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In the image below pivot is placed at the center of mass of the stick (mass = 0.097 kg). If the block (0.022 kg) is placed at 0.766 m it will rotate the block-stick system around the pivot. Calculate the following. (Istick = 1/12 ML^2)

In the image below pivot is placed at the center of mass of the stick (mass = 0.097 kg). If the block (0.022 kg) is placed at 0.766 m it will rotate the block-stick system around the pivot. Calculate the following. (Istick = 1/12 ML^2)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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In the image below pivot is placed at the center of mass of the stick (mass = 0.097 kg). If the block (0.022 kg) is placed at 0.766 m it will rotate the block-stick system around the pivot. Calculate the following. (Istick = 1/12 ML^2)

 

 

**Figure Explanation:** The image depicts a horizontal stick positioned on a pivot point, indicated by a red triangle, which represents the center of mass (CM). The stick is 1 meter long, extending from x = 0 m to x = 1 m. A block, labeled "A," is placed at the end of the stick at x = 1 m. The pivot point is located at the middle of the stick at x = 0.5 m. The horizontal axis is labeled as x, and the specific positions \(x = 0.5 \, \text{m}\) (center) and \(x = 1 \, \text{m}\) (block A) are marked. **Calculations:** (a) The stick's moment of inertia = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (b) The block's moment of inertia = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (c) Total I of the system = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (d) Torque (magnitude) = \(\_\_\_\_\_\_\_\_\_\_\_\) N.m (e) Angular acceleration (magnitude) = \(\_\_\_\_\_\_\_\_\_\_\_\) \(\frac{\text{rad}}{\text{s}^2}\)
Transcribed Image Text:**Figure Explanation:** The image depicts a horizontal stick positioned on a pivot point, indicated by a red triangle, which represents the center of mass (CM). The stick is 1 meter long, extending from x = 0 m to x = 1 m. A block, labeled "A," is placed at the end of the stick at x = 1 m. The pivot point is located at the middle of the stick at x = 0.5 m. The horizontal axis is labeled as x, and the specific positions \(x = 0.5 \, \text{m}\) (center) and \(x = 1 \, \text{m}\) (block A) are marked. **Calculations:** (a) The stick's moment of inertia = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (b) The block's moment of inertia = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (c) Total I of the system = \(\_\_\_\_\_\_\_\_\_\_\_\) kg.m² (d) Torque (magnitude) = \(\_\_\_\_\_\_\_\_\_\_\_\) N.m (e) Angular acceleration (magnitude) = \(\_\_\_\_\_\_\_\_\_\_\_\) \(\frac{\text{rad}}{\text{s}^2}\)
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