Question

A homogeneous cube of mass M and side L is located in a horizontal plane fixed in one of the lower vertex as show as figure. A projectile of mass m and initial Velosidad v inelastically collides at an h height and stopping at the vertical of the center of Mass. Determine the minimum required speed of the projectile to make the cube made a rotation of 90 degrees around the fixed vertex.

(M= 25Kg; m=0.25kg; L = 0.5 m; h= 0.4m)

Answer 1

x_1 = squareroot 2 L/2 = L/ squareroot 2 and Torque about O point due to mg is = z_1 z_1 = x_1 xmg cos 45 z_1 = x_1 mg cos 45 ofter collision m get inside the cube \r\n Torque z_2 due to Mass m z_2 = x_2 times F_2 = > x_2 (mv) - (0)/dt (sin theta_1) st almostequalto 1 z_2 = squareroot L^2 + h^2 mv sin theta_1 for 90 degrees Rotation z_1 > z_1 sin theta_1 squareroot L^2 + h^2 mv > x_1 mg cos 45 Now sin theta_1 = h/ squareroot h^2 + L^2 x_1 = squareroot L^2/4 + L^2/4 = L/ squareroot 2 \r\n h/ squareroot h^2 + L^2 squareroot L^2 + h^2 mv > L/ squareroot 2 Mg cos 45 h mv = > L/ squareroot 2 Mg 1/ squareroot 2u = > L Mg/2mh IF M = 25 kg, h = 0.4m m