Static Friction for round objects gives… rolling without slipping
Step 1: Find the acceleration
Analysis: Method 1 …. Energy conservation:
Since mechanical energy is conserved for the rolling object,
1
1
KE  PE  0  I 2  mv 2   mgy   0 .
2
2
2
For a solid sphere, I  mR 2 and for rolling without slipping, v   R leading to
5
12
1 2
1 2 1 2
10
10
2 2
2
gy 
gd sin 
 mR   mv  mgy  mv  mv  mgy  v 
25
2
5
2
7
7

So, upon using constant acceleration kinematics,
v 2  v02
v  v  2ax  a 

2x
2
2
0
10 gd sin 
5
7
 g sin 
2d
7
Analysis: Method 2 …. Force (more challenging):
Here, we look at the free body diagram and also consider the
torque… easiest to measure about the center of the rolling
sphere.
Analysis:
Fnet x
Fnet y
 max
 ma y
n  mg cos   0
mg sin   f s  ma
n  mg cos 
 net
 I
 f s r  I
Note:  friction   f s r as it is a clockwise torque.
a  r … the minus sign is since the angular acceleration is negative and the
acceleration is positive in the x-direction.
2
Lastly, for a solid sphere, I  mr 2
5
Through substitution from our analysis…
mg sin   f s  ma  mg sin  
Hence, mg sin  
I
2
 ma  mg sin   ma  ma .
r
5
7
5
ma  a  g sin 
5
7
Rolling without slipping:
A Friction Experiment…
a
5
g sin 
7
Michael Faleski
michaelfaleski@delta.edu
MIAAPT Fall Meeting 2008
Rolling without Slipping
Step 2: Determine the coefficient of friction
Use torque and previous result:
2
5
 a 
 r 
 friction   f s r  I    s nr   mr 2       s nr
Solving for the coefficient using the result that a 
5
g sin  , we have for the solid sphere….
7
2
 2  5

 m  g sin     s mg cos   s  tan  crit
7
 5  7

(This question has been asked on previous AP Physics C exams)
What if the object is slipping down the incline?
What is the acceleration?
Analysis: Forces
Fnet x
 max
mg sin   f k  ma
Fnet y
 ma y
n  mg cos   0
mg sin    k n  ma
n  mg cos 
Through substitution,
mg sin    k n  ma  mg sin    k mg cos  ma leading to the acceleration as
a  g sin    k cos  
A Friction Experiment…
Michael Faleski
michaelfaleski@delta.edu
MIAAPT Fall Meeting 2008
“Correcting” for ball moving through photogate
Assume a distance of X from the first gate and total
distance 1.0000 m between gates…
Recast the motion into 2 constant acceleration problems…
Problem 1: The ball moves down the incline for a distance of
X starting from rest.
Problem 2: The ball is timed for the known distance that it moves down
the incline.
Goal: Determine the common acceleration for the two problems
x0
x
v0
v
a
0m
X
0 m/s
V
A
t
where
t M is
x0
x
v0
X
1.0000 + X
V
v
a
A
t
tM
the measured time for the ball by experiment.
Mathematical Analysis (constant acceleration kinematics):
2
2
2
Problem 1: v  v0  2ax  x0   V  0  2 AX
(1)
1
1 2
2




x

x

v

t

a

t

1
.
000

Vt

AtM (2)
0
0
M
Problem 2:
2
2
Two equations… two unknowns… solve for V first by substituting
expression for acceleration from (1) into (2)… result:
 t M2  2
1V2  2
1.000  Vt M   t M   V  t M V  1.00  0
2  2X 
 4X 
A Friction Experiment…
Michael Faleski
michaelfaleski@delta.edu
MIAAPT Fall Meeting 2008
Use the quadratic equation to solve for V:
 t M2
 t M  t  4
 4X
V
 t M2 

2
4
X


2
M

 1


2X
tM

1

1

1



X


1

1

1



X


(Keep the positive root since this assumes that “down the incline” is the
positive direction… already assumed in description of position variables
(from 0 to X and then from X to 1.0000 + X)
V
2X
tM

V2
2 
1  


A


1

2
X
1

1

2 
And so,

2X
X  
tM 

(3)
For my assumption… X  0.0030 m
1.792
2.0000
A  2 instead of the “uncorrected” A 
tM
t M2
Control X experimentally and Eq. (3) will give the
expression to compute the acceleration.
A Friction Experiment…
Michael Faleski
michaelfaleski@delta.edu
MIAAPT Fall Meeting 2008