Question

Flywehel show in figure is rotating with a constant angular velocity of 30 rad/s. Pin C is fastened to the wheel at a point 250 mm from its center and slides in the slot of arm AB. For the instant shown: Determine: R I m 250 mm 300 mm (along bar AB (the slot) 3) OXAE , and 700 mm 600 mm Show all steps please

Answer 1

Consider triangle OCA

$OC = 250cos\theta i +250sin\theta j$

$AC = lcos\beta i +lsin\beta j$

$OA =250cos\theta i +250sin\theta j - (lcos\beta i +lsin\beta j)$

(-300 700) j sinbj - (lcosBi lsinsj) CoS

coSD

$\Rightarrow 0 = 250 cos\theta - l cos\beta$

$\Rightarrow -1000 = 250sin\theta -lsin\beta$

Here $\theta = 0\Rightarrow 250 = l cos\beta$

I- V25010002103 0.8mm

250 B75.96 cos6 1030.78

Differentiate $\Rightarrow 0 = 250 cos\theta - l cos\beta$

$-250 sin\theta \dot \theta + l sin\beta * w_{AB} -\dot l cos\beta= 0$

substituting given values:

250sin0 * θ + 1030. 78sin 75.96" * w -Icos75.96" 0

$999.99 w_{AB} - 0.2426\dot l = 0$

Differentiate $\Rightarrow -1000 = 250sin\theta -lsin\beta$

$\Rightarrow 0 = 250cos\theta * \dot \theta-lcos\beta * w_{AB} - \dot l sin\beta$

substituting given values:

$\Rightarrow 0 = 250cos0^0 * 30-1030.78cos75.96^0 * w_{AB} - \dot l sin75.96^0$

$250.07 w_{AB} + 0.9701\dot l = 7500$

Solving the equations:

$999.99 w_{AB} - 0.2426\dot l = 0$ and $250.07 w_{AB} + 0.9701\dot l = 7500$

$\Rightarrow w_{AB} = 1.7652rad/s;v_c = \dot l = 7276.13mm/s$

Differentiate

$-250 sin\theta \dot \theta + l sin\beta * w_{AB} -\dot l cos\beta= 0$

$-250 sin\theta \ddot \theta -250 cos\theta \dot \theta^2+ l sin\beta * \alpha_{AB}+ l cos\beta * w_{AB}^2 -\ddot l cos\beta+\dot l sin\beta*w_{AB}= 0$

substituting given values:

$-250 sin\theta *0 -250 cos0* 30^2+ 1030.78* sin75.96 * \alpha_{AB}+ 1030.78 cos75.96 * 1.7652^2 -\ddot l cos75.96+7276.13 sin75.96*1.7652= 0$

$1030.78* sin75.96 * \alpha_{AB}-\ddot l cos75.96 = 250 cos0* 30^2 - 7276.13 sin75.96*1.7652 - 1030.78 cos75.96 * 1.7652^2$

$999.99 \alpha_{AB}-0.2426\ddot l = 211760.6743$

Differentiate

$0 = 250cos\theta * \dot \theta-lcos\beta * w_{AB} - \dot l sin\beta$

$250cos\theta * \ddot \theta-250sin\theta * \dot \theta^2-\dot l cos\beta * w_{AB}+l sin\beta * w_{AB}^2 -lcos\beta * \alpha_{AB} - \ddot l sin\beta -\dot l cos\beta * w_{AB} = 0$

substituting given values:

$250cos\theta * 0-250sin0 * \dot \theta^2-7276.13 *cos75.96 * 1.7652+1030.78* sin75.96 * 1.7652^2 -1030.78* cos75.96 * \alpha_{AB} - \ddot l sin75.96 -7276.13* cos75.96 *1.7652 = 0$

$-1030.78* cos75.96 * \alpha_{AB} - \ddot l sin75.96 = 7276.13 *cos75.96 * 1.7652 - 1030.78* sin75.96 * 1.7652^2+7276.13* cos75.96 *1.7652$

$-250.07 \alpha_{AB} - 0.9701 \ddot l = 3115.9$

solving the two equations:

$999.99 \alpha_{AB}-0.2426\ddot l = 211760.6743$ and $-250.07 \alpha_{AB} - 0.9701 \ddot l = 3115.9$

$\alpha_{AB} = 198.566rad/s^2 ;a_c = \ddot l = -54397.734 mm/s^2$