A student sits on a rotating stool holding two 4.0 kg objects. When his arms are
ID: 1698525 • Letter: A
Question
A student sits on a rotating stool holding two 4.0 kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation, and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg·m2 and is assumed to be constant. The student then pulls the objects horizontally to 0.40 m from the rotation axis.(a) Find the new angular speed of the student.
(b) Find the kinetic energy of the student system before and after the objects are pulled in.
Explanation / Answer
From the law of conservation of angular momentum I1w1 = I2w2 w2 = (I1w1)/I2 ---------------(1) I1 = I + mr^2 = 3 kg.m^2 + 2(4Kg)(1.0M)^2 = (I dont have the calculator so compute it) I2 = I + mr^2 = 3 Kg.m^2 + 2(4Kg)(0.40m)^2 = (Calculate) w = 75 rad/s (this is given) Put all the values in equation (1) and you will get the answer. For part b Initial energy, Ki = (1/2) I1 w1^2 (put the values) Final kinetic energy Kf = (1/2) I2 w2^2 (put the values)
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