Problem 8.88 A student sitting on a frictionless rotating stool has rotational i

Question

Problem 8.88 A student sitting on a frictionless rotating stool has rotational inertia 0.88kg*m^2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 7.00 rad/s and has negligible mass. The student extends her arms until her hands, each holding a 5.5 kg mass, are 0.73m from the rotation axis. Part A Ignoring her arm mass, what's her new rotational velocity? Part B Repeat if each arm is modeled as a 0.73 m long uniform rod of mass of 4.6 kg and her total body mass is 57 kg .

Explanation / Answer

We need to use the conservation of momentum. Rotational momentum is defined as[P = I omega] We need to find the new moment of inertia when she extends her arms. [I_e = I_b + I_a + I_w]where (I_e) is the moment of inertia with her arms extended, (I_b) is the moment of inertia of her body, (I_a) is the moment of inertia of her arms, and (I_w) is the moment of inertia of the weights she is holding. With the information given, I don't see any way to solve for (I_b) (we don't know her body's mass distribution about her axis of rotation). For the sake of this example, we will assume it is the same as when her arms are tucked. [I_e = I_b + {m_a L^2 over 3} + m_w L^2][I_e = 0.96 + {4.8 (0.72)^2 over 3} + 4.4 (0.72)^2] Now, back to conservation of momentum[I omega = I_e omega']where (I) is the moment of inertia of the girl with her arms tucked, (omega) is the angular velocity with her arms tucked, (I_e) is the moment of inertia of the girl when her arms are extended, and (omega') is her new angular velocity after extending her arms.

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