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

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

                       $h_c = 42.86 \ cm$    

       

     

                         

Answer 2

We have that for the Question  it can be said that the right height of release will be

• $h=42.86cm$

From the question we are told

A solid sphere of radius R is placed at a height of 30 cm on a 15 degree slope. It is released and rolls, without slipping, to the bottom.

a) From what height should a circular hoop of radius R be released on the same slope in order to equal the sphere's speed at the bottom?

b) Can a circular hoop of different diameter be released from a height of 30 cm and match the sphere's speed at the bottom? If,so what is the diameter? If not, why not?

Generally the equation for the potential energy of the sphere is mathematically given as

$mgh=1.2Iw^2+1/2mv^2\\\\Where\\\\inertia of the spere\\\\I=\frac{2}{5}mr^2\\\\and\\\\angular velocity \\\\w=v/r\\\\Therefore$

$mgh=1/2(\frac{2}{5}(v/r)^2mr^2)^2+1/2mv^2\\\\mgh=1/2mv^2[7/5]\\\\v^2=\frac{10gh}{7}$

Generally the equation for the Inertia of circle of the sphere is

mathematically given as

$I=mr^2\\\\Therefore\\\\mgh=mv^2\\\\h=\frac{v^2}{g}$

Therefore

$h=\frac{\frac{10gh}{7}}{g}\\\\h=42.86cm$

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