anonymous
  • anonymous
Two identicle masses in space connected by a massless cord rotate about their center of mass. Cut the cord and both masses take off in straight lines with their last instantainious velocity. Plainly the system initially had angular momentum. Where does it go when the cord is cut? How is it conserved?
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Whether the particles go in a circle or fly away in a rectilinear motion, the angular momentum is the same.
anonymous
  • anonymous
Because the masses are connected by a cord and always present the same face to each other, each mass is already rotating about its axis at the same angular velocity as the system. The moment of inertia of each mass is much smaller than that of the system. In order for angular momentum to be conserved by rotation of the individual masses their angular velocities would have to increase by a lot. is this the case?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
No it isn't. Your question clearly mentions point-masses, that do not have rotational angular momentum. In this problem, only orbital angular momentum is at stake. |dw:1366300044926:dw| The angular momentum is 2mvR when the masses are connected. It is also 2mvR when they are moving away.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Ok then, what is carrying the angular momentum? What is spinning or being held in a circle? How does the angular momentum manifest itself in the new system? Thanks for your efforts at trying to explain this to me. I appreciate it.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
You do not need to spin or to be held in a circle to have angular momentum relative to a point. Just just need to have : - non zero linear momentum - with line of motion not passing through the point about which the angular momentum is expressed.
anonymous
  • anonymous
Ok, that is a new concept to me. It was never covered in any physics course I have taken-Thanks
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Watch Walter Lewin's presentation. This is how angular momentum is defined.
mos1635
  • mos1635
|dw:1366572987757:dw| IS that the circle that hamiltdw was looking for??
mos1635
  • mos1635
angular momentum = 2*m*R'*Vy angular momentum = 2*m*(R^2+d^2)^1/2 *V*sinθ angular momentum = 2*m*(R^2+d^2)^1/2 *V*(R/(R^2+d^2)^1/2) angular momentum = 2*m*R= constant
mos1635
  • mos1635
|dw:1366573655047:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.