A cylinder with its mass concentrated toward the center has a moment of inertia of 0.2MR².If this cylinder is rolling without slipping along a level surface with a linear speed v,what is the ratio of its rotational kinetic energy to its linear kinetic energy? A)1/l0 B)1/5 C)1/2 D)1/1

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Quizplus · Accepted Answer

The rotational kinetic energy ($K_{rot}$) of an object is given by $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. The linear kinetic energy ($K_{lin}$) is given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the linear velocity. For a cylinder rolling without slipping, $v = R\omega$, where $R$ is the radius. Given $I = 0.2mR^2$, substituting $R\omega = v$ into the rotational kinetic energy formula gives $K_{rot} = \frac{1}{2}(0.2mR^2)\left(\frac{v}{R}\right)^2 = \frac{1}{2} \cdot 0.2mv^2 = 0.1mv^2$. Therefore, the ratio of rotational to linear kinetic energy is $K_{rot}/K_{lin} = (0.1mv^2) / (0.5mv^2) = 1/5$.

A Cylinder with Its Mass Concentrated Toward the Center Has