Two small objects each of mass m = 0.2 kg are connected by a lightweight rod of

Question

Two small objects each of mass m = 0.2 kg are connected by a lightweight rod of length d = 0.9 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 26 m/s and v2 = 63 m/s and are subjected to external forces whose magnitudes are F1 = 52 N and F2 = 25 N. The distance h = 0.2 m, and the distance w = 0.6 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 = 0.17 s, what is the total (linear) momentum total of the system? total = kg·m/s

Explanation / Answer

a) linear momentum ptotal of the system:
........................
P = P1 + P2 = m v1 + m v2 = (0.2 · 26) + (0.2 · 56) = 16.4 kg·m/s along positive X-axis
...........
b) Vcm = P / M = 16.4 / 0.4 = 41 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] = - [0.2 · 26 · (0.9+0.2)] + [0.2 · 63 · 0.2] =
= 5.72 + 2.52 = - 8.24 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) = (0.4, 0.6) m

Then, taking into account the cross product Rcm X Vcm:

Ltrans = M Vcm · (d/2+h) = 0.4 · 41 · (0.4) = - 6.56kg·m²/s along negative Z-axis.

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

Then:

Lrot = Ltot - Ltrans = (- 8.24) - (- 6.56) = 1.68 kg·m²/s along positive Z-axis

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

[It is easy to verify that this result is correct. As Vcm = 41, top mass has a speed of 26 - 41 = - 15 m/s (negative X-axis) and bottom mass has a speed of 63 - 41 = 22 m/s (positive X-axis). The lightweight rod will turn with a angular velocity given by = 0.6/(d/2) = 0.6/0.45 = 1.33 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 =1.07 kg·m²/s as obtained with the previous procedure]

f) F · t = P

where F is the net force:
.......
F = F1 + F2 = 52 - 25 = 27 N (along positive X-axis)

27 · 0.17 = 4.59 kg·m/s

We must add the previous P calculated in a), then

Ptot =16.4 + 4.59 = 20.99 kg·m/s (along X-axis)

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