A horizontal circular platform (m = 108.1 kg, r = 3.25m) rotates about a frictionless vertical axle. A student (m = 85.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 2.5 rad/s when the student is at the rim. Find omega when the student is 2.29m from the center.
moment of inertia is the sum of the platform and the student (mr^2).
I1*w1=I2*w2
To find the final angular velocity (ωf) when the student is 2.29m from the center of the platform, we need to use the conservation of angular momentum. The formula for angular momentum (L) is:
L = Iω
where:
L = angular momentum
I = moment of inertia
ω = angular velocity
In this case, the moment of inertia (I) of the system is the sum of the moment of inertia of the platform (Iplatform) and the moment of inertia of the student (Istudent). The formula for the moment of inertia of a solid disk rotating about its central axis is:
I = (1/2) m r^2
where:
m = mass
r = radius
First, let's calculate the initial angular momentum (Li) when the student is at the rim of the platform. Given that the platform's mass is 108.1 kg and the radius is 3.25 m, we have:
Iplatform = (1/2) m r^2
= (1/2) (108.1 kg) (3.25 m)^2
Next, we can calculate the angular momentum of the student:
Istudent = (1/2) m r^2
= (1/2) (85.3 kg) (3.25 m)^2
Since the student is at the rim of the platform, the total initial angular momentum is the sum of the angular momentum of the platform and the student:
Li = Iplatform * ω + Istudent * ω
Given that the initial angular velocity (ω) is 2.5 rad/s, we can calculate Li.
Now, let's calculate the final moment of inertia (If) when the student is 2.29m from the center. To do this, we use the same formula as before, but substitute the new radius:
If = (1/2) m r^2
= (1/2) (85.3 kg) (2.29 m)^2
Finally, we can find the final angular velocity (ωf) by rearranging the angular momentum formula and solving for ωf:
ωf = L / If
Plugging in the values of Li and If into the equation, we can calculate the final angular velocity.