A bowling ball of mass 6.0 kg and radius 9.0 cm rolls without slipping down a lane at 4.0 m/s. The total kinetic energy of the bowling ball is

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Quizplus · Accepted Answer

The total kinetic energy of the bowling ball consists of both translational kinetic energy and rotational kinetic energy. The translational kinetic energy ($KE_{trans}$) is given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. The rotational kinetic energy ($KE_{rot}$) is given by $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a solid sphere, $I = \frac{2}{5}mr^2$, and $\omega = \frac{v}{r}$, where $r$ is the radius.First, calculate the translational kinetic energy:$$KE_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times 6.0 \, \text{kg} \times (4.0 \, \text{m/s})^2 = 48 \, \text{J}$$Then, calculate the rotational kinetic energy:$$KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mr^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{2} \times \frac{2}{5} \times 6.0 \, \text{kg} \times (0.09 \, \text{m})^2 \times \left(\frac{4.0 \, \text{m/s}}{0.09 \, \text{m}}\right)^2 = 19.2 \, \text{J}$$Finally, add the translational and rotational kinetic energies to get the total kinetic energy:$$KE_{total} = KE_{trans} + KE_{rot} = 48 \, \text{J} + 19.2 \, \text{J} = 67.2 \, \text{J}$$Rounding to the nearest whole number gives 67 J.

A Bowling Ball of Mass 6