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Exam Results
PHYSICS 220
Lecture 14
Equilibrium and Dynamics
Mean 98.26 = 65.6% Aimed for 60 - 65%
which gives maximum in information
> m + std = 127 very good score
< m - std = 69.3 is a poor score
Lecture 14
Purdue University, Physics 220
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Lecture 13
Purdue University, Physics 220
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Exercise
A 50 kg diver stands at the end of a 4.6
m diving board. Neglecting the weight of
the board, what is the force on the pivot
1.2 meters from the end?
1) Draw FBD
2) Choose Axis of rotation
3) ! = 0
Rotational EQ
F1 (1.2) – mg (4.6) = 0
F1 = 4.6 (50 *9.8) / 1.2
F1 = 1880 N
4) F = 0
Translational EQ
F1 – F2 – mg = 0
F2 = F1 – mg = 1390 N
Lecture 14
Purdue University, Physics 220
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Lecture 14
Purdue University, Physics 220
F1
F2
mg
4
Exercise
Find Center of Mass
Not in equilibrium
A 75 kg painter stands at the center of a 50 kg 3 meter
plank. The supports are 1 meter in from each edge.
Calculate the force on support A.
Equilibrium
pivot
1 meter
1 meter
pivot
B
A
d
W=mg
Center of
mass
2) F = 0
Center of
mass
Torque about pivot " 0
FA
1) Draw FBD
FA + FB – mg – Mg = 0
FB
0.5meter
3) Choose pivot
mg Mg
4) ! = 0
Torque about pivot = 0
1 meter
-FA (1) sin(90)+ FB (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0
FA = 0.5 mg + 0.5 Mg = 612.5 N
Lecture 14
Purdue University, Physics 220
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Lecture 14
Purdue University, Physics 220
Problem
Question
How far to the right of support B can the painter
stand before the plank tips?
If the painter moves to the right, the force exerted by
support A
A) Increases
B) Unchanged
1 meter
1 meter
C) Decreases
B
A
Just before board tips, force from A becomes zero
1 meter
1 meter
A
6
FB
1) Draw FBD
2) F = 0
B
FB – mg – Mg = 0
3) Choose pivot
mg
4) ! = 0
0.5meter
x
Mg
FB (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = 0
Lecture 14
Purdue University, Physics 220
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0.5 m = x M -> x = 0.3 meter
Lecture 14
Purdue University, Physics 220
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Demo 1Q - 13
Spool on a Rough Surface
• Consider all of the forces acting: tension T and
friction f.
A large spool will move left or right
when the string is pulled depending
on the angle #
– Using FNET = 0 in the x direction:
T cos! " f = 0
f = T cos!
Using !NET = 0 about the CM axis:
T
T
aT = bf
aT ! bf = 0
#
#
Solving:
a
cos! =
b
a
b
a
y
b
f
x
f
Lecture 14
Spool on a Rough Surface
Purdue University, Physics 220
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Spool on a Rough Surface
• There is another (slick) way to see this:
• Consider the torque about the point of contact
between the spool and the ground. We know the
net torque about this (or any other) point is zero.
• So we can use geometry to get the same result
cos ! =
– Since both Mg and f act through this point, they do not
contribute to the net toque.
T
– Therefore the torque due to T must
#
also be zero.
– Therefore T must act
a
y
along a line that passes
b Mg
through this point!
T
a
b
#
a
#
b
x
f
Lecture 14
Purdue University, Physics 220
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Lecture 14
Purdue University, Physics 220
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Falling Weight & Pulley
Rotational Dynamics
•
!=I$
– Torque is amount of twist provide by a force
• Signs: positive = CCW
– Moment of Inertial like mass. Large I means hard
to start or stop from spinning.
– Angular acceleration is defined as the rate at
which the angular velocity changes.
• Problems Solving:
– Draw free body diagram
– Pick an axis
– Compute torque due to each force
Lecture 14
Purdue University, Physics 220
A mass m is hung by a string that is wrapped
around a pulley of radius R attached to a heavy
flywheel. The moment of inertia of the pulley +
flywheel is I. The string does not slip on the
pulley.
Starting at rest, how long does it take for the
mass to fall a distance L.
I
$
R
T
m
mg
a
L
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Lecture 14
Falling Weight & Pulley
Purdue University, Physics 220
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Falling Weight & Pulley
• For the hanging mass use F = ma
• Using 1-D kinematics we
can solve for the time
required for the weight to
fall a distance L:
mg - T = ma
1 2
at
2
t=
2L
a
I
$
TR = I
Purdue University, Physics 220
T
a
R
• Now solve for a using the above
equations.
m
mg
m
a
mg
L
2
L
Lecture 14
R
• Realize that a = $R
T
a
I
$
! = TR sin(90) = I$
R
1
y = y0 + v0t + at 2
2
L=
• For the flywheel use ! = I$
! mR $
a=#
&g
" mR 2 + I %
16
Lecture 14
Purdue University, Physics 220
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Rolling
Rolling
• An object with mass M, radius R, and moment of
inertia I rolls without slipping down a plane
inclined at an angle # with respect to horizontal.
What is its acceleration?
• Static friction f causes rolling. It is an unknown,
so we must solve for it.
• First consider the free body diagram of the object
and use FNET = Macm :
In the x direction
• Consider CM motion and rotation about
the CM separately when solving this
problem
• Now consider rotation about the CM
and use ! = I$ realizing that
I
M
Mg sin # - f = Macm
R
#
Lecture 14
Purdue University, Physics 220
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Lecture 14
Rolling
• We have two equations:
f = I!
f=I
f =I
f
R
x
! = Rf and acm = $R
a
Rf = I cm
R
M
y
Mg
#
#
acm
R2
Purdue University, Physics 220
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Energy Conservation
Mg sin # - f = Ma
• Friction causes an object to roll, but if it rolls w/o
slipping friction does NO work!
a cm
R2
• We can combine these to eliminate f:
• No dissipated work, energy is conserved
2
" MR sin ! %
acm = g $
'
# MR 2 + I &
• Need to include both translation and rotation
kinetic energy.
I
For a sphere:
acm
Lecture 14
a
"
%
$ MR 2sin ! ' 5
= g$
' = 7 gsin !
2
$ MR 2 + MR 2 '
#
&
5
Purdue University, Physics 220
M
KE = ! m vcm2 + ! I %2
R
#
20
Lecture 14
Purdue University, Physics 220
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Rolling Race (Hoop vs Cylinder)
Demo 1Q - 04
Translational + Rotational KE
• Consider a cylinder with radius R and mass M, rolling w/o
slipping down a ramp. Determine the ratio of the
translational to rotational KE.
A hoop and a cylinder of equal mass roll down a
ramp with height h. Which has greater speed at
the bottom of the ramp?
A) Hoop
B) Same
C) Cylinder
Translational: KET = ! M Vcm2
Rotational:
use
I=
1
MR 2
2
KER = ! I %2
and ! =
Vcm
R
KER = ! (! M
R2)
(Vcm
I = ! MR2
I = MR2
/R)2
= " M Vcm2
H
= ! KET
Cylinder will get to the bottom first because inertia for a cylinder is less than
that for a hoop type object.
Lecture 14
Purdue University, Physics 220
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Lecture 14
Purdue University, Physics 220
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