A student sits on a freely rotating stool holding two dumbbells, each of mass 3.
ID: 2264876 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.01 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.93 m from the axis of rotation and the student rotates with an angular speed of 0.746 rad/s. The moment of inertia of the student plus stool is 2.80 kg
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.01 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.93 m from the axis of rotation and the student rotates with an angular speed of 0.746 rad/s. The moment of inertia of the student plus stool is 2.80 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position0.302 m from the rotation axis (Figure b). Find the new angular speed of the student Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.Explanation / Answer
initial moment of inertia of the system, I1 = 2.8 + 2*[m*r^2] => I1 = 2.8 + 2*[3.01* 0.93^2] = 8 kg-m^2 final moment of inertia, I2 = 2.8 +2*[m*r^2] => I2 = 2.8 + 2*[3.01* 0.302^2] = 3.349 kg-m^2 a). from angular momentum conservation; I1*w1 = I2 *w2 => 8* 0.746 = 3.349* w2 => w2 = 1.782 rad/s b). K(before) = 0.5*I1*w1^2 = 0.5*8*0.746^2 = 2.226 J K(after) = 0.5*I2*w2^2 = 0.5*3.349*1.782^2 = 5.317 J
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