Two small objects each of mass m = 0.5 kg are connected by a lightweight rod of

Question

Two small objects each of mass m = 0.5 kg are connected by a lightweight rod of length d = 1.8 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 38 m/s and V2 = 50 m/s and are subjected to external forces whose magnitudes are F1 = 73 N and F2 = 31 N. The distance h = 0.2 m, and the distance w 0.7 m. The system is moving in outer space. z (out of page) 1m U2 - m = | m/s (c) What is the total angular momentum LA of the system relative to point A? LA = 0,0,-43 kg. m2/s (d) What is the rotational angular momentum Lrot of the syste? Lrot (e) What is the translational angular momentum Itrans of the system relative to point A? trans = | 0,0,-36 | kg.m2/s

Explanation / Answer

a) linear momentum ptotal of the system:
........................
P = P1 + P2 = m v1 + m v2 = (0.5*38) + (0.5*50) = 44 kg·m/s along positive X-axis
...........
b) Vcm = P / M = 44/1.0 = 44 m/s along positive X-axis

c) total angular moment relative A:
....................................
L = L1 + L2 = (r1 x mv1) + (r2 X mv2)
Taking into account the sin of the angles in the cross product and the geometry it results:

r1 sin1 = d+h
r2 sin2 = h

then L (along negative Z-axis):
Ltot = [mv1 (d+h)] + [mv2 h] = - [0.5*38* (1.8+0.2)] + [0.5*50*0.2] =
= -43 kg·m²/s

e) (Let's do first the translational angular momentum).
.....................
Ltrans = Rcm X MVcm

We need Rcm
..................
Rcm = m r1 + m r2 / M

as the two masses are equal, Rcm is easy to calculate because it is in the rod center d/2 with coordinates:

Rcm = (d/2+h, w) = (1.1, 0.7) m
Then, taking into account the cross product Rcm X Vcm:
Ltrans = M Vcm · (d/2+h) = 1.0*44*1.1 = -48.4 kg·m²/s along negative Z-axis.

d) ............
Ltot = Lrot + Ltrans

Then:
Lrot = Ltot - Ltrans = (-43) - (-48.4) = 5.4 kg·m²/s along positive Z-axis

(that seems correct because v2 > v1 and the system turns counter-clockwise with respect to its cm (along positive Z-axis)

[It is easy to verify that this result is correct. As Vcm = 44, top mass has a speed of 38 - 44 = -6 m/s (negative X-axis) and bottom mass has a speed of 50 - 40 = 6 m/s (positive X-axis). The lightweight rod will turn with a angular velocity given by = 6/(d/2) = 6/0.9 = 6.7 rad/s. Now Lrot = I, where the inertia moment I of the rod with respect to its center is: I = 2m(d/2)² = 0.81 kgm².
Then Lrot = I = 5.4 kg·m²/s as obtained with the previous procedure]

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