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Physics 201 – Lecture 19
Lecture 19
Example: Rotating Rod
Goals:
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• 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
A uniform rod of length L=0.5 m and mass m=1 kg is free to
rotate on a frictionless pin passing through one end as in
the Figure. The rod is released from rest in the horizontal
position. What is
(A) its angular speed when it reaches the lowest point ?
(B) its initial angular acceleration ?
(C) initial linear acceleration of its free end ?
L
m
Physics 201: Lecture 19, Pg 1
Physics 201: Lecture 19, Pg 2
Example: Rotating Rod
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Example: Rotating Rod
A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
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(B) its initial angular acceleration ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and put in the Force on a FBD
2. The hinge changes everything!
A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
(C) initial linear acceleration of its free end ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and put in the Force on a FBD
2. The hinge changes everything!
L
L
m
Σ F = 0 occurs only at the hinge
m
but τz = I αz = r F sin 90°
mg
a=αL
mg
at the center of mass and
(ICM + m(L/2)2) αz = (L/2) mg
and solve for αz
Physics 201: Lecture 19, Pg 3
Physics 201: Lecture 19, Pg 4
Connection with Center-of-mass motion
Example: Rotating Rod
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A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate
on a frictionless hinge passing through one end as shown. The rod
is released from rest in the horizontal position. What is
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(A) its angular speed when it reaches the lowest point ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and use the Work-Energy Theorem
2. The hinge changes everything!
L
m
mg
L/2
If an object of mass M is moving linearly at velocity
VCM without rotating then its kinetic energy is
2
K T = 12 MVCM
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W = mgh = ½ I ω2
If an object with moment of inertia ICM is rotating in place
about its center of mass at angular velocity ω then its
kinetic energy is
K R = 12 I CMω 2
W = mgL/2 = ½ (ICM + m (L/2)2) ω2
and solve for ω
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If the object is both moving linearly and rotating then
2
K = 12 I CMω 2 + 12 MVCM
mg
Physics 201: Lecture 19, Pg 5
Physics 201: Lecture 19, Pg 6
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Physics 201 – Lecture 19
But what of Rolling Motion
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Now consider a cylinder rolling at a constant speed.
VCM
CM
CM
VCM
The cylinder is rotating about CM and its CM is moving at
constant speed (VCM). Thus its total kinetic energy is
given by :
2
K TOT = 12 I CMω 2 + 12 MVCM
Physics 201: Lecture 19, Pg 7
Physics 201: Lecture 19, Pg 8
Concept question
Motion
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Again consider a cylinder rolling at a constant speed.
Both with
|vt| = |vCM |
Rotation only
vt = ωR
CM
CM
CM
VCM
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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
Sliding only
VCM
2VCM
B.
C.
D.
E.
Physics 201: Lecture 19, Pg 9
Physics 201: Lecture 19, Pg 10
Question: The Loop-the-Loop … with rolling
Example : Rolling Motion
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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
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M
M
M
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
Disk has radius R
M
h
θ
Mgh = ½ Mv2 + ½ ICM ω2
M
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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
M
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and
M
v?
v =ωR
ball has mass m &
r <<R
U=mg2R
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
h?
v = 2(gh/3)½
R
Physics 201: Lecture 19, Pg 11
Physics 201: Lecture 19, Pg 12
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Physics 201 – Lecture 19
Example: Loop-the-Loop with rolling
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
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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
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Work energy (∆K = W)
Kf = KCM + KRot = mg (h-2R) (at top)
Kf = ½ mv2 + ½ 2/5 mr2 ω2 = mg (h-2R) & v=ωr
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
h?
R
= gR
Just a little bit more….
Physics 201: Lecture 19, Pg 13
Physics 201: Lecture 19, Pg 14
Example: Loop-the-Loop … Energy Conservation
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If rolling then ball has both rotational and CM motion!
E= U + KCM + KRot = constant = mgh (at top)
E= mg2R + ½ mv2 + ½ 2/5 mr2 ω2 = mgh & v=ωr
E= mg2R + ½ mgR + 1/5 m v2 = mgh
h = 5/2R+1/5R
Ub=mgh
ball has mass m &
r <<R
U=mg2R
v
Tangential
= gR
h?
R
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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 α?
1. ΣFx = m acm = mg sin θ - fs
m Racm = mgR sin θ - Rfs
mg
θ
-mR2 α= mgR sin θ + Iα -mR2 α – Iα = mgR sin θ
α = -mgR sin θ / (mR2 + I) = -2g sin θ / 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 α?
1. Στ = -R mg sin θ = I α
2. I = mR2 + Icm
α = -mgR sin θ / (mR2 + Icm)
α = 2g sin θ / 3R
θ
fs
Physics 201: Lecture 19, Pg 15
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N
ΣFy = 0 = N – mg cos θ
(not used)
2. Στ = -R fs = Icm α
3. α = - acm / R
h is just a little bit more….
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= gR
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N
θ
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. Στ = -rF = I α
N
2. I = mR2 + Icm = 3/2 mR2
F
α = -rF / (mR2 + Icm) = -2Fr / 3R
mg
fs
mg
fs
N
θ
fs
F
mg
1. Στ = 0 = I α
α= 0
Physics 201: Lecture 19, Pg 17
1. Στ = rF = I α
2. I = mR2 + Icm
N
α = 2Fr / 3R
fs
F
mg
Physics 201: Lecture 19, Pg 18
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Physics 201 – Lecture 19
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Modified Atwood’s Machine (More torques)
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
µs is the coefficient of static friction at the contact
point, what is the maximum force that can be applied
before slipping occurs?
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.
v What are the tensions in the strings ?
N
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Σ Fy = m1 ay= -m1g +T1
T1
mass 2
T1
m1g
Σ Fx = m2 ax = –T
Σ τ = Iα = R T1 – R T
m1 a = -m1g +T1
ay = ax= -αR = a
m2 a = –T
-I a / R = R T1 – R T
Physics 201: Lecture 19, Pg 20
Modified Atwood’s Machine (More torques)
Angular Momentum
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.
v What are the tensions in the strings?
N
2. m2 a = –T
T1
3. -I a / R = R T1 – R T
T1
m1g
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-I a /
m2g
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= m1 a +m1g + m2 a
is conserved if
T1= m1 a + m1g
r
∑ FExt
r
v
=
r
dp
=0
dt
r
τ = r × F = Iα
r r r
r
L ≡ r × p = Iω
r
∑ τ Ext
r
r v dL
=r×F =
=0
dt
Physics 201: Lecture 19, Pg 22
For Thursday
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Are the rotational equivalents?
angular momentum
Physics 201: Lecture 19, Pg 21
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r
is conserved if
a = – m1g / (m1 + ½ M + m2)
T= - m2 a
r
r
p ≡ mv
momentum
Solve for a
R2
We have shown that for a system of
particles,
T
T
1. m1 a = – m1g +T1
m2g
pulley
Physics 201: Lecture 19, Pg 19
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T
T
mass 1
N
1. ΣFx = m acm = F - fs
F
2. ΣFy = 0 = N – mg
mg
3. Στ = -R fs = Icm α
fs
- acm = - fs mR2 / Icm
4. fs ≤ µs N = µs mg (maximum)
5. acm = - α R
m acm = fs mR2 / Icm = F - fs
fs (1+ mR2 / Icm ) = µs mg (1+ mR2 / Icm ) = F
F = µs mg (1+ mR2 / ( ½ mR2 ) ) = 3 µs mg
Read all of chapter 10 (gyroscopes will not be tested)
HW9
Physics 201: Lecture 19, Pg 23
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