The mass of the plate is calculated in three sections:
\(m1 = σ × 5 = 10 kg,\)
\(m2 = σ × 1 = 2 kg,\)
\(m3 = σ × 6 = 12 kg.\)
Using the coordinates of each center of mass, we calculate the combined center of mass:
\[m_1x_1 + m_2x_2 = m_3x_3.\]
\[10 \cdot 1.5 + 2 \cdot (1.5) = 12 \cdot x_1 \implies x_1 = 1.5 \, \text{cm}.\]
Similarly:
\[m_1y_1 + m_2y_2 = m_3y_3.\]
\[10 \cdot 1 + 2 \cdot (1.5) = 12 \cdot y_1 \implies y_1 = 0.9 \, \text{cm}.\]
The ratio of $x_1$ to $y_1$ is:
\[\frac{x_1}{y_1} = \frac{1.5}{0.9} = \frac{15}{9}.\]
Thus:
\[n = 15.\]
Final Answer: $n = 15$.
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: