The white car car A collides with the yellow car car B the a

The white car (car A) collides with the yellow car (car B) the as shown. After impact, the two cars are locked together as one body. Due to the location of the collision point, the impact causes the two locked cars to spin clockwise.

The weight of car A is 3,000 lb, its speed at impact is 10 mph, and its radius of gyration for this motion is 3 ft. The weight of car B is 3,500 lb, its speed at impact is 15 mph, and its radius of gyration for this motion is 2.5 ft.

The lateral distance between the centers of gravity of the two cars is 1.6 ft, and the longitudinal distance is 14 ft, as shown in the sketch.

Determine the angular velocity of the 2 cars as they start to spin after collision. Ignore friction.

Solution

Y-axis on Gb

X-axis on Ga

Calculate the center of Gravity for the two bodies combined;

X= (MaXa+MbXb)/(Ma+Mb)

=(3000*14+3500*0)/(3000+3500)

=84/13 ft

Y= (MaYa+MbYb)/(Ma+Mb)

=(3000*0+3500*1.6)/(3000+3500)

=11.2/13 ft

Now, Angular momentum before collision: Ra*MaVa + Rb*MbVb

=(11.2/13)*3000*(10*1.47) + (1.6-11.2/13)*3500*(15*1.47) {velocity is converted in ft/s by multiplying 1.47} =493920/13+740880/13 =1234800/13 kg-ft²/s

Now, Moment of inertia of the two car combined about the center of gravity(using parallel axis theorem);

M.I.(I)= M_a*K_a² + M_aD_a² + M_b*K_b² +M_b*D_b²   =3000*3² +3000{(84/13)²+(9.6/13)²} + 3500*2.5² + 3500{(11.2/13)²+(98/13)²} = 27000+ 3000*(7148.16/169) + 21875 + 3500*(9719.44/169) =48875+(55497520/169) =63757395/169 kg-ft²

^{Now, using conservation of angular momentum}

Initial angular momentum= final angular momentum

1234800/13=I*W

=> W=124800/13*I

=>W=1234800/{13*63757395/169}

=>W=(12134800*169)/13*63757395)

=>W=0.25177 radian/sec