Q7.27     Prove the result that the velocity v of translation of a rolling body (like a ring, disc,
               cylinder or sphere) at the bottom of an inclined plane of a height h  is given by

               v^{2}=\frac{2gh}{(1+k^{2}/R^{2)}}
             using dynamical consideration (i.e. by consideration of forces and torques). Note k is
             the radius of gyration of the body about its symmetry axis, and R  is the radius of the
             body. The body starts from rest at the top of the plane.

Answers (1)

Consider the given situation :

                                              Rotational motion,  20195

The total energy when the object is at the top (potential energy)  =  mgh.

Energy when the object is at the bottom of the plane : 

                                                                             E_b\ =\ \frac{1}{2}I\omega^2\ +\ \frac{1}{2}mv^2

Put    I\ =\ mk^2       and      v\ =\ \omega r, we get :

                

                                                                        E_b\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )

By the law of conservation of energy we can write :

                                            mgh\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )

or                                             

                                              v\ =\ \frac{2gh}{\left ( 1\ +\ \frac{k^2}{r^2} \right )}                                        .

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