Asked by Alie
A spherically symmetric object with radius of .7m and mass of 1.6kg rolls without slipping accross a horizontal surface with velocity of 1.7m/s. It then rolls up an invline with an angle of 28degrees and comes to rest a distance d of 2.3m up the incline, before rolling back down. Find the moment of inertia of this object about an axis through its center of mass.
Isn't I just= mr^2?
Isn't I just= mr^2?
Answers
Answered by
bobpursley
No, not for a solid sphere.
You can find it from the experiment.
Conservation of energy:
final energy= energy at top of incline
1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28
but w=v*r, so
1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28
Solve for I.
You can find it from the experiment.
Conservation of energy:
final energy= energy at top of incline
1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28
but w=v*r, so
1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28
Solve for I.
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.