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Rigid Body Rotation and Angular Momentum.
Please refer to figure 9.29 p.358.
From the definition:
I =
r ⊥^2 dm
we observe that if an object is more concentrated around the line has less moment of inertia.(Of course this conclusion is not right if they have different masses)
a) The cylinder’s mass (object A in the figure) is the most concen-
trated near its axis ⇒ Object A has the smallest moment of inertia.
b) The cylinder with a hole has greater I than a cylinder (of the
same mass) but slightly less than a hoop. With the assumption that it’s a thin cylinder we assume it’s very close to I of a hoop:
Ithin cylinder ≈ M R^2.
The cube has moment of inertia Icube = 121 M (a^2 +b^2 ) = 23 M R^2 which is less than the thin cylinder. So, the thin cylinder (object B in the figure) has the largest moment of inertia.
c) The sphere is more concentrated near its axis than even the
cylinder so it will replace solid cylinder as the smallest moment of inertia.
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The key to this problem is to set up the energy conservation equation having in mind that no non-conservative force is present here:
K 1 + U 1 = E 1 = E 2 = K 2 + U 2 (1)
U of the disk will not change due to rotation around around its cen- ter. There are two parts contributing to the kinetic energy: move- ment of center of mass of the mass m, and of rotation of the disk:
K =
mv^2 +
Idisk ω^2 (2)
Idisk =
mR^2
Mass m is attached to the disk so it rotates with the same ω as the disk: v = Rω
Hence,
K 2 =
mR^2 ω^2
We will measure the gravitation potential energy with respect to the height of the horizontal line passing through the center of the disk.
E 1 = 0 + 0 + 0 = 0
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E 2 =
mR^2 ω^2 − mgR
E 1 = E 2 ⇒
mR^2 ω^2 − mgR = 0
ω =
4 g 3 R
2
Plugging in the numbers given in the problem, you’ll get V = 2. 81 m/s.
b) Please refer to figure 1 for the diagram of forces.
Using the rotational analogue of Newton’s second law: ∑ τz = Iαz
we get
T 2 R − T 1 R = IP α = IP ω˙ = IP
v ˙ R
= IP
a R
T 2 − T 1 =
IP
R^2
a (4)
Writing Newton’s law
Fy = may (Let’s omit the subscript y af- terwards)
aR = −aL = a (5)
T 1 − mR g = mR a =⇒ T 1 = mR (a + g) (6)
T 2 − mL g = −mL a =⇒ T 2 = mL (g − a) (7)
Combining (4) with (6) and (7):
mL (g − a) − mR (g + a) =
IP
R^2
a
a = g
mL − mR mL + mR + I RP 2
NOTE: The pulley does not contribute to the imbalance, but only increases the inertia.
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c) In the case IP = 0 we get the familiar result of :
a 0 = g
mL − mR mL + mR
d) Let’s denote
IP
R^2
mP 2 The result of part b shows that, the pulley does not contribute to the imbalance, but only increases the inertia. This suggests the 1st configuration shown in the figure 2. There are of course other possibilities:
By writing a as:
a = g
(mL + mP /4) − (mR + mP /4) (mL + mP /4) + (mR + mP /4)
Which is the familiar result of two masses for ideal pulley problem with masses: m 1 = mL + m 4 P , m 2 = mR + m 4 P. You can set up the 2nd configuration shown in the figure 3.
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Figure 1: 9.80 Force Diagram
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Figure 2: 9.80 part d #
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