Let's break down the problem step by step using the information from the image: - **Given:** - Angular velocity of wheel, \( \omega = 12 \, \text{rad/s} \) - Angular acceleration of wheel, \( \alpha = 6 \, \text{rad/s}^2 \) - Radius from O to B, \( r_{OB} = .3 \, \text{m} \) - Length from B to C, \( L_{BC} = 1.2 \, \text{m} \) - **Find:** - (a) Angular acceleration of link BC (\( \alpha_{BC} \)) - (b) Acceleration of point C --- ## Step 1: Find Linear Velocity and Acceleration of Point B ### Velocity of B \[ v_B = r_{OB} \cdot \omega = .3 \times 12 = 3.6 \, \text{m/s} \] ### Acceleration of B Total acceleration has two components: - **Tangential:** \( a_{t,B} = r_{OB} \cdot \alpha = .3 \times 6 = 1.8 \, \text{m/s}^2 \) - **Centripetal:** \( a_{c,B} = r_{OB} \cdot \omega^2 = .3 \times 12^2 = .3 \times 144 = 43.2 \, \text{m/s}^2 \) --- ## Step 2: Motion of Link BC Let’s assume the piston at C moves horizontally (since the slot is vertical). ### Instantaneous Center of Rotation Point C moves horizontally, so the **instantaneous center** is at B. Let \( \omega_{BC} \) be the angular velocity of link BC. \[ v_C = v_B \cos\theta \] But for small angles (nearly vertical), \( v_C \approx v_B \). Let's move on to acceleration. --- ## Step 3: Acceleration of Point C Acceleration of C (\( a_C \)) has two components: - Due to acceleration of B (\( a_B \)) - Due to angular acceleration of BC and angular velocity of BC \[ a_C = a_B + a_{C/B} \] Where \( a_{C/B} = \alpha_{BC} \times r_{CB} - \omega_{BC}^2 \times r_{CB} \) But first, let's find \( \omega_{BC} \): \[ \omega_{BC} = \frac{v_B}{L_{BC}} = \frac{3.6}{1.2} = 3 \, \text{rad/s} \] --- ## Step 4: Angular Acceleration of Link BC (\( \alpha_{BC} \)) Use relative acceleration: \[ a_C = a_B + a_{C/B} \] But since C moves horizontally, only the x-component matters. Tangential acceleration at C due to BC: \[ a_{C/B,\text{tan}} = \alpha_{BC} \cdot L_{BC} \] Centripetal acceleration at C due to BC: \[ a_{C/B,\text{rad}} = (\omega_{BC})^2 \cdot L_{BC} \] Set up the equation for the horizontal (x) direction: \[ a_{C,x} = a_{B,x} + a_{C/B,\text{tan},x} - a_{C/B,\text{rad},x} \] Given that at the instant, all motion is aligned, so: \[ a_{C} = a_{B,\text{tan}} + \alpha_{BC} \cdot L_{BC} - (\omega_{BC})^2 \cdot L_{BC} \] But since point B moves in a circle, its tangential acceleration is \( 1.8 \) m/s², and its centripetal is perpendicular (ignored for horizontal motion at C). #### Plug in values: \[ a_{C} = 1.8 + \alpha_{BC} \cdot 1.2 - 9 \times 1.2 \] \[ a_{C} = 1.8 + 1.2\alpha_{BC} - 10.8 \] \[ a_{C} = 1.2\alpha_{BC} - 9 \] If **C is constrained to move horizontally**, and since all motion at the instant is in that direction: ### Final Step: Express Answers #### (a) **Angular acceleration of BC** \[ \boxed{\alpha_{BC} = \frac{a_C + 9}{1.2}} \] #### (b) **Acceleration of C** If only the acceleration due to B's tangential is considered (as the centripetal is perpendicular), then: \[ a_C = 1.8 \, \text{m/s}^2 \quad \text{(if only tangential considered)} \] But if you want total (as above): \[ a_C = 1.2\alpha_{BC} - 9 \] --- ## **Summary of Answers** ### (a) **Angular acceleration of link BC:** \[ \boxed{\alpha_{BC} = \frac{a_C + 9}{1.2}} \] ### (b) **Acceleration of piston C:** \[ \boxed{a_C = 1.2\alpha_{BC} - 9} \] > **If you have more details about the relative motion at the instant or a value for \( a_C \), you can substitute them to get a numerical answer.** --- **If you have any specific angle or further data, let me know!**