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Question

A uniform solid cylinder of mass m and radius R is released from the top of a fixed rough inclined plane of inclination $$\theta$$. The coefficient of friction between the cylinder and the inclined plane is $$\mu =(1/4)\tan \theta$$. Find total kinetic energy of the cylinder at the bottom.
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Solution

$${ v }^{ 2 }{ u }^{ 2 }+2as$$
$$V=\sqrt { 2\times \frac { 3gsin\theta  }{ 4 } \times l } =\sqrt { \frac { 3 }{ 2 }  } glsin\theta $$
net $$\frac { 1/2\times m\times 3/2glsin\theta  }{ 3mglsin\theta /4 } $$
$$\mu mgcos\theta \times R=\frac { { mR }^{ 2 } }{ 2 } \times \gamma $$
$$\mu gcos\theta =\frac { { R }^{  } }{ 2 } \gamma $$
$$2\mu gcos\theta /R=\gamma $$
$$\omega ={ \omega  }^{ 0 }+\gamma t$$
$$\frac { 2\mu cos\theta  }{ R } \times \sqrt { \frac { 8l }{ 3gsin\theta  }  } $$
$$\frac { 2\times sin\theta g }{ cos\theta \times 4R } \times \sqrt { \frac { 8l }{ 3gsin\theta  }  } $$


Physics

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