Safarl File Edit View History Bookmarks Window Help 72% Mon 9:47 AM aE 11.8: Rot

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Safarl File Edit View History Bookmarks Window Help 72% Mon 9:47 AM aE 11.8: Rotational and Total Argular 4 112 027 Two small objects each of mass m - 0.7 kg are connected by a lightweight rod of length d-0.7 m (see the figure). At a particular instant they have velocities whose magnitudes are vs-36 m/s and v2 - so m/s and are subjected to external forces whose magnitudes are Fs -73 N and F2-29 N. The distance h 0.4 m, and the distance w-0.8m. The system is moving in outer space (out of page) (a) What is the total (linear) momentum total of this system? total kg·m/s (b) What is the velocity om of the center of mass? mys (e) What is the total angular momentum IA of the system relative to point A? IA kg-m2/s

Explanation / Answer

a) linear momentum ptotal of the system:
........................
P = P1 + P2 = m v1 + m v2 = (0.7 * 36) + (0.3 * 50) = 40.2 kg·m/s along positive X-axis
...........
b) Vcm = P / M = 40.2 / 1.4 = 28.7 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] = - [0.7 * 36 · (.7+0.4)] + [0.3 * 50 * 0.4] = - 33.72 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) = (0.75, 0.8) m

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

Ltrans = M Vcm · (d/2+h) = 1.4 *28.7  * (0.75) = - 30.13 kg·m²/s along negative Z-axis.

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

Then:

Lrot = Ltot - Ltrans = (- 33.72 ) - (- 30.13)= -3.59 kg·m²/s along -ve Z-axis

[It is easy to verify that this result is correct. As Vcm =28.7 , top mass has a speed of 36 - 28.7 = 7.3 m/s (+ve X-axis) and bottom mass has a speed of 50 - 28.7 = 21.3 m/s (positive X-axis). The lightweight rod will turn with a angular velocity given by =21.3/(d/2) = 21/0.35 = 60 rad/s. Now Lrot = I, where the inertia moment I of the rod with respect to its center is: I = 2m(d/2)² = 0.171 kgm².
Then Lrot = I = 10.29 kg·m²/s

f) F · t = P

where F is the net force:
.......
F = F1 + F2 = 73 - 29 = 44 N (along positive X-axis)

44 * 1.5 = 66 kg·m/s

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

Ptot = 40.2 + 66= 106.2 kg·m/s (along X-axis)

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