Knowing that crank AB rotates about point A with a con stant

Knowing that crank AB rotates about point A with a con- stant angular velocity of 900 rpm clockwise, determine the acceleration of the piston P when = 65°. 6 in in.

Solution

Solution:-

_AB = 900 rpm = 94.248 rad/s

= 65

AB = 2 inch = 50.8 mm

BD = 6 inch = 152.4 mm

We know that

Sin/(50.8) = sin65/(152.4)

= 17.57 degree

V_{B =} r_AB * _AB = 0.05 * 94.248 = 4.71 m/s

a_B =r_AB * (_AB)² = 0.05 * (94.248)² = 441.1 m/s²

Rod BD

(V_D)_x = (V_B)_x +(V_D/B)_x

0 = 4.712cos65 – (0.152)* _BD*cos(17.57)

_BD = 13.74 rad/s

(a_D/B)_t = 0.152 * a_BD

(a_D/B)_n = 0.152 * (13.74)² = 28.7 m/s²

a_D = a_B + a_D/B

a_D = a_B + (a_D/B)_t + (a_D/B)_n

Horizontal components:-

0 = 441.1cos25 – (0.152)*a_B* cos (17.57) – 28.7 sin(17.57)

a_B = 1699 m/s²

vertical components:-

a_D = 441.1* sin25 - (0.152) *1699 *sin(17.57) + 28.7 *cos (17.57)

Acceleration( a_D ) = 89.9 m/s²      Answer