The linear acceleration along the inclined plane can be computed by
\(a=\frac{g sin\theta}{1+\frac{K^{2}}{R^{2}}}\)
For a thin uniform circular ring, axis passing through its center is I = MR².
\(\therefore K^{2}=R^{2} \Rightarrow \frac{K^{2}}{R^{2}}=1\)
And the angle of inclination, \(\theta=45 ^{0}\)
\(\Rightarrow (sin 45^{0}=\frac{1}{\sqrt{2}})\)
Hence, \(a=\frac{g+\frac{1}{\sqrt{2}}}{1+1}\)
\(a=\frac{g}{2\sqrt{2}}\)ms^-2.

As the disc is freely rotating, with the insect on it, the angular momentum of the system is conserved.
\(L=I\omega\)=constant

When the insect moves towards the center (from A to 0), the moment of inertia (I) increases. Thus, the angular velocity (co) increases. When it moves away from center (from 0 to B), the moment of inertia (I) decreases. Thus, the angular velocity decreases.

\(\omega_{1}=30 rpm=2\pi \times \frac{30}{60} rad s^{-1}=\pi \)rad s^-1
\(\omega_{2}=720 rpm=2\pi \times \frac{720}{60} rad s^{-1}=24\pi \)rad s^-1
Change in kinetic energy,
\(\Delta KE=\frac{1}{2}I (\omega^{2}_{2}-\omega^{2}_{1})\)
\(I=\frac{2\times \Delta KE}{(\omega^{2}_{2}-\omega^{2}_{1})}=\frac{2\times 1000}{(24\pi)^{2}-(\pi)^{2}}\)
\(I=\frac{2000}{2\pi \times 23 \pi}\) Remember: a² - b² = (a + b)(a - b)
I≈0.35 kg m² and π²=10

Let R be the radius of the ring and M be the total mass of the complete ring.
Let m be the mass of the section removed from the ring then, mass of the incomplete ring is M-m
Let us introduce a positive integer (n), such that, \(n\theta=360^{0} \), or \(n=\frac{360^{0}}{\theta}\)

mass of incomplete ring=M - m
\(m=\frac{M}{360}\times \theta\)
∴ Mass of complete ring = \(M-\frac{M}{360}\times \theta\)
Mass of incomplete ring = \(M-\frac{M}{n}=M(\frac{n-1}{n})\)
For example, (a) when \(\theta=60^{0}; n=\frac{360^{0}}{60^{0}}=6\)
∴ n-1 =5
Mass of incomplete ring = \(\frac{5}{6}M\)
(b)when \(\theta=30^{0}; n=\frac{360^{0}}{30^{0}}=12\)
n-1=11
Mass of incomplete ring = \(\frac{11}{12}M\)
The moment of inertia of the incomplete ring is, I=\(M \frac{(n-1)}{n}R^{2}.\)

The center of mass of an equilateral triangle lies at its geometrical center G. The positions of the mass m₁, m₂ and m₃ are at positions A, B and C as shown in the Figure.

From the given position of the masses, the coordinates of the masses mi and m₁ are easily marked as (0,0) and (1,0) respectively.
To find the position of ma the Pythagoras theorem is applied. 
As the ΔDBC is a right angle triangle,
BC² = CD² + DB²
CD² = BC² - DB²
\(CD^{2}=1^{2}-(\frac{1}{2})^{2}=1-(\frac{1}{4})=\frac{3}{4}\)
\(CD=\frac{\sqrt{3}}{2}\)
The position of mass m₃ is \((\frac{1}{2},\frac{\sqrt{3}}{2})\) or \((0.5,0.5 \sqrt{3})\)
X coordinate of center of mass
\(x_{CM}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}\)
\(x_{CM}=\frac{\sqrt{3}}{4}\)m
∴ The coordinates of center of mass G (x_CM,y_CM) is \((\frac{7}{12},\frac{\sqrt{3}}{4})\).