A hoop of mass 50 kg rolls without slipping.If the center-of-mass of the hoop has a translational speed of 4.0 m/s,the total kinetic energy of the hoop is
A)0.20 kJ
B)0.40 kJ
C)1.1 kJ
D)3.9 kJ
E)None of these is correct.

Question

Quizplus · Accepted Answer

The total kinetic energy of a rolling object (without slipping) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy ($K_{trans}$) is given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity of the center of mass. The rotational kinetic energy ($K_{rot}$) about the center of mass for a hoop is given by $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the hoop about its center ($I = mr^2$ for a hoop) and $\omega$ is the angular velocity. Since the hoop rolls without slipping, $v = r\omega$, where $r$ is the radius of the hoop.Given: $m = 50\,kg$, $v = 4.0\,m/s$1. Calculate $K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times (4.0)^2 = 400\,J$2. To find $K_{rot}$, we use $v = r\omega$ to express $\omega$ as $\omega = \frac{v}{r}$. Then, $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times (mr^2) \times \left(\frac{v}{r}\right)^2 = \frac{1}{2} \times 50 \times 4^2 = 400\,J$Thus, the total kinetic energy $K_{total} = K_{trans} + K_{rot} = 400\,J + 400\,J = 800\,J$, which is not listed among the options provided.

A Hoop of Mass 50 Kg Rolls Without Slipping