A thin, uniform, 3.80-kg bar, 80.0 cm long, has very small 2.50-kg balls glued on at either end (?Fig. P10.57?). It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the bar. (a) Find the angular acceleration of the bar just after the ball falls off. (b) Will the angular acceleration remain constant as the bar continues to swing? If not, will it increase or decrease? (c) Find the angular velocity of the bar just as it swings through its vertical position.

Solution 63P Step 1 of 6: Given data, Mass of bar, M=3.8 kg Length of rod, L=80cm=0.8 m Mass of the ball, m=2.5 kg To find the new center-of-mass relative to where the axle passes through the bar. L That is, x = m2 …………….1 cm M+m The moment of inertia of the bar and ball: I = Ibar+ Iball I = ML2 + m( ) 2 12 2 ML2 mL2 I = 12 + 4 ……………….2 Step 2 of 6: The torque applied on the bar will be due to the weight of the bar and ball. That is, = Fr Using = I, r =x cm and F=(M+m)g I = (M + m)g(x cm Where is torgue, is angular acceleration, I is moment of inertia. Step 3 of 6: (a) Find the angular acceleration of the bar just after the ball falls off. Using equation 1 and 2 in above equation, ML 2 mL 2 m 2 [ 12 + 4 ] = (M + m)g( M+m ) L2 M L 4 [3 + m] = gm 2 L[ M +] = g 2 3m 2g = M L[3m+] Substituting g=9.8m/s , L=0.8m, M=3.8 kg and m=2.5 kg 2 = 2(9.(3.8 kg) (0.8m)3(2.5 kg) = 16.3 rad/s 2 2 Therefore, the angular acceleration of the rod is 16.3 rad/s .