A uniform solid sphere is rolling on a horizontal
surface. The ratio of rotational kinetic energy to
the total kinetic energy of the sphere is
(142:7
(2) 5:7
(3) 1:6
(4) 3:8
Answers
Answered by
1
Answer:
2/5ma^2=I
mgh=(1/2mv^2)+(1/2I w^2)
mgh=(1/2mv^2)+(1/2×2/5ma^2 ×V^2/a^2)
So mgh=7/10 mv^2=whole kinetic Energy
Rotaional Kinetic Energy=1/5 mv^2
so R/W=(1/5)/(7/10)=2:7
Answered by
0
2:7
Explanation:
Rotational kinetic energy of a solid sphere rolling is given by
K =Iω²
where, I = MR² is the moment of inertia of the solid sphere
ω = angular speed of the solid sphere
∴ K = (MR²)ω²
K = Mv² ...................equation(1)
Total kinetic energy of the solid sphere is given by
E =Mv² + Iω²
E = Mv² + Mv² [∵Iω²= K = Mv²........using eq(1)]
E = Mv² ................eq(2)
Dividing eq(2) by eq(1)
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