Two small objects each of mass m 02 kg are connected by a l

Two small objects each of mass m = 0.2 kg are connected by a lightweight rod of length d = 0.8 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 39 m/s and v2 = 65 m/s and are subjected to external forces whose magnitudes are F1 = 47 N and F2 = 26 N. The distance h = 0.3 m, and the distance w = 0.5 m. The system is moving in outer space.

(a) What is the total (linear) momentum total of this system? total = kg·m/s

(b) What is the velocity cm of the center of mass? cm = m/s

(c) What is the total angular momentum A of the system relative to point A? A = kg·m2/s

(d) What is the rotational angular momentum rot of the system? rot = kg·m2/s

(e) What is the translational angular momentum trans of the system relative to point A? trans = kg·m2/s

(f) After a short time interval t = 0.23 s, what is the total (linear) momentum total of the system?

Solution

a)

Total Linear momentum of the system is

P_total=mV₁+mV₂ = 0.2(39+65)

P_total=20.8 Kg-m/s

b)

The velocity of the center of mass is

V_cm=P_total/(m+m) =20.8/2*0.2

V_cm=52 m/s

c)

The total angular momentum A of the system relative to point A is

L_total=-mV₁(d+h)-mV₂h =-0.2[39*(0.8+0.3)+65*0.3]

L_total=-12.48 Kg-m²/s

d)

the rotational angular momentum rot of the system is

L_rot=-mV₁(h+d/2)-mV₂)(h-d/2)

L_rot = -0.2*39*(0.3+0.4)-0.2*65*(0.3-0.4) =-4.16 Kg-m²/s

e)

the translational angular momentum trans of the system relative to point A

L_trans=L_total-L_rot = -12.48+4.16 =-8.32 Kg-m²/s

f)

the total (linear) momentum total of the system

P_system=mV₁+F₁t+mV₂-F₂t =0.2(39+65)+(47-26)*0.23

P_system=25.63 Kg-m/s