A 1.7-kg hoop rolls across the floor without slipping.The hoop has a diameter of 1.1 m.The ratio the translational to rotational kinetic energy of the hoop is
A)0.50.
B)0.75.
C)1.0.
D)1.8.
E)Insufficient information is provided to determine this ratio.

Question

Quizplus · Accepted Answer

The ratio of translational to rotational kinetic energy for a hoop rolling without slipping is 1.0. This is because the translational kinetic energy ($K_{trans} = \frac{1}{2}mv^2$) and the rotational kinetic energy ($K_{rot} = \frac{1}{2}I\omega^2$) are equal when considering a hoop, where $I$ (moment of inertia for a hoop) is $mr^2$ and $\omega$ (angular velocity) is related to $v$ (linear velocity) by $v = r\omega$. Thus, $K_{trans} = K_{rot}$, making their ratio 1.0.

A 17-Kg Hoop Rolls Across the Floor Without Slipping