Answer :
A solid body rotating about an axis has rotational kinetic energy that is proportional to both its moment of inertia and its angular speed of rotation about the axis.
The formula is: K=(1/2)I[tex]w^{2}[/tex]
The most straightforward form of rotational motion is a circle rotating around an axis. The rotational kinematics equations then resemble the linear kinematics equation:
θ [tex]= w_{i} t + \frac{1}{2} \alpha t^{2}[/tex]
[tex]w_{f} = w_{i} + \alpha t[/tex]
where α is the angular acceleration,
ω is the angular speed,
θ is the angle traveled and
t is the time.
Given data:
Mass of the disk, [tex]m_{d}[/tex]=3.3 kg
The radius of the disk, R=0.26m
Mass of the rod, [tex]m_{r}[/tex]=1.2 kg
Length of the rod, L=0.72 m
a) Think about the equation used to calculate the rod's moment of inertia.
[tex]I_{rod}[/tex] = [tex]\frac{1}{3} m_{r} L^{2}[/tex]
Calculate the moment of inertia of each rod about the axis of rotation.
[tex]I_{rod}[/tex] = [tex]13 * (1.2) * (0.72)^{2}[/tex]
[tex]I_{rod}[/tex] = [tex]0.207 kg m^{2}[/tex]
b) A disk's moment of inertia around an axis that passes through its center and is perpendicular to the disk is:
[tex]I_{disk}[/tex] = [tex]\frac{1}{2} m_{d} R^{2}[/tex]
We have,
[tex]I_{disk}[/tex] = [tex]\frac{1}{2} (3.3) (0.26)^{2}[/tex]
[tex]I_{disk}[/tex] = [tex]0.112 kg m^{2}[/tex]
c) The entire ceiling fan's moment of inertia.
[tex]I_{fan} = 5 I_{rod} + I_{disk}[/tex]
[tex]I_{fan} = 5 (0.207) + 0.112\\I_{fan} = 1.147 kg m^{2}[/tex]
d) Here we are given the angle covered and the time for it.
We can find the angular acceleration from the formula:
θ = [tex]\frac{1}{2} \alpha t^{2}[/tex]
We have:
θ = 13 rev
θ = 13(2π) rad
θ = 26π rad ≈ 81.68 rad
Therefore we have:
[tex]\alpha = 2[/tex]θ / [tex]t^{2}[/tex] = [tex]\frac{2(26\pi )}{3.1^{2} }[/tex] = 17 rad / [tex]s^{2}[/tex]
e) To find the final angular velocity of the fan we have:
[tex]w_{f} = w_{i} + \alpha t[/tex]
Substitute [tex]w_{i}[/tex] = 0, [tex]\alpha[/tex] = 17 rad / [tex]s^{2}[/tex], and t=3.1s in the above expression:
[tex]w_{f} = 0 + 17 (3.1)[/tex]
[tex]w_{f} = 52.7 rad / s[/tex]
f) Consider the formula used to calculate the final rotational kinetic energy of the fan.
K = [tex]\frac{1}{2} I_{fan} w_{f}^{2}[/tex]
Substitute the values.
K = [tex]\frac{1}{2} (1.147) (52.7)^{2}[/tex]
K = 1593 J
g) The spinning energy of the fan decreases by half when it is adjusted to a lower speed.
This can give us angular speed.
[tex]K_{f} = \frac{K}{2}[/tex]
[tex]K_{f} = \frac{1593}{2}[/tex]
[tex]K_{f} = 796.5 J[/tex]
Calculate the final angular speed of the fan.
[tex]K_{f} = \frac{1}{2} I_{fan} w_{f}^{2}[/tex]
[tex]w_{f} = \sqrt{\frac{2K_{f}}{I_{fan} } }[/tex]
Substitute the values.
[tex]w_{f} = \sqrt{\frac{2(796.5)}{1.147} }[/tex]
[tex]w_{f} = 37.3 rad / s[/tex]
h) The angular acceleration is calculated when the fan slows down by
[tex]w_{f} = w_{i} + \alpha t[/tex]
[tex]\alpha[/tex] = [tex]\frac{w_{f} - w_{i} }{t}[/tex] = [tex]\frac{37.3 - 52.7}{3.1}[/tex] = − 4.97 rad / [tex]s^{2}[/tex]
Therefore, the magnitude of the angular acceleration is:
|[tex]\alpha[/tex]| = 4.97 rad / [tex]s^{2}[/tex]
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