A solid cylinder rolls up an …

A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, {tex}\theta = 30 ^ { \circ }{/tex}
Height reached by the cylinder = h

1. Energy of the cylinder at point A will be purely kinetic due to the rotation and translational motion. Hence, total energy at A
   = KE_rot + KE_trans
   =   {tex}\frac { 1 } { 2 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 }{/tex}

   The energy of the cylinder at point B will be purely in the form of gravitational potential energy  = mgh
   Using the law of conservation of energy, we can write:
   {tex}\frac { 1 } { 2 } I \omega ^ { 2 } +\frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
   Moment of inertia of the solid cylinder, {tex}I = \frac { 1 } { 2 } m r ^ { 2 }{/tex}
   {tex}\therefore \frac { 1 } { 2 } \left( \frac { 1 } { 2 } m r ^ { 2 } \right) \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
   {tex}\frac { 1 } { 4 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
   But we have the relation, {tex}v = r \omega{/tex}
   {tex}\therefore \frac { 1 } { 4 } v ^ { 2 } + \frac { 1 } { 2 } v ^ { 2 } = g h{/tex}
   {tex}\frac { 3 } { 4 } v ^ { 2 } = g h{/tex}
   {tex}\therefore h = \frac { 3 } { 4 } \frac { v ^ { 2 } } { g }{/tex}
   {tex}= \frac { 3 } { 4 } \times \frac { 5 \times 5 } { 9.8 } = 1.91 \mathrm { m }{/tex}

   To find the distance covered along the inclined plane
   In {tex}\Delta A B C{/tex}:
   {tex}\sin \theta = \frac { B C } { A B }{/tex}
   {tex}\sin 30 ^ { \circ } = \frac { h } { A B }{/tex}
   {tex}A B = \frac { 1.91 } { 0.5 } = 3.82 \mathrm { m }{/tex}
   Hence, the cylinder will travel 3.82 m up the inclined plane.

2. {tex}v = \left( \frac { 2 g h } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
   {tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
   For the solid cylinder, {tex}K ^ { 2 } = \frac { R ^ { 2 } } { 2 }{/tex}
   {tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { 1 } { 2 } } \right) ^ { \frac { 1 } { 2 } }{/tex}
   {tex}= \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } }{/tex}
   The time taken to return to the bottom is:
   {tex}t = \frac { A B } { v }{/tex}
   {tex}= \frac { A B } { \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } } } = \left( \frac { 3 A B } { 4 g \sin \theta } \right) ^ { \frac { 1 } { 2 } }{/tex}
   {tex}= \left( \frac { 11.46 } { 19.6 } \right) ^ { \frac { 1 } { 2 } } = 0.7645{/tex}
   So the total time taken by the cylinder to return to the bottom is (2 {tex}\times{/tex} 0.764)= 1.53 s.as time of ascend is equal to time of descend for the following problem.