Certainly! Below is the transcription and explanation suitable for an educational website.---**ME Assignment #9****Item 1**The system described is used to intermittently operate a piece of equipment at the end of a lifter arm. The lifter is in contact with the highest point of the cam at the instant shown, and the lifter is constrained to move only in the vertical direction. The cam has a radius of \( r = 4.5 \) inches and rotates about point \( O \), which is offset from the center by a distance \( c = 3.0 \) inches. The cam's angular position \( \theta \) is measured from the positive \( x \)-axis and is positive in the clockwise direction. The angular velocity \( \omega \) and acceleration \( \alpha \) are also positive clockwise.**Figure Explanation**The figure depicts a mechanical system involving a cam and lifter mechanism. Key components and dimensions are labeled as follows:- **Cam with Radius \( r \):** The cam rotates around a point \( O \) with a radius \( r = 4.5 \) inches.- **Offset \( c \):** The rotation point \( O \) is offset from the cam's physical center by a distance \( c = 3.0 \) inches.- **Angular Position \( \theta \):** The angle \( \theta \) is formed between the positive \( x \)-axis and the position of the cam, measured in a clockwise direction.- **Angular Velocity \( \omega \):** The speed of rotation of the cam, moving in a clockwise direction.- **Angular Acceleration \( \alpha \):** The rate of change of angular velocity, also measured clockwise.- **Vertical Lifter Arm:** Attached to the cam, moving vertically as the cam rotates.- **Tension Spring (\( w \)):** Applying a force to the system, influencing the movement of the lifter.This mechanical setup is typically used in applications where precise vertical movement is needed, powered by rotational input. **Part D - Acceleration of the Lifter**Find the linear acceleration of the lifter with respect to point O when the angle is \( \theta = 34^\circ \), the angular velocity is \( \omega = 2.2 \, \text{rad/s} \), and the angular acceleration is \( \alpha = -3.0 \, \text{rad/s}^2 \).Express your answer to two significant figures and include the appropriate units. Enter a positive value if the acceleration is upward and a negative value if the acceleration is downward.*View Available Hint(s)*\[a = \, \boxed{\phantom{001}\text{in/s}^2}\]Submitted answer: \(a = 0.658 \, \text{in/s}^2\)**Feedback:** Incorrect; Try Again.**Navigation Buttons:**- Submit- Previous Answers- Request Answer*Navigational Instructions for other parts are visible but not transcribed.***Note:** There are no graphs or diagrams in the image provided.

Write the expression of linear distance. S=r+csinθ Differentiate the above expression to calculate the linear acceleration of lifter. dSdt=0+ccosθdθdt=cωcosθAnd,a=d2Sdt2=c-ωsinθ×dθdt+cosθ×dωdt=c-ω2sinθ+αcosθ=3-2.22sin34°+-3cos34°a=-15.58 in/s2Differentiate the above expression to calculate the linear acceleration of lifter. dSdt=0+ccosθdθdt=cωcosθAnd,a=d2Sdt2=c-ωsinθ×dθdt+cosθ×dωdt=c-ω2sinθ+αcosθ=3-2.22sin34°+-3cos34°a=-15.58 in/s2 Therefore, the the linear acceleration of lifter is 15.58 in/s2 downward.