Consider the rods plus discs system shown in the Figure. The rod has a length $$ L=1.40 \mathrm{~m} $$, and mass $$ M=0.900 \mathrm{lg} $$. The large disc has a radius of $$ R_{L}=0.270 \mathrm{~m} $$ and mass $$ m_{L}=1.90 \mathrm{~kg} $$. The second disk has a radius of $$ R_{s}=0.0900 \mathrm{~m} $$ and mass $$ m_{s}=3.50 \mathrm{~kg} $$. It is glued to the rod at position $$ x=1.12 \mathrm{~m} $$. The system behaves as a compound pendutum that oscillates about pivot point $$ P $$. The pivot is at $$ P $$ where the tip of rod meets with one end of the disk. Assume that the discs and rod have uniform mass distributions. (a) Find how far below $$ P $$ is the center of mass of the rods and disks system. $$ y_{\mathrm{cm}}=1.12 \mathrm{~m} $$ (b) Find the moment of inertia of the rod, as it freely rotates about the pivot $$ P $$. $$ I_{\mathrm{Rp}}=0.588 \mathrm{~kg}^{\wedge} 2 \quad \mathrm{x} $$ (c) Find the moment las inertia of the large disk, about its center of mass. $$ I_{\mathrm{L}_{\mathrm{m}}}=0.0692 \mathrm{k} \quad x $$ Find the moment of inertia of the small disk, about its center of mass. $$ I_{\mathrm{k} \mathrm{~m}}=3,7932 \mathrm{~kg} \quad \mathrm{x} $$ (d) Find the moment of inertia of the large disk, as it freely rotates about $$ P $$. $$ I_{\mathrm{Lp}}=3.7932 \mathrm{~kg} \quad \mathrm{x} $$ Find the moment of inertia of the small disk, as it freely rotates about $$ P $$. $$ I_{\mathrm{sp}}=4.4046 \mathrm{~kg} \quad \mathrm{x} $$ (e) Find the moment of inertia of the whole system as it freely rotates about the pivot point $$ P $$. $$ I=879 \mathrm{~kg} \quad x $$

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Let's solve the problem step by step, extracting the known conditions and calculating the required values.

### Known Conditions:
- Length of the rod, $$ L = 1.40 \, \text{m} $$
- Mass of the rod, $$ M = 0.900 \, \text{kg} $$
- Radius of the large disk, $$ R_L = 0.270 \, \text{m} $$
- Mass of the large disk, $$ m_L = 1.90 \, \text{kg} $$
- Radius of the small disk, $$ R_s = 0.0900 \, \text{m} $$
- Mass of the small disk, $$ m_s = 3.50 \, \text{kg} $$
- Position of the small disk on the rod, $$ x = 1.12 \, \text{m} $$

### (a) Center of Mass of the System
To find the center of mass $$ y_{cm} $$ of the system, we can use the formula for the center of mass of a system of particles:

$$
y_{cm} = \frac{\sum m_i y_i}{\sum m_i}
$$

Where $$ m_i $$ are the masses and $$ y_i $$ are their respective distances from the pivot point $$ P $$.

1. **Rod's center of mass**: The center of mass of the rod is at $$ \frac{L}{2} = \frac{1.40}{2} = 0.70 \, \text{m} $$.
2. **Large disk's center of mass**: The large disk is at the pivot point, so its distance is $$ 0 \, \text{m} $$.
3. **Small disk's center of mass**: The small disk is at $$ x = 1.12 \, \text{m} $$.

Now, we can calculate:

$$
y_{cm} = \frac{M \cdot 0.70 + m_L \cdot 0 + m_s \cdot 1.12}{M + m_L + m_s}
$$

Substituting the values:

$$
y_{cm} = \frac{0.900 \cdot 0.70 + 1.90 \cdot 0 + 3.50 \cdot 1.12}{0.900 + 1.90 + 3.50}
$$

Calculating the numerator and denominator:

$$
y_{cm} = \frac{0.630 + 0 + 3.92}{6.30} = \frac{4.55}{6.30} \approx 0.722 \, \text{m}
$$

### (b) Moment of Inertia of the Rod about Pivot $$ P $$
The moment of inertia $$ I_{Rp} $$ of a rod about one end is given by:

$$
I_{Rp} = \frac{1}{3} M L^2
$$

Substituting the values:

$$
I_{Rp} = \frac{1}{3} \cdot 0.900 \cdot (1.40)^2
$$

Calculating:

$$
I_{Rp} = \frac{1}{3} \cdot 0.900 \cdot 1.96 \approx 0.588 \, \text{kg} \cdot \text{m}^2
$$

### (c) Moment of Inertia of the Large Disk about its Center of Mass
The moment of inertia $$ I_{L_{cm}} $$ of a disk about its center is given by:

$$
I_{L_{cm}} = \frac{1}{2} m_L R_L^2
$$

Substituting the values:

$$
I_{L_{cm}} = \frac{1}{2} \cdot 1.90 \cdot (0.270)^2
$$

Calculating:

$$
I_{L_{cm}} = \frac{1}{2} \cdot 1.90 \cdot 0.0729 \approx 0.0692 \, \text{kg} \cdot \text{m}^2
$$

### Moment of Inertia of the Small Disk about its Center of Mass
The moment of inertia $$ I_{s_{cm}} $$ of the small disk is:

$$
I_{s_{cm}} = \frac{1}{2} m_s R_s^2
$$

Substituting the values:

$$
I_{s_{cm}} = \frac{1}{2} \cdot 3.50 \cdot (0.0900)^2
$$

Calculating:

$$
I_{s_{cm}} = \frac{1}{2} \cdot 3.50 \cdot 0.0081 \approx 0.014175 \, \text{kg} \cdot \text{m}^2
$$

### (d) Moment of Inertia of the Large Disk about Pivot $$ P $$
Using the parallel axis theorem:

$$
I_{Lp} = I_{L_{cm}} + m_L d^2
$$

Where $$ d $$ is the distance from the pivot to the center of the large disk, which is $$ R_L = 0.270 \, \text{m} $$.

Calculating:

$$
I_{Lp} = 0.0692 + 1.90 \cdot (0.270)^2
$$

Calculating:

$$
I_{Lp} = 0.0692 + 1.90 \cdot 0.0729 \approx 0.0692 + 0.13851 \approx 0.20771 \, \text{kg} \cdot \text{m}^2
$$

### Moment of Inertia of the Small Disk about Pivot $$ P $$
Using the parallel axis theorem:

$$
I_{sp} = I_{s_{cm}} + m_s d^2
$$

Where $$ d = 1.12 \, \text{m} $$.

Calculating:

$$
I_{sp} = 0.014175 + 3.50 \cdot (1.12)^2
$$

Calculating:

$$
I_{sp} = 0.014175 + 3.50 \cdot 1.2544 \approx 0.014175 + 4.3914 \approx 4.405575 \, \text{kg} \cdot \text{m}^2
$$

### (e) Moment of Inertia of the Whole System about Pivot $$ P $$
The total moment of inertia $$ I $$ is the sum of the individual moments of inertia:

$$
I = I_{Rp} + I_{Lp} + I_{sp}
$$

Calculating:

\[
I = 0.588 + 0.20771 + 4.405575 \approx 5.201285 \, \
The center of mass of the system is 0.722 meters below pivot point $$ P $$. The moment of inertia of the rod about pivot $$ P $$ is 0.588 kg·m². The moment of inertia of the large disk about its center of mass is 0.0692 kg·m². The moment of inertia of the small disk about its center of mass is approximately 0.0142 kg·m². The moment of inertia of the large disk about pivot $$ P $$ is approximately 0.2077 kg·m². The moment of inertia of the small disk about pivot $$ P $$ is approximately 4.4056 kg·m². The total moment of inertia of the system about pivot $$ P $$ is approximately 5.2013 kg·m².

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