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?
To determine the new angular momentum in terms of the initial angular momentum, we'll consider the conservation of angular momentum.
The angular momentum of a rotating object is given by the equation:
Angular Momentum = (moment of inertia) x (angular velocity)
In this case, we have a hoola-hoop of radius R spinning about its symmetry axis. The moment of inertia for a rotating hoop is given by:
Moment of Inertia = mass x radius^2
Let's denote the initial angular momentum as L1. So, initially, the angular momentum (L1) can be expressed as:
L1 = (moment of inertia of the hoop) x (initial angular velocity)
Now, when we double the mass and halve the angular velocity, we need to find the new angular momentum, denoted as L2.
The moment of inertia of the hoop with double mass is given by:
Moment of Inertia2 = (2 x mass) x radius^2 = 2 x (mass x radius^2)
The new angular velocity is halved, so it becomes (initial angular velocity) / 2.
The new angular momentum (L2) can be expressed as:
L2 = (moment of inertia2) x (new angular velocity)
Substituting the expressions for moment of inertia2 and the new angular velocity, we have:
L2 = 2 x (mass x radius^2) x (initial angular velocity / 2)
Simplifying the expression:
L2 = mass x radius^2 x (initial angular velocity)
Therefore, we can see that the new angular momentum (L2) is equal to the initial angular momentum (L1). In other words, doubling the mass and halving the angular velocity does not change the angular momentum.