A loop rolls down on an inclined plane .the fraction ot its kinetic energy that is associated with only the rotational motion is
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Rotational kinetic energy is
(KE)r =1/2 Iw2 =1/2{1/2MR2}
=1/4 MR2w2 where(I=1/2MR2)
Where Mis the mass of the disc and R radius ,
Translational kinetic energy is
(KE)r = 1/2 Mv2 =1/2 m(Rw)2 =1/2MR2 w2 where (V=Rw)
Total energy KE =(KE)r +(KE)r =3/4Mr2 W2
Hence (KE)r/KE =2/3
(KE)r =1/2 Iw2 =1/2{1/2MR2}
=1/4 MR2w2 where(I=1/2MR2)
Where Mis the mass of the disc and R radius ,
Translational kinetic energy is
(KE)r = 1/2 Mv2 =1/2 m(Rw)2 =1/2MR2 w2 where (V=Rw)
Total energy KE =(KE)r +(KE)r =3/4Mr2 W2
Hence (KE)r/KE =2/3
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2:3 is the correct answer
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