QuestionFebruary 2, 2026

006 (part 1 or 2) 10.0 points A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.8 m from the axis of rotation, and he rotates with angular speed of 0.68rad/sec . The moment of iner- tia of the student plus the stool is 5kgm^2 and is assumed to be constant. The student

006 (part 1 or 2) 10.0 points A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.8 m from the axis of rotation, and he rotates with angular speed of 0.68rad/sec . The moment of iner- tia of the student plus the stool is 5kgm^2 and is assumed to be constant. The student
006 (part 1 or 2) 10.0 points A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.8 m from the axis of rotation, and he rotates with angular speed of 0.68rad/sec . The moment of iner- tia of the student plus the stool is 5kgm^2 and is assumed to be constant. The student

Solution
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Answer

0.84 \, \text{rad/sec} Explanation 1. Calculate Initial Moment of Inertia Use I_{initial} = I_{student} + 2 \cdot m \cdot r^2. Here, m = 1 \, \text{kg}, r = 0.8 \, \text{m}, and I_{student} = 5 \, \text{kg m}^2. So, I_{initial} = 5 + 2 \cdot 1 \cdot (0.8)^2 = 6.28 \, \text{kg m}^2. 2. Apply Conservation of Angular Momentum Angular momentum is conserved: I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final}. Solve for \omega_{final} when arms are pulled in. 3. Calculate Final Moment of Inertia Assume arms are pulled in to r_{final} = 0.2 \, \text{m}. I_{final} = I_{student} + 2 \cdot m \cdot r_{final}^2 = 5 + 2 \cdot 1 \cdot (0.2)^2 = 5.08 \, \text{kg m}^2. 4. Solve for Final Angular Speed \omega_{final} = \frac{I_{initial} \cdot \omega_{initial}}{I_{final}} = \frac{6.28 \cdot 0.68}{5.08} = 0.84 \, \text{rad/sec}.

Explanation

1. Calculate Initial Moment of Inertia<br /> Use $I_{initial} = I_{student} + 2 \cdot m \cdot r^2$. Here, $m = 1 \, \text{kg}$, $r = 0.8 \, \text{m}$, and $I_{student} = 5 \, \text{kg m}^2$. So, $I_{initial} = 5 + 2 \cdot 1 \cdot (0.8)^2 = 6.28 \, \text{kg m}^2$.<br />2. Apply Conservation of Angular Momentum<br /> Angular momentum is conserved: $I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final}$. Solve for $\omega_{final}$ when arms are pulled in.<br />3. Calculate Final Moment of Inertia<br /> Assume arms are pulled in to $r_{final} = 0.2 \, \text{m}$. $I_{final} = I_{student} + 2 \cdot m \cdot r_{final}^2 = 5 + 2 \cdot 1 \cdot (0.2)^2 = 5.08 \, \text{kg m}^2$.<br />4. Solve for Final Angular Speed<br /> $\omega_{final} = \frac{I_{initial} \cdot \omega_{initial}}{I_{final}} = \frac{6.28 \cdot 0.68}{5.08} = 0.84 \, \text{rad/sec}$.
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