Question

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, where M = 14.2 kg is the mass of the pulley and R=1.5 m is its radius ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of r=R=1.5 m during the block's descent, and the distance of the block at initial position to the floor is 8.0 m. What is the angular velocity in round per minute when the block drops to the floor? Use g = 10 m/s2.

Answer 1

Answer:

The angular velocity is  $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

    The mass of the block is  $m = 61.2kg$

     The of the pulley is  $M = 14.2 kg$

      The radius of the pulley is  $R = 1.5m$

       The radius  of the cord around the pulley is  $r = 1.5 m$

       The distance of the block to the floor is  $d = 8.0 m$

         

From the question we are told that the moment of inertia of the pulley is

          $I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value  

         $I = \frac{1}{2} * 14.2 * (1.5)^2$

         $I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

              $ma = mg -T$

         =>  $T = mg -ma$

Now when the pulley is rotated that  torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

                  $\tau = I \alpha$

     Here $\alpha$ is the angular acceleration

           Here $\tau$ is the torque which can be equivalent to

              $\tau = T r$

  Substituting this above

            $Tr = I \alpha$      

Substituting for T

         $(mg - ma ) r = I\ r \alpha$

Here $a$ is the  linear acceleration which is mathematically represented as

           $a = r\alpha$

    $(mg - m(r\alpha ) ) r = I\ r \alpha$

     $mgr = I\alpha + m(r\alpha ) r$

    $mgr = \alpha [ I + mr^2]$

   making $\alpha$ the subject

          $\alpha = \frac{mgr}{I -mr ^2}$          

   Substituting values

            $\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

             $\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

             $a = 5.854 * 1.5$

                $a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

         $v = \frac{d}{t}$

Substituting this into the formula acceleration

        $a = \frac{d}{t^2}$

making t the subject

         $t = \sqrt{\frac{d}{a} }$

substituting value

      $t = \sqrt{\frac{8}{8.78}}$

     $t = 0.9545 \ s$

Now the linear velocity is

       $v = \frac{8}{0.9545}$

       $v = 8.38 m/s$

The angular velocity is  

       $w = \frac{v}{r}$

So

       $w = \frac{8.38}{1.5}$

        $w = 5.59 rad/s$

Generally 1 radian is equal to  0.159155 rounds or turns

        So  5.59 radian is  equal to x

Now x is mathematically obtained as

         $x = \frac{5.59 * 0.159155}{1}$

            $= 0.8892 \ rounds$

 Also

      60  second =  1 minute

So   1 second  = z      

Now z is mathematically obtained as

         $z = \frac{ 1}{60}$

            z $= 0.01667 \ minute$

Therefore

              $w = \frac{0.8892}{0.01667}$

              $w = 53.35 \ rounds /minute$