A light, rigid rod of length I = 1.00 m joins two particles, with masses = 4.00

Question

A light, rigid rod of length I = 1.00 m joins two particles, with masses = 4.00 kg and m2 = 3.00 kg, at its ends. The combination rotates in the xy plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 5.20 m/s. A 1.30-kg particle moves in the xy plane with a velocity of v = (4.30 - 3.20 ) m/s. Determine the angular momentum of the particle about the origin when its position vector is A uniform solid disk of mass m = 3.03 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 5.90 rad/s. Calculate the magnitude of the angular momentum of the disk when the axis of rotation passes through its center of mass, What is the magnitude of the angular momentum when the axis of rotation passes through a point midway between the center and the rim?

Explanation / Answer

Angular momentum of particle=mxp=mrv
total angular momentum=4*0.5*5.2+3*0.5*5.2=18.2
direction: positive z direction

Angular momentum =rxp
r=1.5i+2.2j
p=1.3(4.3i-3.2j)=5.59i-4.16j
L=-18.538j

a)moment of inertai of Disk=0.5*MR^2
=0.5*3.03*0.2^2=0.0606
w=8.9 rad/s
L=Iw=0.0606*8.9=0.53934

b)0.53934

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