If you arent going to do all 4 parts then dont do it at all.. Two small objects

Question

If you arent going to do all 4 parts then dont do it at all..

Two small objects each of mass m 0.8 kg are connected by a lightweight rod of length d1.5 m (see the figure). At a particular instant they have velocities whose magnitudes are V1 27 m/s and v2-67 m/s and are subjected to external forces whose magnitudes are F1 68 N and F2 30 N. The distance h0.4 m, and the distance W 0.8 m. The system is moving in outer space Fi z (out of page) 2 F2 AX (a) What is the total (linear) momentum total of this system? total = | ? 75.2.0.0 > (b) What is the velocity Vcm of the center of mass? -cm-1 ? 47,0,0 > m/s kg-m/s

Explanation / Answer

From the given data

m = 0.8 kg each

d = 1.5 m

light weight rod

v1 = 27 m/s

v2 = 67 m/s

F1 = 68 N

F2 = 30 N

h = 0.4 m

w = 0.8 m

a. total linear momentum = along x direction (as velocities are along x axis)

hence

P = <m(v1 + v2), 0,0> = <0.8(27 + 67), 0, 0> = <75.2,0,0> kg m/s

b. vcm = along x direction, as there are no velocities in any other direction

hence

vcm = <mv1 + mv2, 0, 0>/<2m> = <(v1 + v2)/2, 0, 0> = <47,0,0> m/s

c. total anhular moemntum relative to point A = along z direction

La = <0,0,-m(v1*(d + h) + v2*h)> = <0,0,-0.8(27(1.9) + 67*0.4)> = <0,0,-62.48> kg m^2/s

d. rotational angyular momentum = Lrot

Lrot = along z axis

hence

Lrot = <0,0,-(mv1*d/2 - mv2*d/2)> = <0,0,(0.8*(67 - 27 )*1.5/2)> = <0,0,24> kg m^2/s

e. translational angular momentum about point A

Ltr = along -z axis

Ltr = <0,0,-2m*vcm*(h + d/2)> = <0,0,-2*0.8*47*(0.4 + 1.5/2)> = <0,0, -86.48> kg m^2/s

f. after dt = 0.18 s

Ptotal = Pi + Fnet*dt = <75.2,0,0> + <(F1 - F2)*0.18, 0, 0> = <82.04, 0, 0, > kg m/s

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