### Problem Statement: Rolling Cylinder on an Inclined Plane#### Description:A cylinder of radius \( r = 0.2 \) meters is about to start rolling down an inclined plane. The acceleration of the center of the cylinder is \( 9.81 \sin(\theta) \) meters per second squared (m/s\(^2\)). The angular acceleration is \( 9.81 \sin(\theta) / r \) radians per second squared (rad/s\(^2\)). Assume \( \theta = 0.8 \) rad.#### Objective:Calculate the magnitude of acceleration of the point on the cylinder furthest from the inclined plane in meters per second squared (m/s\(^2\)).#### Hint:Since the cylinder is about to start rolling, the velocity of the center of the cylinder and the angular velocity of the cylinder are still zero at that instant.#### User Interaction:- **User Answered**: 0.2739 (This is the user's input).- **Correct Answer**: 14.075 (with a margin of error of +/- 2%).### Explanation:1. **Acceleration of the Center of the Cylinder**: - Given acceleration of the center: \( a_c = 9.81 \sin(\theta) \) m/s\(^2\). - With \( \theta = 0.8 \) rad, calculate \( a_c \).2. **Angular Acceleration**: - Given angular acceleration: \( \alpha = 9.81 \sin(\theta) / r \) rad/s\(^2\).3. **Acceleration of the Furthest Point**: - The acceleration of the point on the cylinder furthest from the inclined plane combines linear and rotational motion. - This point experiences both tangential and centripetal components.#### Calculation:1. Compute the given values with the provided formulae: Let's calculate \( a_c \) first: \[ a_c = 9.81 \sin(0.8) \text{ m/s}^2 \quad \text{(Use sin(0.8) value from a calculator or table)} \]2. Compute the angular acceleration \( \alpha \): \[ \alpha = \frac{9.81 \sin(0.8)}{0.2} \text{ rad/s}^2 \]

When there is a change in angular velocity of a rotating body, angular acceleration is associated…

A cylinder of radius r = 0.2 m is about to start rolling down an inclined plane. The acceleration of the center of the cylinder is 9.81 sin(0) m/s?. The angular acceleration is 9.81 sin(0)/r rad/s?. Assume 0 = 0.8 rad. What is the magnitude of acceleration of the point on the cylinder furthest from the inclined plane in m/s?. Hint: Since it is about to start rolling, the velocity of the center of the cylinder and angular velocity of the cylinder are still zero at that instant. You Answered 0.2739 Correct Answer 14.075 margin of error +/- 2%

A cylinder of radius r = 0.2 m is about to start rolling down an inclined plane. The acceleration of the center of the cylinder is 9.81 sin(0) m/s?. The angular acceleration is 9.81 sin(0)/r rad/s?. Assume 0 = 0.8 rad. What is the magnitude of acceleration of the point on the cylinder furthest from the inclined plane in m/s?. Hint: Since it is about to start rolling, the velocity of the center of the cylinder and angular velocity of the cylinder are still zero at that instant. You Answered 0.2739 Correct Answer 14.075 margin of error +/- 2%

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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