Consider a hoola-hoop of radius R and it is spinning about its symmetry axis at a constant angular velocity. Now, I double its mass and halve its angular velocity. What is the new angular momentum in terms of the initial angular momentum?

L = I w = m R^2 w

L2 = 2 m R^2 w/2

which is the same

To find the new angular momentum in terms of the initial angular momentum, we need to understand the relationship between angular momentum and the variables involved.

Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω). Mathematically, it can be expressed as L = Iω.

For a hoop of radius R spinning about its symmetry axis, the moment of inertia can be calculated using the formula I = MR², where M is the mass and R is the radius of the hoop.

Let's denote the initial mass as M₀ and the initial angular velocity as ω₀. The initial angular momentum (L₀) can be expressed as L₀ = I₀ω₀, where I₀ = M₀R².

Now, according to the problem, we double the mass and halve the angular velocity. After this change, the new mass becomes 2M₀ and the new angular velocity becomes 0.5ω₀.

To find the new angular momentum (L), we need to calculate the new moment of inertia (I) using the new mass (2M₀) and radius (R). So, I = (2M₀)R².

Thus, the new angular momentum (L) can be expressed as L = Iω = (2M₀R²)(0.5ω₀).

Therefore, the new angular momentum (L) in terms of the initial angular momentum (L₀) is L = (2M₀R²)(0.5ω₀) / (I₀ω₀) = 2MR²/(2R²) = 2M.

So, the new angular momentum is simply twice the initial mass (2M₀).