Question

(30 Pints) The 10-kg bar is released from the horizontal position when it is rested. The coefficient of restitution between the bar and bumper B is 0.5, and the duration of the impact is 5 ms. Determine the maximum angle of inclination of the bar subsequent to the impact and the average force exerted by the bumper during the impact. (you can assume that g 10 m/s?) 12 m 1.6m

Answer 1

According to the above given Question by the following solution as given that here implies bar in Horizontal position is denoted with Ac^1 Where at vertical position is denoted with Ao the point 'o' and o re the position of rightarrow the bar of horizontal and vertical positions Respectively And A o^1 = A o = 0.8 m Now implies when 'n' being relaxed from horizontal position the bar a chines the vertical position the potential energy convert into kinetic energy suppose just before impact the velocity of the bar is v_1 the potential energy = kinetic energy mgh = 1/2 mv^2_1 V_1 = squareroot 2gh V_1 = 2gh = squareroot 2 times 10 times 0.8 m/s = 4 m/s m = mass of the bar 10kgs h = height of cg the horizontal position w.r. to the vertical position = 0.8m g = 10m/s^2 note the (point E) of bar which collides with the bumper is at B