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A disc rolls straight on a flat floor without slipping. The magnitude of the angular velocity ω= 2.0 rad/s, the angular acceleration α=1.0 rad/s2. The radius of the disc is r=1.0 m. Determine f) the direction of the acceleration of the point A_

A disc rolls straight on a flat floor without slipping. The magnitude of the angular velocity ω= 2.0 rad/s, the angular acceleration α=1.0 rad/s2. The radius of the disc is r=1.0 m. Determine f) the direction of the acceleration of the point A_

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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  1. A disc rolls straight on a flat floor without slipping. The magnitude of the angular velocity ω= 2.0 rad/s, the angular acceleration α=1.0 rad/s2. The radius of the disc is r=1.0 m. Determine

    f) the direction of the acceleration of the point A________





     
     
     

      A.

    left
      B.

    up
      C.

    right
      D.

    down
## Motion of a Rolling Disc **Problem Statement:** A disc rolls straight on a flat floor without slipping. The magnitude of the angular velocity (ω) is 2.0 rad/s, and the angular acceleration (α) is 1.0 rad/s². The radius (r) of the disc is 1.0 m. Determine the direction of the acceleration of point A. **Diagram Explanation:** The diagram shows a disc rolling on a flat surface. The disc’s center is labeled G, and point A is located on the rim of the disc. Arrows indicate the directions of angular velocity (ω) and angular acceleration (α). The radius of the disc (r) is also marked. **Calculation Requirements:** We need to determine the direction of the acceleration of point A, taking into account the given ω, α, and r. **Options:** A. Left B. Up C. Right D. Down Answer choices provided require understanding the motion dynamics of a rolling disc and analyzing the combined effects of linear and rotational movements. **Graph/Diagram in Detail:** - The disc is shown in a side view with a shaded area indicating its rolling motion. - Arrows along the disc’s circumference represent the direction of the angular velocity (ω) (counterclockwise) and angular acceleration (α) (clockwise) relative to the disc’s movement. - Point G represents the disc's center of mass. - The radius (r = 1.0 m) connects point G to point A on the disc’s perimeter. **Solution Approach:** To determine the direction of the acceleration at point A, analyze the following: - **Tangential Acceleration (at)**: Caused by angular acceleration (α). For a point on the disc’s rim, \( at = r \cdot α \). - **Centripetal (Radial) Acceleration (ar)**: Caused by angular velocity (ω). For a point on the disc’s rim, \( ar = r \cdot ω^2 \). - Combine these vectors to find the net acceleration direction at point A, considering contributions from both the tangential and centripetal components. By examining these components and their directions, one can deduce the correct direction of the acceleration at point A. **Answer:** Upon careful analysis of the dynamics involved: - The tangential acceleration is upwards along the rim due to the disc’s angular acceleration
Transcribed Image Text:## Motion of a Rolling Disc **Problem Statement:** A disc rolls straight on a flat floor without slipping. The magnitude of the angular velocity (ω) is 2.0 rad/s, and the angular acceleration (α) is 1.0 rad/s². The radius (r) of the disc is 1.0 m. Determine the direction of the acceleration of point A. **Diagram Explanation:** The diagram shows a disc rolling on a flat surface. The disc’s center is labeled G, and point A is located on the rim of the disc. Arrows indicate the directions of angular velocity (ω) and angular acceleration (α). The radius of the disc (r) is also marked. **Calculation Requirements:** We need to determine the direction of the acceleration of point A, taking into account the given ω, α, and r. **Options:** A. Left B. Up C. Right D. Down Answer choices provided require understanding the motion dynamics of a rolling disc and analyzing the combined effects of linear and rotational movements. **Graph/Diagram in Detail:** - The disc is shown in a side view with a shaded area indicating its rolling motion. - Arrows along the disc’s circumference represent the direction of the angular velocity (ω) (counterclockwise) and angular acceleration (α) (clockwise) relative to the disc’s movement. - Point G represents the disc's center of mass. - The radius (r = 1.0 m) connects point G to point A on the disc’s perimeter. **Solution Approach:** To determine the direction of the acceleration at point A, analyze the following: - **Tangential Acceleration (at)**: Caused by angular acceleration (α). For a point on the disc’s rim, \( at = r \cdot α \). - **Centripetal (Radial) Acceleration (ar)**: Caused by angular velocity (ω). For a point on the disc’s rim, \( ar = r \cdot ω^2 \). - Combine these vectors to find the net acceleration direction at point A, considering contributions from both the tangential and centripetal components. By examining these components and their directions, one can deduce the correct direction of the acceleration at point A. **Answer:** Upon careful analysis of the dynamics involved: - The tangential acceleration is upwards along the rim due to the disc’s angular acceleration
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