A long, thin rod with moment of inertia I=2 kg•m2 is free to rotate about an axis passing through the midpoint of the rod. The rod begins rotating from rest at time t= 0 s, accelerating constantly so that it has a rotational velocity of 4𝜋 rad/ s after rotating through two complete revolutions. What is the rod's angular momentum at this point?
To find the angular momentum of the rod at this point, we can use the formula:
Angular Momentum (L) = Moment of Inertia (I) * Angular Velocity (ω)
Given that the moment of inertia of the rod is I = 2 kg•m^2 and the angular velocity is ω = 4π rad/s, we can substitute these values into the formula:
L = I * ω
L = 2 kg•m^2 * 4π rad/s
L = 8π kg•m^2 rad/s
Therefore, the rod's angular momentum at this point is 8π kg•m^2 rad/s.
The angular momentum of an object is given by the equation:
L = I * ω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In this case, the moment of inertia (I) is given as 2 kg•m2 and the angular velocity (ω) is given as 4𝜋 rad/s.
Plugging these values into the equation, we can calculate the angular momentum:
L = 2 kg•m2 * 4𝜋 rad/s
To simplify the calculation, we can multiply the moment of inertia (I) by the angular velocity (ω):
L = 8𝜋 kg•m2/s
Therefore, the rod's angular momentum at this point is 8𝜋 kg•m2/s.
angular momentum = moment of inertia * angular velocity
= 2 kg m^2 * 4 pi rad/s
= 8 pi