Question

It takes 10 min for a 3.1-Mg flywheel to coast to rest from an angular velocity of 300 rpm. Knowing that the radius of gyration of the flywheel is 1 m, determine the average magnitude of the couple due to kinetic friction in the bearing. (Round the final answer to one decimal place.) The average magnitud hof the couple due to kinetic friction in the bearing is N.m.

Answer 1

Ans) We know,

  $\omega$ = $\omega_o$ + $\alpha$ t ...........................(1)

where, $\omega$ = final angular velocity = 0

  $\omega_o$ = initial angular velocity

  $\alpha$ = angular acceleration

t = time = 10 min or 600 sec

Since, flywheel comes to rest ($\omega$ = 0), we can say that a deceleration is acting on flywheel

Also, $\omega_o$ = 2$\pi$N/ 60

where, N = rotational speed = 300 rpm

=> $\omega_o$ = 2$\pi$ (300) / 60

=> $\omega_o$ = 31.415 rad/sec

Putting values in equation 1,

=> 0 = 31.415 + $\alpha$ (600)

=> $\alpha$ = - 0.0523 rad/

5

Now, moment (M),

M = I $\alpha$

where, I = moment of inertia = m $k^2$

   m = 3.1 Mg or 3100 kg

k = Radius of gyration = 1 m

=> I = 3100 ($1^2$) = 3100 kg $m^2$

=> M = 3100 kg$m^2$ x (- 0.0523 rad/)

5

=> M = -162.13 kg $m^2$ /
5
or -162.13 N.m

Hence, average magnitude of couple due to kinetic friction in bearing is -162.13 N.m