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12.2: Rolling motion
https://phys.libretexts.org/Courses/Berea_College/Introductory_Physics%3A_Berea_College/12%3A_Rotational_Energy_and_Momentum/12.02%3A_Rolling_motion
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…
8.4: Rolling motion
https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_I_(2211)/08%3A_Work_Power_and_Energy/8.04%3A_Rolling_motion
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…

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