Let's break down and **solve each question step-by-step** using the provided diagram and information. --- ## **1. Length of the Offset** > **Question:** How long is the offset? - **Offset** is the lateral distance between the piston axis and the crankshaft axis. - **Given:** The diagram shows the crank and connecting rod both aligned, with no lateral offset. - **Answer:** **0** --- ## **2. Value of Q and β** > **How much is Q?** > **How much is β?** - **Q** is typically a constant related to offset (if present). - **Given:** The mechanism is symmetric, so Q = **1**. - **β** is the angle of offset, which is **0** since there is no offset. - **Answers:** - **Q = 1** - **β = 0** --- ## **3. Stroke of the Mechanism** > **How long is the stroke of the mechanism in inches?** - **Stroke** is the total distance the piston travels up and down. - **Given:** The connecting rod is 4" long. The crank radius (distance from crank center to crank pin) is 2". - **Stroke = 2 × Crank radius = 2 × 2" = 4"** - **Answer:** **4** --- ## **4. Midpoint Piston Displacement** > **How much is the midpoint piston displacement in inches?** - **Midpoint displacement** is the piston position when the crank is at 90° (midway between TDC and BDC). - **Displacement (y) = r × (1 - cosθ) + l × (1 - sqrt(1 - (r/l)² sin²θ))** - **At midpoint:** θ = 90° ⇒ cosθ = 0, sinθ = 1 Let’s plug in the numbers: - r = 2 in - l = 4 in So, - y = 2 × (1 - 0) + 4 × (1 - sqrt(1 - (2/4)² × 1²)) - y = 2 × 1 + 4 × (1 - sqrt(1 - 0.25)) - y = 2 + 4 × (1 - sqrt(0.75)) - sqrt(0.75) ≈ 0.866 - y = 2 + 4 × (1 - 0.866) - y = 2 + 4 × 0.134 - y = 2 + 0.536 - y ≈ **2.54 inches** **Answer:** **2.54 inches** --- ## **5. Crank Angle from Positive X-axis** > **How much is the crank angle from the positive direction of the X-axis?** - **By convention:** At midpoint displacement (piston halfway down), the crank is at 180° from the positive X-axis (if X-axis is to the right). - **Answer:** **180°** --- ## **6. Angular Velocity of the Crank** > **How much is the angular velocity of the crank in rad/s? What is its direction?** ### **Step 1: Find the angular velocity** Given: - Piston speed, \( V_p = 5 \) in/s (downward) - Crank radius, \( r = 2 \) in - Let \( \theta = 270^\circ \) (midpoint downward stroke, crank is pointing downward) The velocity of piston in a slider-crank mechanism is: \[ V_p = r \omega \sin\theta + \dfrac{r^2 \omega \sin2\theta}{2l} \] At θ = 270°: - \(\sin\theta = -1\) - \(\sin2\theta = \sin540^\circ = \sin180^\circ = 0\) \[ V_p = r \omega \sin\theta = 2 \omega \times (-1) = -2\omega \] But piston speed is downward and considered positive (so -2ω = -5 in/s): \[ -2\omega = -5 \implies \omega = 2.5 \text{ rad/s} \] ### **Step 2: Direction** - The crank rotates **counterclockwise (CCW)** for the piston to move down. **Final Answers:** - **Angular velocity = 2.5 rad/s** - **Direction = CCW** --- ## **7. Angular Velocity of the Connecting Rod** > **How much is the angular velocity of the connecting rod in rad/s? What is its direction?** The angular velocity of the connecting rod (\( \omega_{rod} \)) is: \[ \omega_{rod} = \frac{r \omega \cos\theta}{l} \] At θ = 270°: - \(\cos\theta = 0\) - So, \[ \omega_{rod} = \frac{2 \times 2.5 \times 0}{4} = 0 \] **Direction:** Since the value is zero, **no direction** is needed. --- ## **Summary Table** | Question | Answer | |--------------------------------------------------------------------------|---------------------| | How long is the offset? | 0 | | How much is Q? | 1 | | How much is β? | 0 | | How long is the stroke of the mechanism in inches? | 4 | | How much is the midpoint piston displacement in inches? | 2.54 | | How much is the crank angle from the positive direction of the X-axis? | 180 | | How much is the angular velocity of the crank in rad/s? | 2.5 | | What is its direction? | CCW | | How much is the angular velocity of the connecting rod in rad/s? | 0 | | What is its direction? | No direction | --- **Let me know if you want a detailed derivation for any specific part!**