Physics - Rotation Question with Solution | TestHub

$\text{KE}=\frac{1}{2}{\mathit{\text{Mv}}}^2+\frac{1}{2}\mathit{\text{I}}{\omega }^2$

$\text{KE}=\frac{1}{2}{\mathit{\text{Mv}}}^2+\frac{1}{2}\mathit{\text{I }}{\left(\frac{\mathit{\text{v}}}{2\mathit{\text{R}}}\right)}^2$...(1)

Consider solid cylinder of radius$2R$and mass$M\text{'}$, from which a cylinder of radius$R$of mass$\mathit{\text{M}}\text{"}=\frac{\mathit{\text{M}}\text{'}}{4}$is taken out from a common axis.

Mass of hollow cylinder$M = M\text{'}-M\text{'}\text{'}$

$\mathit{\text{M}}=\mathit{\text{M}}\text{'}-\frac{\mathit{\text{M}}\text{'}}{4}=\frac{3\mathit{\text{M}}\text{'}}{4} \Rightarrow \mathit{\text{M}}=\frac{3\mathit{\text{M}}\text{'}}{4} \Rightarrow \mathit{\text{M}}\text{'}=\frac{4\mathit{\text{M}}}{3}\text{ \& }\mathit{\text{M}}\text{"}=\frac{\mathit{\text{M}}}{3}$

${\mathit{\text{I}}}_{2\text{R}}={\mathit{\text{I}}}_{\text{R}}+\mathit{\text{I}} \Rightarrow \mathit{\text{I}}={\mathit{\text{I}}}_{2\text{R}}-{\mathit{\text{I}}}_{\text{R}} \Rightarrow \mathit{\text{I}}=\frac{1}{2}\mathit{\text{M}}\text{'}{\left(2\mathit{\text{R}}\right)}^2-\frac{1}{2}\mathit{\text{M}}\text{"}{\mathit{\text{R}}}^2$

$\mathit{\text{I}}=\left[\frac{1}{2}\times \frac{4\mathit{\text{M}}}{3}\times 4{\mathit{\text{R}}}^2-\frac{1}{2}\times \frac{\mathit{\text{M}}}{3}{\mathit{\text{R}}}^2\right] \Rightarrow \mathit{\text{I}}=\frac{5}{2}{\mathit{\text{MR}}}^2$... (2)

$\text{KE}=\frac{13}{16}{\mathit{\text{Mv}}}^2$