Cylinder Rolling Down a Slope

As an object rolls down a slope gravitational potential energy is changed into kinetic energy, If the body is not of negligible size the body will also have energy because it is rotating about it;s own centre. This is because the body has inertia  
\[I\]
  and such a body rotating with angular velocity  
\[\omega\]
  has rotational energy  
\[\frac{1}{2} I \omega^2\]
.
A body starting from rest at the top of a slop, in moving a vertical distance  
\[h\]
, will gain kinetic energy  
\[\frac{1}{2}mv^2\]
  and rotational energy  
\[\frac{1}{2}I \omega^2\]
  and we can use conservation of energy to write

\[mgh= \frac{1}{2}mv^2+ \frac{1}{2}I \omega^2\]

For a solid cylinder about an axis about the line of symmetry through its centre  
\[I=mr^2\]
  and we can use this and the relationship  
\[v= \omega r\]
  to write the above conservation of energy equation as
\[mgh= \frac{1}{2}mv^2+ \frac{1}{2}\frac{1}{2}mr^2 )\frac{v}{r}_ \omega^2= \frac{3}{4} mv^2\]

For a slope inclined at an angle  
\[\\]
  to the horizontal, as it travels a distance  
\[x\]
  it falls a distance  
\[h=x sin \\]
. The equations becomes  
\[mg x sin \ = \frac{3}{4}mv^2 \rightarrow xg sin \ = \frac{3}{4}v^2\]
.
Differentiating with respect to  
\[t\]
  gives  
\[vg sin \theta = \frac{3}{2}va \rightarrow g sin \theta =\frac{3}{2}a \]
 

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