A uniform wheel of radius R is set into rotation about its axis at an angular speed $\omega.$ This rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling.
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A uniform wheel of radius R is set into rotation about its axis (case-I) at an angular speed $\omega.$ 
This rotating wheel is now placed on a rough horizontal. Because of its friction at contact, the wheel accelerates forward and its rotation decelerates. As the rotation decelerates the frictional force will act backward. 
If we consider the net moment at A then it is zero. Therefore the net angular momentum before pure rolling & after pure rolling remains constant Before rolling the wheel was only rotating around its axis. Therefore Angular momentum $=\ell\omega=\Big(\frac{1}{2}\Big)\text{MR}^2\omega\ \dots(1)$ After pure rolling the velocity of the wheel let v Therefore angular momentum $=\ell_{\text{cm}}\omega+\text{m}(\text{v}\times\text{R})$ $=\Big(\frac{1}{2}\Big)\text{m}\text{R}^2\Big(\frac{\text{v}}{\text{R}}\Big)+\text{mvR}=\frac{3}{2}\text{mvR}\ \dots(2)$ Because, Eq. (1) and (2) are equal Therefore, $\frac{3}{2}\text{mvR}=\frac{1}{2}\text{mR}^2\omega$ $\Rightarrow\text{V}=\omega\frac{\text{R}}{3}$
This rotating wheel is now placed on a rough horizontal. Because of its friction at contact, the wheel accelerates forward and its rotation decelerates. As the rotation decelerates the frictional force will act backward. If we consider the net moment at A then it is zero. Therefore the net angular momentum before pure rolling & after pure rolling remains constant Before rolling the wheel was only rotating around its axis. Therefore Angular momentum $=\ell\omega=\Big(\frac{1}{2}\Big)\text{MR}^2\omega\ \dots(1)$ After pure rolling the velocity of the wheel let v Therefore angular momentum $=\ell_{\text{cm}}\omega+\text{m}(\text{v}\times\text{R})$ $=\Big(\frac{1}{2}\Big)\text{m}\text{R}^2\Big(\frac{\text{v}}{\text{R}}\Big)+\text{mvR}=\frac{3}{2}\text{mvR}\ \dots(2)$ Because, Eq. (1) and (2) are equal Therefore, $\frac{3}{2}\text{mvR}=\frac{1}{2}\text{mR}^2\omega$ $\Rightarrow\text{V}=\omega\frac{\text{R}}{3}$
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