Question

The 600 mm long homogeneous slender bar OA has a mass of 8 kg and is supported in the horizontal position by the cable at A and the corner hinge at O. The cable at A is suddenly released. Just before hitting the incline, the bar has an angular velocity of ω= 5 rad / sec. Calculate the angle θ. Neglect the curvature of the path of the bar mass center when the bar is falling and assume that the mass center follows a vertical path.

Answer 1

Given 0 L = 600 mm = o.6m. bar sec. m - Org w (Enitial Angular velocity) = o rand eve (final Angular velocity) s rove when cable is suddenly released then it roture torque (T) see اہم 0 du T Io when inertia ng im =w dw de TT cit moment of about o a a= Angular accelertation dw. da dw do at do do dt Now Ta ŽML?. wew do Also, Torque (T) force x perpendicular distance To mg.d Lws & il Now equate equation equation © and こ ing wl mca. wdw do demo mg. I coso wqw 3g 2 L coso.do TT 11 Now integrate Equation in u Z I wow a 3g coso do 21 ci initial angle to 3 - from the horizontal final an ungle a

2 5 3g (sino). W2 21 2. 3x9,81 sind g ? (25) ( 9 = 9.01 m/see?) 25 3x9.01 sino - .6 Answer 3 30.64 PLEASE Rate the Answer guys.

Answer 2

= 600 mm bar 0 A Clar 0 I sino Length of bar=L by Energy conservation between position (1) and Positiori (2) mg 1 sine 12 m v2 refererice - CO W O o of L sino 2 sino = Va 2 Im mg 2 sino = 2 Lw2 4g sino = 0.3822 o 220470 0.645? 4 x 9.81 reforence (2)

Answer 3

Answer 4

= 600 mm bar 0 A Clar 0 I sino Length of bar=L by Energy conservation between position (1) and Positiori (2) mg 1 sine 12 m v2 refererice - CO W O o of L sino 2 sino = Va 2 Im mg 2 sino = 2 Lw2 4g sino = 0.3822 o 220470 0.645? 4 x 9.81 reforence (2)