## Physics: Principles with Applications (7th Edition)

As the cylinder rolls down the incline, the initial potential energy will be converted into translational and rotational kinetic energy. We can use conservation of energy to solve this question: $KE_t + KE_r = PE$ $\frac{1}{2}mv^2+ \frac{1}{2}I\omega^2 = mgh$ $\frac{1}{2}mv^2+ \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = mgh$ $\frac{1}{2}mv^2+ \frac{1}{4}mv^2 = mgh$ $\frac{3}{4}v^2 = gh$ $v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{(4)(9.80~m/s^)(7.20~m)}{3}}$ $v = 9.70~m/s$ Therefore, the translational speed at the bottom of the incline is 9.70 m/s.