The moment of inertia of a solid disc about its central axis is given by I = (1/2)MR².

The moment of inertia I of a disc about its axis is (1/2)MR². Therefore, angular momentum L = Iω = (1/2)MR²ω.

The disk has a smaller moment of inertia than the ring, resulting in greater acceleration down the incline.

Using the formula α = (ω - ω₀)/t, we have α = (0 - 10)/5 = -2 rad/s², so the deceleration is 2 rad/s².

Using the formula α = (ω - ω₀)/t, we have α = (0 - 30 rad/s) / 10 s = -3 rad/s².

Angular deceleration α = (final angular velocity - initial angular velocity) / time = (0 - 10) / 5 = -2 rad/s².

Rotational kinetic energy K.E. = (1/2)Iω² = (1/2)(2)(10)² = 100 J.

New angular momentum L' = I'ω' = (2I)(ω/2) = L.

The rotational kinetic energy is given by KE = 0.5 I ω² = 0.5 * 3 * (4)² = 24 J.

The rotational kinetic energy is given by KE = (1/2)Iω² = (1/2)(5 kg·m²)(4 rad/s)² = 40 J.

For a satellite in a stable orbit, the gravitational force provides the necessary centripetal force, hence they are equal.

The solid cylinder has a lower moment of inertia, thus it accelerates faster and reaches the ground first.

The solid cylinder has a smaller moment of inertia compared to the hollow cylinder, thus it accelerates faster and reaches the bottom first.

The solid sphere has a smaller moment of inertia compared to the hollow sphere, allowing it to accelerate faster down the incline.

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. The total kinetic energy is the sum of translational and rotational kinetic energy. Thus, v = √(5gh/7).

Using τ = Iα, we have α = τ/I = 15/3 = 5 rad/s².

According to Newton's second law for rotation, τ = Iα. If τ is doubled, α also doubles.

The torque τ = Mg(L/2) and moment of inertia I = (1/3)ML². Using τ = Iα, we find α = 3g/2L.

Using conservation of energy, potential energy at the top converts to rotational kinetic energy at the bottom. The angular speed ω = √(3g/L).

Using the formula ω = ω₀ + αt, where ω₀ = 0, α = 2 rad/s², and t = 5 s, we get ω = 0 + 2*5 = 10 rad/s.

Using the formula ω = ω₀ + αt, we have ω = 10 rad/s + (2 rad/s²)(5 s) = 20 rad/s.

The total kinetic energy is the sum of translational and rotational kinetic energy. K.E. = (1/2)Mv² + (1/2)(Iω²) where I = (1/2)MR² for a solid cylinder.

The linear velocity v of the center of the wheel is given by v = Rω.

Using the formula τ = Iα, we have α = τ/I = 10 Nm / 2 kg·m² = 5 rad/s².

Angular momentum L = Iω, so if I is doubled and ω remains constant, L also doubles.

Linear velocity v = rω, so if r is halved and ω remains constant, v also halves.

Angular momentum L = mvr, where v is the tangential speed. L = 2 kg * 4 m/s * 3 m = 24 kg·m²/s.