Physics, asked by thulasinarayan123, 11 months ago

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 ArjunPartha
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 babundrachoubay123
0

2:7

Explanation:

Rotational kinetic energy of a solid sphere rolling is given by

K =\frac{1}{2}Iω²

where, I = \frac{2}{5}MR² is the moment of inertia of the solid sphere

ω = angular speed of the solid sphere

∴ K = \frac{1}{2}\times \frac{2}{5}(MR²)ω²

K =  \frac{1}{5}Mv²  ...................equation(1)

Total kinetic energy of the solid sphere is given by

E =\frac{1}{2}Mv² + \frac{1}{2}Iω²

E = \frac{1}{2}Mv² + \frac{1}{5}Mv²       [∵\frac{1}{2}Iω²= K = \frac{1}{5}Mv²........using eq(1)]

E = \frac{7}{10}Mv²  ................eq(2)

Dividing eq(2) by eq(1)

\frac{E}{K} = \frac{2}{7}

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