Two small objects each of mass m-0.6 kg are connected by a ightweight rod of len

Question

Two small objects each of mass m-0.6 kg are connected by a ightweight rod of length d- 1.8 m (see the figure). At a particular instant they have velocities whose magnitudes are v-40 m/s and v2-69 mý's and are subjected to external forces whose magnitudes are F1 = 52 N and F2 = 34 N. The distance h = 0.3 m, and the distance w = 0.6 m. The system is moving in outer space. (out of page) (a) What is the total (lineer) mometum total , |kg-m/s of this system? (b) What is the velocity of the center of mass? (e) What is the total angular momentum ZA of the system reletive to point A? (d) What is the rotational angular momentum Lot of the system? oxkg-m2/s (e) What is the translational angular momentum trams of the system relative to point A? (f) After a short time interval at-0.21 what is the total linear) momentum total of the system? total kg·m/s

Explanation / Answer

linear momentum ptotal of the system:

P = P1 + P2

= m v1 + m v2 = (0.6 * 40) + (0.6 · 69)

= 65.4 kg·m/s along positive X-axis

b) Vcm = P / M

= 65.4 / 1.2

= 54.5 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):

L(tot) = [mv1 (d+h)] + [mv2 h] = [0.6*40* (1.8+0.3)] + [0.6 *69* 0.3] =
= -62.82kg·m²/s

d)

the center of mass of the system is at

Xcm= m1r1+m2r2/m1+m2

So here m1=m2,

So Xcm=d/2

Now moment of inertia of the objects about center of mass,

M(d/2)2

0.486

and we know that angular momentum is

m(r*v)

So angular momentum due to V1 will be in negative z-direction while due to V2 it will be in postive z direction so net angular momentum will be,

L=0.6*[(69*0.9)-(40*.9)]

=15.66 along the positive z direction

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