Let's break down and solve the problem step by step. --- ## **Given** - Particle \( P \) moves along a fixed circle of radius \( R \): \( x^2 + y^2 = R^2 \) - The slot guiding \( P \) moves horizontally: \( x = b \sin(\omega t) \) - Parameters: \( R = 50 \) mm, \( b = 40 \) mm, \( \omega = 5\pi \) rad/s --- ## **Step 1: Show Unit Vectors \( \mathbf{i} \) and \( \mathbf{j} \) in a Sketch** - \( \mathbf{i} \): Unit vector along \( x \)-axis (to the right). - \( \mathbf{j} \): Unit vector along \( y \)-axis (upwards). You can sketch these on the provided diagram as arrows from the origin: - \( \mathbf{i} \) points to the right (horizontal). - \( \mathbf{j} \) points up (vertical). --- ## **Step 2: Find Cartesian Components of Velocity and Acceleration** ### **A. Position of P** Given: - \( x = b \sin(\omega t) \) - \( x^2 + y^2 = R^2 \Rightarrow y = \sqrt{R^2 - x^2} \) So, \[ \mathbf{r}_P = x\,\mathbf{i} + y\,\mathbf{j} \] \[ x = b \sin(\omega t) \] \[ y = \sqrt{R^2 - b^2 \sin^2(\omega t)} \] --- ### **B. Velocity of P** \[ \mathbf{v}_P = \frac{d\mathbf{r}_P}{dt} = \dot{x}\,\mathbf{i} + \dot{y}\,\mathbf{j} \] Where: \[ \dot{x} = b\omega \cos(\omega t) \] \[ y = \sqrt{R^2 - x^2} \] So, \[ \dot{y} = \frac{d}{dt}\left[\sqrt{R^2 - x^2}\right] = \frac{1}{2}\left(R^2 - x^2\right)^{-1/2} \cdot (-2x\dot{x}) = -\frac{x\dot{x}}{\sqrt{R^2 - x^2}} \] Substitute \(x = b\sin(\omega t)\), \(\dot{x} = b\omega\cos(\omega t)\): \[ \dot{y} = -\frac{b\sin(\omega t) \cdot b\omega\cos(\omega t)}{\sqrt{R^2 - b^2\sin^2(\omega t)}} = -\frac{b^2\omega\sin(\omega t)\cos(\omega t)}{\sqrt{R^2 - b^2\sin^2(\omega t)}} \] So, \[ \boxed{ \mathbf{v}_P = b\omega\cos(\omega t)\,\mathbf{i} - \frac{b^2\omega\sin(\omega t)\cos(\omega t)}{\sqrt{R^2 - b^2\sin^2(\omega t)}}\,\mathbf{j} } \] --- ### **C. Acceleration of P** \[ \mathbf{a}_P = \ddot{x}\,\mathbf{i} + \ddot{y}\,\mathbf{j} \] \[ \ddot{x} = -b\omega^2\sin(\omega t) \] For \(\ddot{y}\), differentiate \(\dot{y}\): \[ \dot{y} = -\frac{x\dot{x}}{y} \] So, \[ \ddot{y} = \frac{d}{dt} \left(-\frac{x\dot{x}}{y}\right) = -\frac{\dot{x}\dot{x} + x\ddot{x}}{y} + \frac{x\dot{x}\dot{y}}{y^2} \] Plug in values: - \(\dot{x} = b\omega\cos(\omega t)\) - \(\ddot{x} = -b\omega^2\sin(\omega t)\) - \(x = b\sin(\omega t)\) - \(y = \sqrt{R^2 - b^2\sin^2(\omega t)}\) - \(\dot{y}\) as above Thus, \[ \ddot{y} = -\frac{[\dot{x}^2 + x\ddot{x}]}{y} + \frac{x\dot{x}\dot{y}}{y^2} \] --- ## **Step 3: Evaluate at \( t = \)** - \( \sin() = \), \( \cos() = 1 \) \[ x = \] \[ y = R = 50\,\text{mm} \] \[ \dot{x} = b\omega = 40 \times 5\pi = 200\pi\,\text{mm/s} \] \[ \dot{y} = \quad (\text{since } \sin() = ) \] \[ \ddot{x} = \] \[ \dot{x}^2 = (200\pi)^2 \] \[ x\ddot{x} = \] \[ \ddot{y} = -\frac{(200\pi)^2}{50} \] --- ### **Final Answers at \( t = \):** #### **Velocity:** \[ \boxed{ \mathbf{v}_P = (200\pi)\,\mathbf{i} + \,\mathbf{j}\ \text{mm/s} } \] #### **Acceleration:** \[ \boxed{ \mathbf{a}_P = \,\mathbf{i} - \frac{(200\pi)^2}{50}\,\mathbf{j}\ \text{mm/s}^2 } \] Numeric value for \( \mathbf{a}_P \): \[ (200\pi)^2 = 40000 \pi^2 \approx 40000 \times 9.87 = 394800 \] \[ \frac{394800}{50} = 7896 \] So: \[ \boxed{ \mathbf{a}_P = \,\mathbf{i} - 7896\,\mathbf{j}\ \text{mm/s}^2 } \] --- ## **Summary Table** | Quantity | Expression | Value at \( t = \) | |-----------------|-------------------|-------------------------------| | \( \mathbf{v}_P \) | \( b\omega\,\mathbf{i} \) | \( 200\pi\,\mathbf{i} \) mm/s | | \( \mathbf{a}_P \) | \( -\frac{\dot{x}^2}{R}\,\mathbf{j} \) | \( -7896\,\mathbf{j} \) mm/s\(^2\) | --- ## **Sketch** Include: - \( \mathbf{i} \): right along \( x \) - \( \mathbf{j} \): up along \( y \) - \( \mathbf{v}_P \) to the right at \( t = \) - \( \mathbf{a}_P \) downward at \( t = \) --- **Let me know if you need the sketch drawn out!**