Problem 9-4: Two semi-circular rings of mass mu and mp are rigidly connected to

Question

Problem 9-4: Two semi-circular rings of mass mu and mp are rigidly connected to form an mb 75 kg. unbalanced ring of radius R as shown in the picture below. Take R 1 m, m 60 kg, and (a) Determine the center of mass of the unbalanced ring ia xi t yi + zo k, where (b) Determine the radius of gyration k of the unbalanced ring. Show and interpret k note that zG0 pictorially If the ring is released from rest, then determine its angular acceleration and acceleration of the point B for two cases given below in part (c) and part (d). Take -0.3, and k = 0.25. (c) = 25° [Note: In setting up the rotational equation of motion (ie, the moment equation), take moment about the center of mass of the unbalanced ring.] (d) = 40° (e) For part (c), set up separately the rotational equation of motion of the ring about point of contact A, and show that this will result in the same results as in part (c) lu 84

Explanation / Answer

R = 1 m

mu = 60 kg

mb = 75 kg

a. center of mass of the unbalanced ring = (x,y)

x = 0 from symmetry

y = (mu*2R/pi - mb*2R/pi)/(mu + mb)

y = -0.0707355 m

hence

location of com = (0,-0.0707355,0)

b. radius of gyration of the ring = k

moment of inertia about the center, Ic = (mu + mb)R^2

but Ic = I + my^2

and I = (mu + mb)k^2

(mu + mb)k^2 + (mu + mb)*0.0707355^2 = (mu + mb)R^2

k = 0.9974951 m

c. beta = 25 deg

friciton force = f

from force balance

(mu + mb)gsin(theta) - f = (mu + mb)a

from torque balance

f*(R - 0.0707355) = (mu + mb)k^2 *alpha

but alpha = a/R

hence

f = 144.54929a

hence

a = 2.002131 m/s/s

alpha = 2.002131 rad/s/s

acceleration of point B = 2*a = 4.004 m/s/s

d. for beta = 40

a = 3.04517 m/s/s

alpha = 3.04517 rad/s/s

acceleration of point B = 2*a = 6.0903447 m/s/s

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