A ring and a solid sphere roll down the same | vedclass

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(B) In pure rolling motion,the work done by static friction is zero because the point of contact is instantaneously at rest. By the law of conservatio

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) In pure rolling motion,the work done by static friction is zero because the point of contact is instantaneously at rest. By the law of conservation of energy,the initial potential energy $(PE = mgh)$ is converted into the total kinetic energy $(KE)$ at the bottom of the incline. Since both objects start from rest at the same height $h$,their total kinetic energies at the bottom must be equal,regardless of their moments of inertia. Therefore,$KE_{	ext{ring}} = KE_{	ext{sphere}}$. The ratio of their kinetic energies is $\frac{KE_{	ext{ring}}}{KE_{	ext{sphere}}} = 1$. Given that the ratio is $\frac{7}{x}$,we have $\frac{7}{x} = 1$,which implies $x = 7$.

$A$ ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is . . . . . . .

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