Question

A flywheel of mass 50 kg is rotating at a speed of 6 revolutions per second.
If its mass can be considered to be concentrated at a distance of 0.3m
from the axis, what constant torque is required to stop it in 10 revolutions.​

Answer 1

Answer:

2.7 kg m2

Explanation:

hope the answer will help you

Answer 2

Given :

The mass of a flywheel = m = 50 kg

Initial rotating speed = $\omega _i$ = 6 rps

Final rotating speed = $\omega _f$ = 10 rps

Distance cover = 0.3 meters

To Find :

The constant torque required to stop

Solution :

∵  moment of inertia of flywheel = I = $\dfrac{mr^{2} }{2}$

    where r = radius     and  m = mass

As, Distance cover = circumference of wheel = 2 π r

So,     2 × 3.14 × r = 0.3 m

Or,     6.28 × r = 0.3

∴                   r = $\dfrac{0.3}{6.28}$

i.e   radius = 0.0477  meters

So,  moment of inertia of flywheel = I = $\dfrac{mr^{2} }{2}$

Or,       I = $\dfrac{50\times (0.0477)^{2}}{2}$

Or, Moment of Inertia = I = 0.0568 kg m²

Now,

Angular acceleration = α = $\dfrac{Torque}{Moment of Inetia}$

i.e   α = $\dfrac{T}{I}$

Or, Angular acceleration = α = $\dfrac{T}{0.0568}$ rad/s²         ...........1

Again

from the motion equation

Final rotating speed  = Initial rotating speed  +  angular acceleration × time

I.e    $\omega _f$ = $\omega _i$ + α × t

For t = 1 sec

      10 rps = 6 rps + $\dfrac{T}{0.0568}$ ×  1                ( from eq 1 )

Or,  10 - 6 = $\dfrac{T}{0.0568}$

Or,    $\dfrac{T}{0.0568}$ = 4

∴          T = 4 × 0.0568

i.e  Torque = T = 0.2272   N/m

Hence, The constant torque required to stop flywheel is 0.2272 N/m Answer