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Rotation of a Rigid BodyRotation of a Rigid Body
Readings: Chapter 131
Readings: Chapter 13
How can we characterize the acceleration during rotation?
t l ti l l ti dF
Newton’s second law:
- translational acceleration and
- angular acceleration
F ma=
2
Angular acceleration2v
1v1r
ϕ
1r
Center of rotation
Both points have the same angular velocityϕω Δ
=Both points have the same angular velocity t
ωΔ
1 1v rω= 2 2v rω=Linear acceleration:
11 1
va r
t tωΔ Δ
= =Δ Δ
22 2
va r
t tωΔ Δ
= =Δ Δ
3
t tΔ Δ t tΔ Δ
Both points have the same angular acceleration tωα Δ
=Δ
Rotation of Rigid Body:
Every point undergoes circular motion with the same angular velocity and the same angular acceleration
tωα Δ
=Δ
v rω=tΔ
4
The relation between angular velocity and angular acceleration is the same as the
l i b li l i d lirelation between linear velocity and linear acceleration
Δtωα Δ
=Δ
vat
Δ=Δ
5
The Center of Mass
For Rigid Body sometimes it is convenient to describe the rotation aboutFor Rigid Body sometimes it is convenient to describe the rotation about the special point– the center of mass of the body.
Definition: The coordinate of the center of mass:
1 1 2 2
1 2cm
m x m xx
m m+
=+
Definition: The coordinate of the center of mass:
Rigid body consisting of two particles: 1 2
1 2m m=If then
x x+1 2
2cmx x
x+
=
2 0 0.5 0.5 0.12.0 0.5cmx m⋅ + ⋅
= =+
6
The Center of Mass
Definition: The coordinate of the center of mass:
x mΔ∑ ∫i ii
cm
x m xdmx
M M
Δ= =∑ ∫
i ii
cm
y m ydmy
M M
Δ= =∑ ∫
cmyM M
7
The Center of Mass: Example
21 1
L L
L
Mx dx M xdxxdm L L L∫ ∫∫ 20 00
1 12 2 2
L
cm
xdm L L Lx xM M LM L L
= = = = = =∫ ∫∫
The center of mass of a disk is the center O of the disk
8O
Torque: Rotational Equivalent of Force
9
Torque
Th t ti f th b d i d t i d b th tThe rotation of the body is determined by the torque
sinFrτ φ=
10
Torque
φsinFrτ φ=
090φ =Torque is maximum if
Torque is 0 if 0 00 180orφ φ= =
0τ =
1 3τ τ>
1 4τ τ>
11
1 4τ τ>
Torque sinFrτ φ=
Torque is positive if the force is trying to rotate the body counterclockwise
Torque is negative if the force is trying to rotate the body clockwise
1 0τ >3 0τ < 1F
3F
i 2 0τ <4 0τ > 2F4Faxis 2 0τ <4
The net torque is the sum of the torques due to all applied forces:
24
1 2 3 4netτ τ τ τ τ= + + +
12
Torque: Example sinFrτ φ=Fi d th t tFind the net torque
1F3F 5F
3 030
FF
1m
2m
3m
4m 060
045
045
axis 2F4F
1 2 3 4 5 5F F F F F N= = = = =
1 5 3sin(30) 7.5N mτ = ⋅ = ⋅1 2 3 4 5
2 5 4sin(60) 17.3N mτ = − ⋅ = − ⋅3 5 1sin(45) 3.5N mτ = − ⋅ = − ⋅
4 5 2sin(45) 7.1N mτ = ⋅ = ⋅2 ( ) 4 ( )
5 5 0 0N mτ = ⋅ = ⋅
131 2 3 4 4 7.5 17.3 3.5 7.1 6.2net N mτ τ τ τ τ τ= + + + + = − − + = − ⋅
Torque: Relation between the torque and angular acceleration:
v Δ Δr
va r rt t
ω αΔ Δ= = =Δ Δ
1F
ϕ
F ma mrα= =
2Fr mrτ α= =
2m r Iτ α α= =∑net i ii
m r Iτ α α= =∑2I m r= ∑ moment of inertiai i
iI m r= ∑ - moment of inertia
14
Moment of Inertia
21I MLThin rod about center 2
12I ML=Thin rod, about center
Thin rod, about end 213
I ML=
1Cylinder (or disk), about center 21
2I MR=
Cylindrical loop about center2I MR=
15
Cylindrical loop, about center I MR=
Moment of Inertia: Parallel-axis Theorem
If you know the moment of Inertia about the center of mass (point O)
then the Moment of Inertia about point (axis) P will be
2d 2P OI I Md= +
OP Od
2 2( ) ( )I Δ Δ∑ ∑2 2
2 2 2
( ) ( )
( ) ( ) 2 ( )
P i P i i O O P ii i
i O i i O i O i
I x x m x x x x m
x x d m r r d r r d m
= − Δ = − + − Δ =
⎡ ⎤= − + Δ = − + − + Δ =⎣ ⎦
∑ ∑
∑ ∑2 2 2
( ) ( ) 2 ( )
( ) 2 ( )
i O i i O i O ii i
i O i i O i Oi i
x x d m r r d r r d m
r r m d r r m Md I Md
⎡ ⎤+ Δ + + Δ⎣ ⎦
+ − Δ + − Δ + = +
∑ ∑
∑ ∑16
i i
Parallel-axis Theorem: Example 2P OI I Md= +
1Thin rod, about center of mass
2112
I ML=
22 2 2 21 1 1 1
12 2 12 4 3LI ML M ML ML ML⎛ ⎞= + = + =⎜ ⎟
⎝ ⎠12 2 12 4 3⎝ ⎠
17
net Iτ α=Equilibrium: 0α =
0netτ =
1 2m kg=2 10m kg=
2d m= ?x =Massless rod1n
2n
Two forces (which can results in rotation) acting on the rod
1 1 1n w m g= = 1 1 1n d m gdτ = =1 1 1g
2 2 2n w m g= =1 1 1g
2 2 2n x m gxτ = − = −
Equilibrium: 0τ τ τ= + =18
Equilibrium:1 2 0netτ τ τ= + =
1 2 0m gd m gx− =1
2
0.4m
x d mm
= =
Rotational Energy:
2 2 21 1 2 2 3 3
1 1 1 ...2 2 2rotK m v m v m v= + + + =
2 2 2 2 2 21 1 2 2 3 3
1 1 1 ...2 2 2
m r m r m rω ω ω= + + + =
2 2 2 2 21 1 2 2 3 3
1 1( ...)2 2
m r m r m r Iω ω= + + + =
Conservation of energy (no friction):
2 21 1Mgy Mv I constω+ + =
19
2 2Mgy Mv I constω+ +
Kinetic energy of rolling motion 212rotK Iω=
212rot PK I ω= 2
P cmI I MR= +
1 1 1 1cmv Rω=
2 2 2 2 22
1 1 1 12 2 2 2rot cm cm cm cmK I MR I v Mv
Rω ω= + = +
Cylinder: 212cmI MR= 2 2 21 1 1 3
2 2 2 4rot cm cm cmK Mv Mv Mv= + =
Cylindrical loop: 2cmI MR=
2 2 21 12 2rot cm cm cmK Mv Mv Mv= + =
20Solid sphere: 22
5cmI MR=2 2 21 2 1 7
2 5 2 10rot cm cm cmK Mv Mv Mv= + =