A horizontal circular platform (m = 110.1 kg, r = 3.21m) rotates about a frictionless vertical axle. A student (m = 98.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 3.9 rad/s when the student is at the rim. Find omega when the student is 1.23m from the center.
This is conservation of angular momentum
Iiωi = Ifωf
Note: you must account for I of the platform (which will not change) plus I of the student (which changes as she goes from 3.21 to 1.23)
To find the final angular velocity, ωf, when the student is 1.23 m from the center, we can apply the principle of conservation of angular momentum.
The angular momentum, L1, of the system when the student is at the rim is given by:
L1 = (moment of inertia of the platform + moment of inertia of the student) * ω1
The moment of inertia, I1, of the platform is given by:
I1 = m1 * r1^2
where m1 is the mass of the platform and r1 is the radius of the platform.
Similarly, the moment of inertia, I2, of the student is given by:
I2 = m2 * r2^2
where m2 is the mass of the student and r2 is the distance of the student from the center.
Therefore, the angular momentum, L1, can be written as:
L1 = (m1 * r1^2 + m2 * r2^2) * ω1
When the student moves toward the center, the moment of inertia of the system changes. When the student is 1.23 m from the center, the moment of inertia, If, of the system can be calculated as:
If = m1 * r1^2 + m2 * r2^2
To find the final angular velocity, ωf, we can use the principle of conservation of angular momentum:
L1 = (moment of inertia of the system) * ωf
Therefore, we can write:
(m1 * r1^2 + m2 * r2^2) * ω1 = (m1 * r1^2 + m2 * r2^2) * ωf
Simplifying the equation, we get:
ωf = ω1
Therefore, the final angular velocity, ωf, when the student is 1.23 m from the center, is 3.9 rad/s.