Asked by Lancelot
In the manufacture of a cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. Then cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed omega about the axis of the shaft.
Use the parallel axis theorem
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
to compute the moment of inertia about the new axis. Call it I'. The parallel axis theorem tells you how to relate I' to the moment of inertia about the center of mass, which is
I = (1/2) M R^2
In your case, the parallex axis theorem says that
I' = I + M (R/2)^2 = (3/4)MR^2
Finally, the kinetic energy is
(1/2) I' w^2
where w is the angular velocity, "omega"
I see now, Thanks.
Use the parallel axis theorem
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
to compute the moment of inertia about the new axis. Call it I'. The parallel axis theorem tells you how to relate I' to the moment of inertia about the center of mass, which is
I = (1/2) M R^2
In your case, the parallex axis theorem says that
I' = I + M (R/2)^2 = (3/4)MR^2
Finally, the kinetic energy is
(1/2) I' w^2
where w is the angular velocity, "omega"
I see now, Thanks.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.