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

We can find an expression for the kinetic energy of the disk while it is rolling at a speed of 1.5 m/s. Note that the kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy. $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $KE = \frac{3}{4}Mv^2$ We can use conservation of energy to find the maximum height reached by the disk. The potential energy at the maximum height will be equal to the kinetic energy at the bottom of the incline. $PE = KE$ $Mgh = \frac{3}{4}Mv^2$ $h = \frac{3v^2}{4g}$ $h = \frac{(3)(1.5~m/s)^2}{(4)(9.80~m/s^2)}$ $h = 0.1722~m$ We can use the angle to find the distance $d$ the disk rolls up the incline. $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ $d = \frac{0.1722~m}{sin(15^{\circ})}$ $d = 0.67~m$ The disk rolls a distance of 0.67 meters up the incline before rolling back down. We can find an expression for the kinetic energy of the ring while it is rolling at a speed of 1.5 m/s. Note that the kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy. $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(MR^2)(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}Mv^2$ $KE = Mv^2$ We can use conservation of energy to find the maximum height reached by the ring. The potential energy at the maximum height will be equal to the kinetic energy at the bottom of the incline. $PE = KE$ $Mgh = Mv^2$ $h = \frac{v^2}{g}$ $h = \frac{(1.5~m/s)^2}{(9.80~m/s^2)}$ $h = 0.2296~m$ We can use the angle to find the distance $d$ the ring rolls up the incline. $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ $d = \frac{0.2296~m}{sin(15^{\circ})}$ $d = 0.89~m$ The ring rolls a distance of 0.89 meters up the incline before rolling back down.