Question

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m from the axis of rotation, and he rotates with an angular speed of 0.770 rad/s. The moment of inertia of the student plus stool is 3.25 kg⋅m2 and is assumed to be constant. (Note that this moment of inertia does not include the two 3.09-kg masses.) The student then pulls the masses horizontally to 0.34 m from the rotation axis. (a) Find the new angular speed of the student. Find the kinetic energy of the rotating system (student, stool, and masses) (b) before and (c) after the masses are pulled in. Caution: There are two weights.

Answer 1

Answer:

a

The New angular speed is  $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is  $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is   $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

             The radius $r =1.08m$

              The  mass is $m = 3.09\ kg$

              The  angular speed $w = 0.770 \ rad/sec$

  The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

           New radius  $r_{new} = 0.34m$

             

Generally the conservation of angular momentum can be mathematical represented as

                         $w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

           $w_i$ is the initial  angular speed

          $I_i$ is the initial moment of inertia

           $I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

                       $I = m r^2$

Where I is the moment of inertia

          m is the mass

           r is the radius

So the Initial moment of inertia is given as  

     $I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

     $I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

    $I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final  moment of inertia is given as  

     $I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$      

Substituting these values into equation 1

         $w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$                                                          

Generally Kinetic energy is mathematically represented in term of moment of inertia as

                       $KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

                     $KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be  $I_i = 10.46 \ Kg \cdot m^2$

  The Angular speed would be  $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

              $KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,                      

               $KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be  $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be  $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

                        $KE_f= \frac{1}{2} *(3.96)(2.034)^2$  

                        $= 8.192J$