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title: "A centrifuge rotor rotating at \\( 9200 \\) \\( \\text{rpm} \\) is shut off and is eventually brought uniformly to rest by a frictional torque of \\( 1.20 \\) \\( \\text{N} \\cdot \\text{m} \\). If the mass of the rotor is \\( 3.10 \\) \\( \\text{kg} \\) and it can be approximated as a solid cylinder of radius \\( 0.0710 \\) \\( \\text{m} \\), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \\( \\frac{1}{2} m r^2 \\)."
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url: "https://nerd-notes.com/ubq/29222/"
date_modified: "2025-03-14T03:56:48+00:00"
---

# A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).

A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).

- **A.** 3021 revolutions
- **B.** 2100 revolutions
- **C.** 481 revolutions
- **D.** 300 revolutions
- **E.** 3 revolutions

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/29222/*