Let's break down the problem: **Given:** - Angular acceleration, \( \alpha = 2 \ \text{rad/s}^2 \) - Angular velocity, \( \omega = 3 \ \text{rad/s} \) - \( x = 20 \ \text{mm} \) - \( \dot{x} = 275 \ \text{mm/s} \) - \( \ddot{x} = -60 \ \text{mm/s}^2 \) We are to find the **magnitude of the slip acceleration (\( a_{slip} = a_{rel} \))**. ### Step 1: Formula for Relative Acceleration (\( a_{rel} \)) The slip or relative acceleration of the slider \( A \) with respect to the rod is purely the acceleration of \( A \) along the slot (since the slot itself is moving due to rotation): \[ a_{rel} = \ddot{x} \] But, if the question is referring to the total acceleration of \( A \) relative to the disk (not just \( x \)-direction; i.e., accounting for Coriolis acceleration), then: \[ \vec{a}_A = \vec{a}_O + \vec{a}_{A/O} \] Where: - \( \vec{a}_{A/O} = \ddot{x} \hat{i} + 2\dot{x}\omega\hat{j} \) (Coriolis acceleration term) - \( \vec{a}_O \) is the acceleration of the origin (which is zero if O is fixed). #### So, **slip acceleration** (relative acceleration in the slot) consists of: - **Tangential acceleration**: \( \ddot{x} \) - **Coriolis acceleration**: \( 2\dot{x}\omega \) ### Step 2: Substitute Values \[ a_{slip} = \sqrt{(\ddot{x})^2 + (2\dot{x}\omega)^2} \] Given: - \( \ddot{x} = -60 \ \text{mm/s}^2 \) - \( \dot{x} = 275 \ \text{mm/s} \) - \( \omega = 3 \ \text{rad/s} \) \[ 2\dot{x}\omega = 2 \times 275 \times 3 = 1650 \ \text{mm/s}^2 \] \[ a_{slip} = \sqrt{(-60)^2 + (1650)^2} \] \[ a_{slip} = \sqrt{3600 + 2,722,500} \] \[ a_{slip} = \sqrt{2,726,100} \] \[ a_{slip} \approx 1651 \ \text{mm/s}^2 \] ### Step 3: Compare with Choices None of the options match \( 1651 \ \text{mm/s}^2 \), which suggests the question only wants the magnitude of \( \ddot{x} \) (slip along the slot), or possibly just the Coriolis term. #### Let's check the options for just \( |\ddot{x}| \) and \( 2\dot{x}\omega \): - \( |\ddot{x}| = 60 \ \text{mm/s}^2 \) *(Option a)* - \( 2\dot{x}\omega = 1650 \ \text{mm/s}^2 \) *(not an option)* So the most reasonable answer, based on the options and standard terminology, is: --- ### **Final Answer:** \[ \boxed{60 \ \text{mm/s}^2} \] Option **a**. This matches the magnitude of \( \ddot{x} \), the slip acceleration in the direction of the slot.

Let's carefully analyze the problem step-by-step to determine the magnitude of the slip acceleration \( a_{slip} \). --- ### **Step 1: Understanding the problem and given data** - The slider \(A\) moves along a slot in a rotating disk. - The disk rotates with angular velocity \( \omega = 3\, \text{rad/sec} \). - The angular acceleration of the disk is \( \alpha = 2\, \text{rad/sec}^2 \). (Note: It may or may not be needed depending on the problem statement; since the problem mentions \( \alpha \), we'll consider its impact.) - The slider's position along the slot: \( x = 20\, \text{mm} \). - The velocity of the slider along the slot: \( \dot{x} = 275\, \text{mm/sec} \). - The acceleration of the slider along the slot: \( \ddot{x} = -60\, \text{mm/sec}^2 \). --- ### **Step 2: Determine what is asked** - The problem asks for the **magnitude of the slip acceleration \( a_{slip} \)**. - In kinematics involving a slider in a rotating system, the total acceleration of the slider relative to the fixed frame has two main components: 1. **Acceleration along the slot** (\( \ddot{x} \)) 2. **Coriolis acceleration** due to the movement of the slider in the rotating frame: \( 2 \, \dot{x} \, \omega \) --- ### **Step 3: Identify the relevant acceleration components** The **slip acceleration** (relative to the moving disk) in the direction of the slot includes: \[ a_{slip} = \sqrt{\left(\ddot{x}\right)^2 + \left(2 \, \dot{x} \, \omega \right)^2} \] This is because the total acceleration in the radial direction (along the slot) combines the direct acceleration \( \ddot{x} \) and the Coriolis acceleration component \( 2 \dot{x} \omega \). --- ### **Step 4: Calculate the Coriolis acceleration** Given: - \( \dot{x} = 275\, \text{mm/sec} \) - \( \omega = 3\, \text{rad/sec} \) Calculate: \[ 2 \, \dot{x} \, \omega = 2 \times 275 \times 3 = 1650\, \text{mm/sec}^2 \] --- ### **Step 5: Calculate the total slip acceleration** Using: \[ a_{slip} = \sqrt{ (\ddot{x})^2 + (2 \, \dot{x} \, \omega)^2 } \] Substitute the values: \[ a_{slip} = \sqrt{ (-60)^2 + (1650)^2 } \] \[ a_{slip} = \sqrt{ 3600 + 2,722,500 } \] \[ a_{slip} = \sqrt{ 2,726,100 } \] Calculate the square root: \[ a_{slip} \approx 1651\, \text{mm/sec}^2 \] --- ### **Step 6: Final comparison and conclusion** The options provided are: - a) 60 mm/sec\(^2\) - b) 275 mm/sec\(^2\) - c) 20 mm/sec\(^2\) - d) 0 mm/sec\(^2\) Our calculated total slip acceleration (~1651 mm/sec\(^2\)) is **much larger** than the options, but notice that the question likely refers to **the magnitude of the slip acceleration component along the slot**, which can sometimes be considered as just \( |\ddot{x}| \), especially if the Coriolis component is considered as a secondary effect or the problem asks for the tangential acceleration along the slot. Given the options and the data, the **most reasonable choice** is: \[ \boxed{\textbf{a) } 60\, \text{mm/sec}^2} \] which corresponds to the direct acceleration \( |\ddot{x}| \). --- ### **### **Final Answer:** \[ \boxed{ \text{Magnitude of slip acceleration} \approx \mathbf{60\, \text{mm/sec}^2} } \] **Answer: a) 60 mm/sec\(^2\)** --- ### **Summary:** - The total slip acceleration considering both \( \ddot{x} \) and Coriolis effects is approximately 1651 mm/sec\(^2\). - But the problem's options suggest simplifying or focusing on the direct acceleration component \( \ddot{x} \) (\( 60\, \text{mm/sec}^2 \)), which is the most consistent with the options provided. Let me know if you'd like further clarifications!