Question

The ABC bar has a mass of 2.80kg and is attached to a B-bolt. A 0.93kg D-sphere hits the end of the bar with a vertical velocity v1 of 2.64m / s. If L = 0.67m and the coefficient of restitution between the sphere and the bar is 0.40, determine the angular velocity, in rad / s, of the bar immediately after impact.

L 4 V1 D А С -L-

Answer 1

Solution » 니다 Vi TO D B C given - mass of bar ABC L urt after Impact _ M= 2.8 kg mass of apere D = 0.93 kg Vi = 2.64 m/ L = m. 44 0.67 m B C Coefficient of restitution e=0.40 w v momentum conservation equation - ( About point B) Initial momentum final momentum IW, + I Winital Is Wz t I beam w beam - "O ده here ball D Just before callkion is just like rentating around pornt 'B' in wide of radius 4/4 which have linear velocity vi Vi = 2, (44) W, = 15.76 a kec lo 4vi/ rad -

Similarly alter callkion ball I have linear velbuty V2 (which is a garu srumpetron) w3() W, = 412 ro V2 = L initial angular velauty of bar is zero Winitial And after impact we arrume that bor ABC will have angular velouty of w about 'B' Joint. pont "A will have linear velouty w w xo 4V L coefficient of restitution - velouty of reparation velouty of approach V - V2 Vi - 0 0.4 V-V2 2.64 V-V2 = 1.056 V2 = V-1056 mis

V2= V-1.056 26.) W (%) -1.056 W2 = W124) -11056 (44) W2 = [:L L= 0.67 m 아온서 x 1.056 D2= W-6.304 sadhoo 0.93 → mass moment of Inertia of Ball D Id = m(u)? I, = 0.026 (1989) Kg-m? mass moment of Inerta of Beam ABC about pont B I beam + M(5) 12 2 2 ME? Parallel с A anus theorem В 2.8 (0.6) ; 2-8 (642 (2 니다 4 "I bean 0.1831 kg-m? Ꮗ → putting all these values in eqo ID W. +o IW - 6:304) + I beam Ig @, + Ip¥(6'304) w Ist I bearm = 2: 744 rod/sec a Angular velocity of bearm Just after impact is 2.744 rocks Aug