Question

a solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h .what is the speed of its centre of mass when cylinder reaches its bottom

Answer 1

mgh = 1/2 mv² + 1/2 Iw²
therefore,
v = √4/3(gh)

Answer 2

The speed of its center of mass when cylinder reaches its bottom is $\sqrt{\frac{4 g h}{3}}$

Solution:

Consider a solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h.

Here in this case,

$mgh\quad =\quad \frac { 1 }{ 2 } mv^{ 2 }\quad +\quad \frac { 1 }{ 2 } I\omega ^{ 2 }$

Where,

$\omega \quad =\quad \frac { v }{ r }$

By substituting the value in above equation, we get,

$mgh\quad =\quad \frac { 1 }{ 2 } mv^{ 2 }\left[ 1+\frac { I }{ mr^{ 2 } } \right]$

We know that,

$\left[ 1\quad +\quad \frac { I }{ mr^{ 2 } } \right] \quad =\quad \beta \quad =\quad 1\quad +\quad \frac { m^{ 2 } }{ r^{ 2 } }$

$V\quad =\quad \sqrt { \frac { 2gh }{ \beta } }$

Moment of Inertia of cylinder is,

$I\quad =\quad \frac { mr^{ 2 } }{ 2 }$

$\beta \quad =\quad 1\quad +\quad \frac { \frac { mr^{ 2 }} { 2 }}{ mr^{ 2 } } \quad =\quad 1\quad +\quad \frac { 1 }{ 2 } \quad =\quad \frac { 3 }{ 2 }$

$V\quad =\quad \sqrt { \frac { 2gh }{ 3/2 } }$

$V\quad =\quad \sqrt { \frac { 4gh }{ 3 } }$