Question

Two small objects each of mass m = 0.2 kg are connected by a lightweight rod of length d = 1.5 m (see the figure). At a particular instant they have velocities whose magnitudes are v_1 = 36 m/s and v_2 = 50 m/s and are subjected to external forces whose magnitudes are F_1 = 63 N and F_2 = 26 N. The distance h = 0.4 m, and the distance w = 0.6 m. The system is moving in outer space. What is the total (linear) momentum P_total of this system? What is the velocity v cm of the center of mass? What is the total angular momentum L_A of the system relative to point A? What is the rotational angular momentum L_rot of the system? What is the translational angular momentum I_trans of the system relative to point A? After a short time interval Delta t = 0.22 s, what is the total (linear) momentum P_total of the system?

Answer 1

Mass of the object, m- 0.2 kg Length of the rod, d-1.5 m Magnitude of the velocity, หุ่-36 m/s, v-50 m/s Magnitude of the external force, F 63 N, F, -26 N Distance, h=0.4 m, w=0.6 m a) Total linear momentum of the system, 0.2x(36+50) 17.2kg m/s b) Velocity of the center of mass, m+ m 17.2 0.2+0.2 - 43m/s c)

Total angular momentum about the point A, = myri sin g +myr, sin a Consider the diagram, rsing=dth =1.5+0.4 = 1.9 m 0.4 m Substituting し=(0.2×36x19)-(0.2×50×0.4) =17.68 kg m/s The direction is along the negative z - axis. Hence total angular momentum relative to the point A, L=-17.68kg m2/s d) Consider the expression for total angular momentum,

Translational angular momentum, The distance is given by, 1.5 0.4 Substituting, 1.15x0.4x43 19.78kgm'/s (along -ve z axis)

Now equation (1) implies -17.68-L +-19.78 Hence the rotational angular momentum, L,,-1978-1768 =2.1kgm /s