-0.01 rad/s2 2.0m oint Pis on the rim of a wheel of radius 2.0 m. At timet= 0, the wheel is at rest, and Pis on the x-a. The wheel undergoes a uniform angular acceleration of g.01 radis about the center o. Figure 8.1, the magnitude of the linear acceleration of P. when it reaches the y-axis, is dosest to: DÃ 0 000 mis? O B. 0.088 mis? OCo.083 mis? O D.0.075 mis? DE 0.072 mis?
-0.01 rad/s2 2.0m oint Pis on the rim of a wheel of radius 2.0 m. At timet= 0, the wheel is at rest, and Pis on the x-a. The wheel undergoes a uniform angular acceleration of g.01 radis about the center o. Figure 8.1, the magnitude of the linear acceleration of P. when it reaches the y-axis, is dosest to: DÃ 0 000 mis? O B. 0.088 mis? OCo.083 mis? O D.0.075 mis? DE 0.072 mis?
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter12: Rotation I: Kinematics And Dynamics
Section: Chapter Questions
Problem 62PQ: Problems 62 and 63 are paired. 62. C A disk is rotating around a fixed axis that passes through its...
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![**Transcription for Educational Website:**
**Figure 9.1:**
[Diagram Description]
The diagram shows a wheel with a radius of 2.0 meters, centered at point O. The wheel rotates counterclockwise with an angular acceleration of \( \alpha = 0.01 \, \text{rad/s}^2 \). Point P is marked on the rim of the wheel, positioned initially along the x-axis.
**Problem Statement:**
Point P is on the rim of a wheel with a radius of 2.0 m. At time \( t = 0 \), the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of \( 0.01 \, \text{rad/s}^2 \) about the center O.
In Figure 9.1, the magnitude of the linear acceleration of P, when it reaches the y-axis, is closest to:
- A. \( 0.008 \, \text{m/s}^2 \)
- B. \( 0.064 \, \text{m/s}^2 \)
- C. \( 0.063 \, \text{m/s}^2 \)
- D. \( 0.075 \, \text{m/s}^2 \)
- E. \( 0.072 \, \text{m/s}^2 \)
**Analysis:**
This question involves calculating the linear acceleration of a point on the rim of a wheel as it rotates under specified conditions. The linear acceleration can be determined by analyzing both tangential and centripetal components due to the wheel's motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa01058a6-7004-44e6-9008-71f97b41fb6e%2F5912b1d9-b7c1-42f4-919d-2fedc13be60e%2F8nxfz73_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website:**
**Figure 9.1:**
[Diagram Description]
The diagram shows a wheel with a radius of 2.0 meters, centered at point O. The wheel rotates counterclockwise with an angular acceleration of \( \alpha = 0.01 \, \text{rad/s}^2 \). Point P is marked on the rim of the wheel, positioned initially along the x-axis.
**Problem Statement:**
Point P is on the rim of a wheel with a radius of 2.0 m. At time \( t = 0 \), the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of \( 0.01 \, \text{rad/s}^2 \) about the center O.
In Figure 9.1, the magnitude of the linear acceleration of P, when it reaches the y-axis, is closest to:
- A. \( 0.008 \, \text{m/s}^2 \)
- B. \( 0.064 \, \text{m/s}^2 \)
- C. \( 0.063 \, \text{m/s}^2 \)
- D. \( 0.075 \, \text{m/s}^2 \)
- E. \( 0.072 \, \text{m/s}^2 \)
**Analysis:**
This question involves calculating the linear acceleration of a point on the rim of a wheel as it rotates under specified conditions. The linear acceleration can be determined by analyzing both tangential and centripetal components due to the wheel's motion.
Expert Solution
Step 1
radius = r = 2 m
angular acceleration = = 0.01 rad/s2
when it reaches the y- axis angular displacement= = ==1.57 rad
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