The mechanical energy at the bottom of the incline is thus: \[\begin{aligned} E' = K' + U = K'_{rot}+K'_{trans}+(0)=\frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}Mv_{cm}^2\end{aligned}\] Since the disk is ro...The mechanical energy at the bottom of the incline is thus: \[\begin{aligned} E' = K' + U = K'_{rot}+K'_{trans}+(0)=\frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}Mv_{cm}^2\end{aligned}\] Since the disk is rolling without slipping, its angular speed is related to the speed of center of mass: \[\begin{aligned} \omega = \frac{v_{CM}}{R}\end{aligned}\] The moment of inertia of the disk about its center of mass is given by: \[\begin{aligned} I_{CM}=\frac{1}{2}MR^2\end{aligned}\] We can thus write the mecha…
The mechanical energy at the bottom of the incline is thus: \[\begin{aligned} E' = K' + U = K'_{rot}+K'_{trans}+(0)=\frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}Mv_{cm}^2\end{aligned}\] Since the disk is ro...The mechanical energy at the bottom of the incline is thus: \[\begin{aligned} E' = K' + U = K'_{rot}+K'_{trans}+(0)=\frac{1}{2}I_{CM}\omega^2 + \frac{1}{2}Mv_{cm}^2\end{aligned}\] Since the disk is rolling without slipping, its angular speed is related to the speed of center of mass: \[\begin{aligned} \omega = \frac{v_{CM}}{R}\end{aligned}\] The moment of inertia of the disk about its center of mass is given by: \[\begin{aligned} I_{CM}=\frac{1}{2}MR^2\end{aligned}\] We can thus write the mecha…
