Transcript Ppt

Lecture 19
Goals:
• Specify rolling motion (center of mass velocity to
angular velocity
• Compare kinetic and rotational energies with rolling
• Work combined force and torque problems
• Revisit vector cross product
• Introduce angular momentum
Physics 201: Lecture 19, Pg 1
Connection with Center-of-mass motion

If an object of mass M is moving linearly at velocity
VCM without rotating then its kinetic energy is
2
K T  12 MVCM

If an object with moment of inertia ICM is rotating in place
about its center of mass at angular velocity  then its
kinetic energy is
K R  I CM
1
2


If the object is both moving linearly and rotating then
K  I CM  MV
1
2
2
1
2
2
CM
Physics 201: Lecture 19, Pg 6
But what of Rolling Motion




Consider a cylinder rolling at a constant speed.
Contact point has zero velocity in the laboratory frame.
Center of wheel has a velocity of VCM
Top of wheel has a velocity of 2VCM
2VCM
VCM
CM
Physics 201: Lecture 19, Pg 7
Rolling Motion

Now consider a cylinder rolling at a constant speed.
VCM
CM
The cylinder is rotating about CM and its CM is moving at
constant speed (VCM). Thus its total kinetic energy is
given by :
K TOT  I CM  MV
1
2
2
1
2
2
CM
Physics 201: Lecture 19, Pg 8
Motion

Again consider a cylinder rolling at a constant speed.
Rotation only
vt = R
CM
Both with
|vt| = |vCM |
2VCM
VCM
CM
Sliding only
VCM
CM
Physics 201: Lecture 19, Pg 9
Concept question

For a hoop of mass M and radius R that is rolling
without slipping, which is greater, its translational
or its rotational kinetic energy?
A.
Translational energy is greater
Rotational energy is greater
They are equal
The answer depends on the radius
The answer depends on the mass
B.
C.
D.
E.
Physics 201: Lecture 19, Pg 10
Example : Rolling Motion


A solid cylinder is about to roll down an inclined plane.
What is its speed at the bottom of the plane ?
Use Work-Energy theorem
Disk has radius R
M
h
q
Mgh = ½ Mv2 + ½ ICM 2
and
M
v?
v =R
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
v = 2(gh/3)½
Physics 201: Lecture 19, Pg 11
Question: The Loop-the-Loop … with rolling

To complete the loop-the-loop, how does the height h
compare with rolling as to that with frictionless sliding?
A. hrolling > hsliding
B. hrolling = hsliding
C.hrolling < hsliding
ball has mass m &
r <<R
U=mg2R
h?
R
Physics 201: Lecture 19, Pg 12
Example: Loop-the-Loop with rolling
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Ub=mgh
To complete the loop the loop, how high do we
have to release a ball with radius r (r <<R) ?
Condition for completing the loop the loop: Circular
motion at the top of the loop (ac = v2 / R)
Use fact that E = U + K = constant (or work
energy)
ball has mass m &
r <<R
U=mg2R
h?
R
v
Tangential
Recall that “g” is the source of
the centripetal acceleration
and N just goes to zero is
the limiting case.
Also recall the minimum speed
at the top is
 gR
Physics 201: Lecture 19, Pg 13
Example: Loop-the-Loop with rolling
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Work energy (DK = W)
Kf = KCM + KRot = mg (h-2R) (at top)
Kf = ½ mv2 + ½ 2/5 mr2 2 = mg (h-2R) & v=r
½ mgR + 1/5 mgR = mg(h-2R)  h = 5/2R+1/5R
Ub=mgh
ball has mass m &
r <<R
U=mg2R
v
Tangential
 gR
h?
R
Just a little bit more….
Physics 201: Lecture 19, Pg 14
Torque : Rolling Motion (center of mass point)
 A solid cylinder, with mass m and radius R, is
rolling without slipping down an inclined plane.
 What is its angular acceleration a?
1. SFx = m acm = mg sin q - fs
 m Racm = mgR sin q - Rfs
SFy = 0 = N – mg cos q
(not used)
. St = -R fs = Icm a
3. a = - acm / R
N
fs
q
mg
q
 -mR2 a= mgR sin q + Ia  -mR2 a – Ia = mgR sin q
a = -mgR sin q / (mR2 + I) = -2g sin q / 3R
Physics 201: Lecture 19, Pg 16
Torque : Rolling Motion (contact point)
 A solid cylinder, with mass m and radius R, is
rolling without slipping down an inclined plane.
 What is its angular acceleration a?
1. St = -R mg sin q = I a
2. I = mR2 + Icm
a = -mgR sin q / (mR2 + Icm)
a = 2g sin q / 3R
N
q
mg
fs
q
Physics 201: Lecture 19, Pg 17
Torque : Come back spool (contact point)
 A solid cylinder, with mass m, inner radius r and
outer radius R, being pulled by a horizontal force F
at radius r? If the cylinder does not slip then what
is the angular acceleration?
1. St = -rF = I a
2. I = mR2 + Icm = 3/2 mR2
a = -rF / (mR2 + Icm) = -2Fr / 3R
N
fs
F
mg
1. St = 0 = I a
a= 0
N
F
mg
fs
1. St = rF = I a
2. I = mR2 + Icm
N
a = 2Fr / 3R
fs
F
mg
Physics 201: Lecture 19, Pg 18
Torque : Limit of Rolling (center of mass)
 A solid cylinder, with mass m and radius R, being
pulled by a horizontal force F on its axis of rotation. If
ms is the coefficient of static friction at the contact
point, what is the maximum force that can be applied
before slipping occurs?
N
1. SFx = m acm = F - fs
F
2. SFy = 0 = N – mg
mg
3. St = -R fs = Icm a 
fs
- acm  - fs mR2 / Icm
4. fs ≤ ms N = ms mg (maximum)
5. acm = - a R
m acm = fs mR2 / Icm = F - fs
fs (1+ mR2 / Icm ) = ms mg (1+ mR2 / Icm ) = F
F = ms mg (1+ mR2 / ( ½ mR2 ) ) = 3 ms mg
Physics 201: Lecture 19, Pg 19
Modified Atwood’s Machine (More torques)

Two blocks, as shown, are attached by a massless string
which passes over a pulley with radius R and rotational inertia
I = ½ MR2. The string moves past the pulley without slipping.
The surface of the table is frictionless.
 What are the tensions in the strings ?
N
S Fy = m1 ay= -m1g +T1
T1
mass 2
T1
m1g
S Fx = m2 ax = –T
T
T
mass 1
m2g
pulley
S t = Ia = R T1 – R T
m1 a = -m1g +T1
ay = ax= -aR = a
m2 a = –T
-I a / R = R T1 – R T
Physics 201: Lecture 19, Pg 20
Modified Atwood’s Machine (More torques)

Two blocks, as shown, are attached by a massless string
which passes over a pulley with radius R and rotational inertia
I. The string moves past the pulley without slipping. The
surface of the table is frictionless.
 What are the tensions in the strings?
N
T
1. m1 a = – m1g +T1
2. m2 a = –T
T1
3. -I a / R = R T1 – R T
T1
m1g
T
m2g
Solve for a
-I a / R2 = m1 a +m1g + m2 a
a = – m1g / (m1 + ½ M + m2)
T= - m2 a
T1= m1 a + m1g
Physics 201: Lecture 19, Pg 21
Angular Momentum

We have shown that for a system of
particles,
momentum
is conserved if


p  mv


dp
0
 FExt 
dt



Are the rotational equivalents? t  r  F  Ia


 
angular momentum
L  r  p  I

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is conserved if

t Ext
 
 r F 

dL
dt
0
Physics 201: Lecture 19, Pg 22
For Thursday
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Read all of chapter 10 (gyroscopes will not be tested)
HW9
Physics 201: Lecture 19, Pg 23