## College Physics (4th Edition)

The cylinder travels $~1.53~meters~$ along the incline.
We can find an expression for the total kinetic energy of the hollow cylinder at the bottom of the incline: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$ $KE = Mv^2$ We can use conservation of energy to find the vertical height where the hollow cylinder stops moving: $U_g = KE$ $Mgh = Mv^2$ $h = \frac{v^2}{g}$ $h = \frac{(3.00~m/s)^2}{9.80~m/s^2}$ $h = 0.9184~m$ We can find the distance the cylinder rolls along the incline: $sin~\theta = \frac{h}{d}$ $d = \frac{h}{sin~\theta}$ $d = \frac{0.9184~m}{sin~37.0^{\circ}}$ $d = 1.53~m$ The cylinder travels $~1.53~meters~$ along the incline.