1. Problem

A woman doing yoga weighs 65 kg, and her legs are 85 cm long. The combined mass of both her legs is 11.7 kg, one of which is lifted 90 cm above the ground (as in the image below). Her torso contributes to 50% of her mass and is 60 cm long. Her arms are 4.5 kg each, and they are 90 cm from the fingertips to the shoulder. What is her center of gravity in the warrior 3 pose from the origin?

2. Draw

3. Knowns and Unknowns

Knowns:

• m= 65 kg
• m_L = 11.7 kg
• m_A = 4.5 kg
• m_T = 50% of 65 kg

Unknowns:

• [latex]\overrightarrow{r}[/latex], vector from origin to center of gravity

Assumptions

g= 9.81 m/s²

4.  Approach

Calculate mass and distance from origin, keeping in mind that the mass distribution is different for legs, torso, and arms, but assuming that the mass is evenly distributed within each of these parts. Use the center of gravity equation.

5. Analysis

Point of mass for leg in air – 11.7 kg at 90 cm up and 42.5 cm left from the origin.

Point mass for leg on ground – 11.7 kg at 42.5 cm up and 90 cm left.

Point mass for torso – 32.5 kg at 90 cm up and 115 cm left.

Point mass for both arms – 9 kg at 90 cm up and 190 cm left.

$$\bar{x_m}=\frac{(m_L\cdot0.425m)+(m_L\cdot0.90m)+(m_T\cdot1.15)+(2\cdot m_A\cdot1.90m)}{m}$$

$$\bar{x_m} = \frac{(11.7 kg \cdot 0.425 m ) + (11.7 kg \cdot 0.90 m )+(32.5 kg \cdot 1.15 m )+(2 \cdot 4.5 kg \cdot 1.90 m) }{65 kg}$$

 $$\bar{x_m }=1.08 m $$

$$\bar{y_m} = \frac{(m_L\cdot0. 9m)+(m_L\cdot0.425m)+(m_T\cdot0.9)+(2\cdot m_A\cdot0.90m)}{m}$$

$$\bar{y_m} = \frac{(11.7 kg \cdot 0.90 m ) + (11.7 kg \cdot 0.425 m )+(32.5 kg \cdot 0.90 m )+(2 \cdot4.5 kg \cdot 0.90 m) }{65 kg}$$

$$\bar {y_m} = 0.81 m  $$

in vector form, [latex]\overrightarrow{r}=\begin{bmatrix}1.08m\\0.81m \end{bmatrix}[/latex]

6. Review

The answer makes sense because it is close to where the torso is, and it has the most mass. It is a little lower than the torso, which makes sense as this pose would lower the center of gravity.

1. Problem

You are the engineer for a tech company, and your role is to design a new extended mount for their new phones. Currently, you are trying to figure out what distance the mount can be from the hand “d”.

You can only use the equation for the midpoint of the rod Izz = (1/12)ml2at your disposal, and the pole weighs 15 N. If a hand is held up the pole at the bottom, it can have a mass moment of inertia of 5 kgm2 before breaking (Iz’z’). Figure out what size the extended mount can be (d).

2. Draw

3. Knowns and Unknowns

Knowns:

• W_p =15 N
• g = 9.81 m/s²
• Izz = (1/12)ml²
• Max Iz’z’ =5 kgm²

Unknowns:

• L, length of stick

4. Approach 

Newton’s Second Law, Parallel Axis Theorem, Understanding of Mass Moment of Inertia

5. Analysis 

$$ m = \frac{w_p}{g}$$

$$m = \frac{15N} {9.81 \: m/s^2}$$

$$m = 1.529\: kg $$

$$I_g =\frac{1}{12}\:md^2 $$

$$Iz’z’ = I_g + ml^2$$

$$Iz’z’ = \frac{1}{12}\:md^2+m\times \left(\frac{1}{2}d\right) ^2$$

thus,

$$5kgm^2 = \left (\frac{1}{12}\times (1.529kg)\times d^2 \right)+\left(1.529kg\: \times \frac{1}{4}d^2\right)$$

[latex]d =3.13 m[/latex]

In conclusion, the maximum size the pole can be is 3.13 m without the device breaking.

6. Review 

Applying the derived value of d in

$$Iz’z’ = \frac{1}{3}md^2$$

$$Iz’z’ =\frac{1}{3}\times (1.529kg)\times(3.13m)^2$$

$$Iz’z’ = 4.99999kgm^2 =5.0kgm^2$$

According to the equation, the max distance of 3.13 m does make sense

1. Problem

A thin uniform metal disc with a diameter of 18cm and an initial mass of 50.87kg has a circular hole cut out of it, turning it into a crescent shape, shown in the figure below. The radius of the hole is 6cm. Assuming the mass of the disc is evenly distributed, calculate the mass of the crescent formed and its center of mass.

3. Knowns and Unknowns

Knowns:

• Mass of Initial disc: m₁ = 50.87kg
• Radius of larger disc: r₁ = 18cm/2 = 9cm
• Radius of cut: r₂ = 6cm

Unknowns:

• Mass of crescent: m₂
• Center of mass of crescent: CM_c

4. Approach
Use the mass and radius of the larger disc to find the density of the material. Calculate the area of the crescent
Use the area and density to find the mass of the crescent.
Find the center of mass of the smaller circle.
Use the center of mass of each circle and the mass of each circle to determine the center of mass of the crescent.

5. Analysis
Density of materials:
$$ \rho = \frac{m_1}{2\pi \cdot r_1 ^2} \\ \rho = \frac{50.87kg}{2\pi \cdot 9^2} \\ \rho = 0.1kg/cm^2$$

Area of crescent A_c :
$$ A_c = 2\pi \cdot (r_1 ^2 – r_2 ^2) cm^2 \\ A_c = 2\pi \cdot (9^2 – 6^2) cm^2 \\ A_c = 90\pi cm^2 $$

Mass of crescent:
$$m_2 = \rho \cdot A_c \\ m_2 = 0.1kg/cm^2 \cdot 90\pi cm^2 \\ m_2 = 9\pi kg \approx 28.27kg $$

Center of mass of the cut circle:

Since the edge of the cut circle lines incident to the larger circle, the radius of the cut can be subtracted from the radius of the full disc to find the center of mass of the cut, CM_cut

$$CM_{cut x} = R_1 – R_2 \\ CM_{cut x} = 9cm – 6cm \\ CM_{cut x} = 3cm$$

CM_{cut x} is 3cm in the negative x direction.

CM_{cut y} is equal to 0 as the center of mass of both the disc and the cut is zero in the y-direction.

The center of mass of the full disc is at [0cm, 0cm] as it’s centred at the origin.

Center of mass of the crescent:

The center of mass of the full circle (CM_full) is [0cm, 0cm] and its mass m₁ = 50.87kg.
The center of mass of the cut (CM_cut) is [-3cm, 0cm], and its mass (m_cut) is the mass of the crescent minus the mass of the full circle:  28.27kg – 50.87kg = -22.6kg.

The center of mass of the crescent CM_c is:

$$ CM_{cx}  = \frac{(CM_{full x} \cdot m_1) + (CM_{cut x} \cdot m_{cut})}{m_2}$$

$$ CM_{cx}  = \frac{(0cm \cdot 50.87kg) + (-3cm \cdot -22.6kg)}{28.27kg}$$

$$ CM_{cx} = 2.4cm$$

Both the center of mass of the full circle and the cut in the y-direction are 0cm, giving the center of mass of the crescent in the y-direction also 0cm.

The complete center of mass of the crescent is therefore:
CM_c = [2.4cm, 0cm]

6. Review:

The center of mass of the crescent leans to the side with more mass, as one would expect. The mass and center of mass found are both in the expected units with a reasonable magnitude.

1. Problem

2. Draw

3. Knowns and Unknowns 

Knowns:

w = 2 lbs

Inner radius r₁ = 0.9 in

Outer radius r₂ = 1 in

h=12 in

Unknowns:

Mass moment of inertia

4. Approach 

Finding the mass of the roll, identifying the axis of rotation, and applying the corresponding mass moment of inertia of the cylinder.

[latex]I_{x}= \frac{1}{2} m(r_{1}^2 + r_{2}^2)[/latex]

[latex]I_{y}= I_{z}= \frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][/latex]

5. Analysis 

weight of the roll = 2 lb

mass of  the roll, [latex]m = \frac{2 lb}{32.2 ft/s^2} = 0.062 lb[/latex]

Since we used constant of gravity with a unit of [latex]ft/ s^2[/latex] it is important to convert other values from inches to ft.

[latex]0.9 in = \frac{0.9}{12}=0.075 ft[/latex]

[latex]1 in =\frac{1}{12} = 0.083 ft[/latex]

[latex]12 in = 1 ft[/latex]

now, [latex]I_{x}= \frac{1}{2} m(r_{1}^2 + r_{2}^2)[/latex]

[latex]I_{x} = \frac{1}{2} 0.062(0.075^2 + 0.083^2) = 0.00039 \quad lb ft^2[/latex]

[latex]I_{y}= I_{z}= \frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][/latex]

[latex]I_{y}= I_{z}= \frac{1}{12} 0.062[3(0.075^2 + 0.083^2)+1^2] =0.0054 \quad lb ft^2[/latex]

6. Review

Although the weight of an empty paper roll is exaggerated for easier calculation, the answers make sense. The value of I_y is higher than I_x, which means more force will be needed for rotation in that axis. From this, it is important to note that the roll will rotate along the x-axis. Also note that weight is converted to mass, and units must be the same.