Question

A centrifuge rotor (hollow disk) is rotating at 10,000 rpm is shut off and brought to rest by a constant frictional torque of 1.20 Nm. If the mass of the rotor is 4.37 kg and its radius of gyration is 0.0710 m, through how many revolutions will the rotor turn before coming to rest and how long will it take

Answer 1

Answer:

The no of revolutions rotor turn before coming to rest is 1,601.1943 and time taken is equal to 19.21 seconds

Explanation:

here we know that torque = I×α

     α= angular acceleration

     I = moment of inertia of hollow disc = m×$k^{2}$

  given that m=4.37kg

                    k=0.0710m

           torque=1.2Nm

        $w_{o}=10000\times \frac{\pi}{30} rad /s$

         $\alpha =\frac{torque}{I}$

from the above equation we can calculate the angular acceleration of the hollow disc .

  since  $w^{2}-w_o^{2} =2\alpha$$\theta$

 from this above equation  $\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }$

   no of revolutions = $\frac{\theta}{2\pi}$  = 1,601.1943.

Now to calculate time we know that time = $\frac{2\theta}{w+w_{o}}$

   so upon calculating we will be getting t=19.21 seconds