Question

A 2.66 kg, 22.25 cm diameter turntable rotates at 238 rpm on frictionless bearings. Two 420 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event

Answer 1

Answer:

The value is $w_f = 146 \ rpm$

Explanation:

From the question we are told that

  The mass is  $m = 2.66 \ kg$

  The diameter is  $d = 22.25 \ cm = 0.2225 \ m$

   The angular speed is  $w_i = 238 \ rpm$

   The mass of each of the blocks is $m_b = 420 \ g = 0.420 \ kg$

Generally the radius of the turntable  is mathematically represented as

        $r = \frac{d}{2}$

=>     $r = \frac{0.2225}{2}$

=>     $r = 0.11125$

The moment of inertia of the turntable before the blocks fell is mathematically represented as

         $I_{i} = \frac{1}{2} * m * r^2$

=>       $I_{i} = \frac{1}{2} * 2.66 * (0.11125)^2$

=>       $I_{i} = 0.01646 \ kg\cdot m^2$

The moment of inertia of the turntable after the blocks fell is mathematically represented as

         $I_{f} = \frac{1}{2} * m * r^2 + 2 m_b * r^2$

=>       $I_{f} = \frac{1}{2} * 2.66 * (0.11125)^2 + 2 * 0.420 * 0.11125^2$

=>       $I_{f} = 0.02686 \ kg\cdot m^2$

Generally from the law of angular momentum conservation  

          $L_i = L_f$

Here $L_i$ is the initial  angular momentum of the turntable  before the blocks fell which is mathematically represented as

        $L_i = I_i * w_i$

and  $L_f$ is the initial  angular momentum of the turntable after the blocks fell which is mathematically represented as

       $L_i = I_f * w_f$

So

      $0.01646* 238 = 0.02686 * w_f$

=>    $w_f = 146 \ rpm$