Asked by Justin
What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of MR2, where M is the mass of the object and R is its radius.
Answers
Answered by
Elena
PE=KE=KE(transl) + KE(rot).
mgh = mv²/2 + I ω²/2=
= mv²/2 + I v²/2R².
I= (2R²/v²)(mgh- mv²/2)=
=mR²(2gh/v² -1) =
= mR²(2•9.8•2/36 -1)=
=mR²(1.09-1)=0.09 mR²
mgh = mv²/2 + I ω²/2=
= mv²/2 + I v²/2R².
I= (2R²/v²)(mgh- mv²/2)=
=mR²(2gh/v² -1) =
= mR²(2•9.8•2/36 -1)=
=mR²(1.09-1)=0.09 mR²
Answered by
A.nally P.enetrating Physics
The answer's actually 12.
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