Question

A solid sphere of radius r `r` is rolling on a horizontal surface. The ratio between the rotational kinetic energy and total energy.
A. `(5)/(7)`
B. `(2)/(7)`
C. `(1)/(2)`
D. `(1)/(7)`

Answer 1

Correct Answer - B
The rotation kinetic energy, `K_(r)=(1)/(2)Iomega^(2)`
where, `I=` moment of inertia of solid sphere. The total kinetic energy and translational kinetic energy
`K_(T)=(1)/(2)mv^(2)+(1)/(2) Iomega^(2)`
`=(1)/(2)m omega^(2)r^(2)+(1)/(2) I omega^(2)" "( :. "For pure rolling" v=omegar)`
Now, `(K_(r))/(K_(T))=((1)/(2)I omega^(2))/((1)/(2) omega^(2)(mr^(2)+I))=(I)/(mr^(2)+I)`
As `I=(2)/(5)mr^(2)` (for solid sphere)
`rArr(K_(r))/(K_(T))=((2)/(5)mr^(2))/(mr^(2)+(2)/(5)mr^(2))=((2)/(5)mr^(2))/((7)/(5)mr^(2))=(2)/(5)xx(5)/(7)=(2)/(7)`.