Physics Ch. 9 #65Please see attached image this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station’s angular speed due to the radial movement of the people? **Problem 65:**A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is 82.5 meters. The moment of inertia of the station without people is \(3.00 \times 10^9 \, \text{kg} \cdot \text{m}^2\). Suppose that 500 people, with an average mass of 70.0 kg each, live on the space station.*[Note: No graphs or diagrams are present in the image.]*

The given data we have: Radius of the cylinder: 82.5m Moment of inertia of the station without people: 3.00×109kg.m2 Average mass of people: 70.0kg Total people: 500 The objective is to calculate the percent change in station's angular speed. The moment of inertia with people present on the outward position will be: Moment of inertia =500×70kg×(82.5 m)2 =238.218×106 kg m2 The total moment of inertia will be: Total moment of inertia=238.218×106 kg m2+3.0×109 kg m2 =3.238×109kg m2 Now, according to the conservation of angular momentum, we can write: Initial angular velocity×Initial moment of inertia = final angular velocity×final moment of inertia=>ωi×Ii=ωf×If The initial moment of inertia will be: 3.238×109kg m2 and since people came close to the axis, the moment of inertia due to them will become 0 as the distance becomes 0. So, the final moment of inertia will be: 3.0×109kg m2. Substituting the values in the equation for conservation of angular momentum, we get: ωi×3.0×109=ωf×3.238×109=>ωiωf=3.238×1093.0×109=>ωiωf=1.08 kg m2 Now, the percent change in angular velocity will be given as: %change =(ωiωf-1)×100% =(1.08-1)×100% =8% Therefore, there is an increase in angular velocity by 8%.