Question

The particle of weight W is guided along the circular path with slotted arm guide. If the arm has an angular velocity $\Theta {}'$ and angular acceleration $\Theta {}''$ at the time $\Theta =\Theta _{1}$ , determine the force of the guide over the particle.The movement occurs in the horizontal plane.

Answer 1

I have answered this question but with the values provided :

wieght of particle : 0.5 lb

angular velocity = 4 rad/s

angular acceleration = 8 rad/s ²

I have answered this question but with the values provided: weight of particle: 0.5 lb angular velocity = 4 rad/s angular acceleration = 8 rad/s^2 r = 2 (0.5 cos theta) = 1 cos theta r = -sin theta theta vector r = -cos theta theta^2 - sin theta theta At theta = 30 degree, theta = 4 rad/s and vector theta = 8 rad/s^2 r = 1 cos 30 degree = 0.8660 ft r = -sin 30 degree (4) = -2 ft/s r = -cos 30 degree (4)^2 - sin 30 degree (8) = -17.586 ft/s^2 a_r = r - r theta^2 = -17.586 - 0.8660 (4)^2 = -31.713 ft/s^2 a_theta = r theta + 2r theta = 0.8660 (8) + 2 (-2) (4) = -9.072 ft/s^2 North east arrow + sigma F_r = ma_: -N os 30 degree = 0.5/32.2 (-31.713) N = 0.5686 lb + North west arrow sigma F_A = ma_A: F - 0.5686 sin 30 degree = 0.5/32.2 (-9.072) F = 0.143 lb

answer)