center of mass (=½ Mv2com)
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The Forces of Rolling: Friction and Rolling
A wheel rolls horizontally without sliding
while accelerating with linear acceleration
acom. A static frictional force fs acts on the
wheel at P, opposing its tendency to slide.
The magnitudes of the linear acceleration
acom, and the angular acceleration a can be
related by:
where R is the radius of the wheel.
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The Forces of Rolling: Rolling Down a Ramp
A round uniform body of radius
R rolls down a ramp. The forces
that act on it are the
gravitational force Fg, a normal
force FN, and a frictional force
fs pointing up the ramp.
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Sample problem: Rolling Down a Ramp
b) What are the magnitude and direction of
the frictional force on the ball as it rolls
down the ramp?
Calculations: First we need to determine
the ball’s acceleration acom,x :
A uniform ball, of mass M=6.00 kg and radius
R, rolls smoothly from rest down a ramp at
angle q=30.0°.
a) The ball descends a vertical height h=1.20 m
to reach the bottom of the ramp. What is its
speed at the bottom?
Calculations: Where Icom is the ball’s rotational
inertia about an axis through its center of mass,
vcom is the requested speed at the bottom, and w
is the angular speed there.
Substituting vcom/R for w, and 2/5 MR2 for Icom,
We can now solve for the frictional force:
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q
A wheel and a cube,
both of mass M are
made to race down a
slope.
The wheel rolls without
slipping. The cube
slides without rolling.
Which one gets to the
bottom first?
q
Which one has greater
KE at the bottom?
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Energy analysis of rolling down a ramp
q
q
a) A cubical block with mass M
and side 2R is allowed to slide
without rolling down a
ramp/slide from height h to
height 0. How fast does the
center of mass travel at the
bottom of the slope?
b) A wheel with mass M and
radius R is allowed to roll
without slipping on a ramp of
height h to the level ground at
height 0. How fast does its
center of mass travel at the
bottom of the ramp?
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The Yo-Yo
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Galilean transformations
Consider measurements of the motion of a mass m
made from each of two coordinate systems.
For simplicity let’s assume that corresponding axes are
parallel and the scale units (time and distance) are
the same.
The origins of the two systems are displaced by S so
r  ra  S
If the two systems move at a fixed velocity V with respect to one another,
then V 
dS
dt
Observer a sees the object moving with velocity va
Observer  sees the object moving with velocity v
dra

.
dt
dr dra


V.
dt
dt
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Galilean transformations 2
Observer a measures acceleration aa 
Observer  measures acceleration a 
d 2 ra
dt
2
d 2 r
dt 2
.

d 2 r
dt 2

dV
.
dt
If V is constant then aa  a and
Fa  maa  ma  F and Newton's second law is the same.
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XKCD LINK
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Linearly Accelerating Frame
Fictitious forces
How is Newton’s second law affected by measurement
in accelerating frames?
a  aa  A assuming constant A
Fa  maa
F  maa  mA  Fa  mA
Assuming the frame a is inertial, then the frame  observer
measures an additional force.
F  Fa  F fict
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Example: Hanging mass in a car
A small weight of mass m hangs from a string in a car which accelerates
at rate A=1/4g.
a) What is the static angle of the string from the vertical?
b) What is the tension in the string?
q
T
T cosq  mg  0
T sin q  mA  0
mA A
tan q 

mg g
Tfict
mg
T  m( g 2  A2 )1/2
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Principle of Equivalence
In the example problem, we treated
acceleration A in the same way as we treated
gravitational acceleration.
The Principle of Equivalence states that there is
no way to distinguish locally* between a
gravitational acceleration and an acceleration
of the coordinate system.
*Locally means that we don’t look outside the system
for the cause of an acceleration or gravity.
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Rotating motion and fictitious forces
Consider two reference frames, one rotating about the
z axis, the other inertial. Origins coincide. Z axes
coincide.
What fictitious forces are observed in the rotating frame?
We already have some experience with this from
analyzing forces in the spinning terror ride and centripetal
acceleration.
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Rotating frames
Let's consider the time derivatives of a changing particle position
in inertial and rotating (primed parameters) frames.
Let the x,y,z and x',y',z' coordinates of the particle coincide at t=0.
At a later time r (t  t )  r (t )  r so r  r (t  t )  r (t ).
Since observer' and observer(inertial) started in the same position,
r '  r (t  t )  r '(t )
From the drawing we see that r  r ' r '(t ) - r (t )
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Rotating frames
We can easily calculate the difference between r (t ) and r '(t ) :


r '(t )  r (t )    r t
dr dr '

   r or vinertial  vrot    r
dt
dt
This approach was general and applies to the time derivative of
any vector, so we can apply it to v and a.
so
 dv 
 dv 
a inertial   in    in     vin
 dt in  dt rot
and using vin  vrot    r we have
d

ain  
vrot    r     vrot      r
 dt
 rot




arot  ain  2  vrot      r


From which, F fict   m2  vrot  m    r


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Fictitious forces in rotating frames

F fict  m2  vrot  m    r

with vrot  vin    r


m    r  centrifugal force
m2  vrot  Coriolis force
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