At time t = 6.5 sec, a particle of mass M = 1.5 kg is at the position (x,y,z) =

Question

At time t = 6.5 sec, a particle of mass M = 1.5 kg is at the position (x,y,z) = (4,4,6) m and has velocity (2,1,-2) m/s.

1) Finally a thin rod of mass 7 kg is added between the two particles Since the particles are moving perpendicular to the line separating them and in opposite directions (you should convince yourself that this is true), then the particles + rod system rotates about its center of mass. What the angular frequency of rotation of the rod + particles system? (Note: The particles are moving perpendicular to the line separating them and in opposite directions (you should convince yourself that this is true), then the particles + rod system rotates about its center of mass.)

2) What is the magnitude of the total angular momentum of the system (particles plus rod)?

Explanation / Answer

1)conservation of angular momentum
Lmag initial = sqrt(42^2 + 60^2 + 12^2) = ( 2 mr^2 + 1/12 M L^2) w

sqrt(42^2 + 60^2 + 12^2) = ( 2*1.5*(4^2+4^2+6^2) + 1/12*7*(8^2+8^2+12^2) w
w=0.205

2 )Lmag = sqrt(42^2 60^2 12^2)=74.22

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.