Rotational Motion
Examples
Example-Calculating Moments of Inertia
Moment of Inertia Of a Rigid Rod Rotating Around CM
y
dx
CM
x
0
L
x
dm =
I=
X
mi Ri2
i
=
XM
i
L
dx
L
−x
2
M
L dx
2
3 x=L
M L
=
−x 3L
2
0
x=0
3
3
M L
M
L
M 2
−
=
L
=
−
3L 2
3L
2
12
M
=
L
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Z
L
L
−x
2
2
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Rotational Motion
Examples
Parallel Axis Theorem
Proof
Let ICM be the moment of inertia around an axes going through
the CM.
Choose z axis to be along the this axes.
Choose a second axes that goes through the point (x0 , y0 ).
The distance of point with coordinates (xi , yi , zi ) from the second
axis is R 2 = (xi − x0 )2 + (yi − y0 )2
The moment of inertial with respect to the second axis is
i
X h
(125)
I=
mi (xI − x0 )2 + (yi − y0 )2
i
=
X
mi (xi2 + yi2 ) +
i
X
mi (x02 + y02 ) − 2
i
X
mi (xi x0 + yi y0 ) (126)
i
= ICM + Md 2
where d 2 = x02 + y02 and
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(127)
P
i
mi xi =
P
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i
mi yi = 0
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Rotational Motion
Examples
Parallel Axis Theorem
Example
y
dx
0
CM
x
L
x
dm =
M
L dx, ICM
= 21 ML2
Moment of inertia of a thin rod around one end:
I = ICM
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2
L
1
1
1
+M
=
ML2 + ML2 = ML2
2
12
4
3
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Rotational Motion
Examples
Example: Thin Rod Rotating Around One End
dx
x
L
The net torque on the rod around the fixed axes:
!
X
X
~ri × (mi ~g ) =
~
mi~ri × ~g = (M~rCM ) × ~g = ~rCM × w
~τ =
i
i
(125)
Hence the center of gravity for an object in uniform gravitational
field is its CM.
τ = MgL
2
α=
τ
I
=
MgL
2
1
ML2
3
=
3g
2L
CM = α L = 3 g < g
If the rod is initially at rest: aCM = atan
2
4
ext
aCM = FM . Which other force is acting on the rod?
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Rotational Motion
Examples
Perpendicular Axis Theorem
Proof
Consider a very thin object in the xy plane.
For any point in the object zi ' 0
P
P
Ix = i mi (yi2 + zi2 ) ' i mi yi2
P
P
Similarly Iy = i mi (xi2 + zi2 ) ' i mi xi2
P
P
P
Then Iz = i mi (xi2 + yi2 ) = i mi xi2 + i mi yi2 = Ix + Iy
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Rotational Motion
Examples
Perpendicular Axis Theorem
Example
The moment of inertia of a loop of mass M and radius R for
rotations around an axis that
to its plane and
Pis perpendicular
2
2
going through its CM: Iz = i mi R = MR
Its moment of inertia around any axis that goes through its CM
and is in its plane:
If x and y axes are the two axes in the plane of the loop, due to
symmetry Ix = Iy
By perpendicular axis theorem: Iz = Ix + Iy = 2Ix
Hence Ix = 12 MR 2 .
Moment of inertia for rotation around an axis that goes through the
edge and is in the plane of the loop:
I 0 = ICM + Md 2 = 12 MR 2 + MR 2 = 32 MR 2
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Rotational Motion
Examples
Heavy Pulley
Since the pulley has a mass, the
tensions at each end of the pulley will
be different.
m1
m2
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~1
T
~2
T
m1
m2
~1
w
~1
~ 2 −T
w
z
~2
−T
Let x axis be out of the screen
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Rotational Motion
Examples
Heavy Pulley
~1
T
~2
T
m1
m2
~1
w
~1
~ 2 −T
w
z
~2
−T
The net forces acting on mass m1
and m2 are:
m1
m2
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~ 1T = (T1 − m1 g)ẑ
F
~ 2T = (T2 − m2 g)ẑ
F
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(126)
(127)
172 / 175
Rotational Motion
Examples
Heavy Pulley
~1
T
~2
T
m1
m2
~1
w
~1
~ 2 −T
w
z
~2
−T
The torque acting on the pulley is
~τ = R(T1 − T2 )x̂
m1
m2
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Rotational Motion
Examples
Heavy Pulley
Let ~a1 = ai ẑ and α
~ = α~x . Then
T1 − m1 g = m1 a1
(126)
T2 − m2 g = m2 a2
(127)
R(T1 − T2 ) = Iα
(128)
Unkowns: T1 , T2 , a1 , a2 , and α: 5
unknowns
The remaining two eqns are:
m1
m2
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a1 = −a2
(129)
αR = −a1
(130)
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Rotational Motion
Examples
Heavy Pulley
The solutions of these equations are:
(m2 − m1 )R 2
g
I + (m1 + m2 )R 2
(126)
(m1 − m2 )R
α=
g
(127)
I + (m1 + m2 )R 2
m1 g(I + 2m2 R 2 )
T1 =
(128)
I + (m1 + m2 )R 2
m2 g(I + 2m1 R 2 )
T2 =
(129)
I + (m1 + m2 )R 2
a1 = −a2 =
m1
m2
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