Question

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to the bottom.
a) From what height should a circular hoop of radius R bereleased on the same slope in order to equal the sphere's speed atthe bottom?
b) Can a circular hoop of different diameter be released froma height of 30 cm and match the sphere's speed at the bottom? If,so what is the diameter? If not, why not?

Answer 1

Answer:

The height is  $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is $h_s = 30 \ cm$

            The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

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

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    $I = \frac{2}{5} mr^2$

  The angular velocity $w$ is mathematically represented as

                         $w = \frac{v}{r}$

So the equation for conservation of energy becomes

               $mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

              $mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

             $mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

            $gh_s =[\frac{7}{10} ] v^2$

              $v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             $I = mr^2$

Substituting this into the conservation equation above

              $mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 $mgh_c = mv^2$

                     $gh_c = v^2$

                     $h_c = \frac{v^2}{g}$

Recall that   $v^2 = \frac{10gh_s}{7}$

                    $h_c= \frac{\frac{10gh_s}{7} }{g}$

                      $= \frac{10h_s}{7}$

            Substituting values

                   $h_c = \frac{10(30)}{7}$

                       $h_c = 42.86 \ cm$