## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$KE = 0.375~J$
The kinetic energy of the can is the sum of the translational kinetic energy and the rotational kinetic energy. $KE = KE_{trans}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $KE = \frac{3}{4}Mv^2$ $KE = \frac{3}{4}(0.50~kg)(1.0~m/s)^2$ $KE = 0.375~J$