Question

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0 kg and diameter 75.0 cm . The power is off for 27.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 180 complete revolutions.

At what rate is the flywheel spinning when the power comes back on?

How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on?

How many revolutions would the wheel have made during this time?

Answer 1

Given that :

initial angular velocity, ₀ = 500 rpm

rpm change into rad/s -

₀ = (500) (2 rad) / (60 s) 52.3 rad/s

angular displacement, = (180 rev) (2 rad) 1130.4 rad

time taken, t = 27 sec

using an equation,    = ₀ t + (1/2) t²    { eq.1 }

inserting the values in above eq.

(1130.4 rad) = (52.3 rad/s) (27 s) + (0.5) (27 s)²

(1130.4 rad) - (1412.1 rad) = (364.5 s²)

= -(281.7 rad) / (364.5 s²)

= -0.772 rad/s²

(a) When the power comes back on, then the final angular velocity of flywheel which is given as :

using equation of rotational motion,

_f = ₀ + t                                                                         { eq.2 }

inserting the values in eq.2,

_f = (52.3 rad/s) - (0.772 rad/s²) (27 s)

_f = (52.3 rad/s) - (20.8 rad/s)

_f = 31.5 rad/s

(b) After the beginning of the power failure, time taken by the flywheel to stop if the power had not come back on which will be given as :

using eq.2,      _f = ₀ + t            

where, _f = final angular velocity = 0 rad/s

t = - (52.3 rad/s) / (-0.772 rad/s²)

t = 67.7 sec

(c) During this time, number of revolutions by the wheel which is given as :

using equation 1, we have

= ₀ t + (1/2) t²

= (52.3 rad/s) (67.7 s) + (0.5) (-0.772 rad/s²) (67.7 s)²

= (3540.7 rad) - (1769.1 rad)

= 1771.6 rad

we know that, N = / 2                                                           { eq.3 }

inserting the values in eq.3,

N = (1771.6 rad) / (6.28)

N = 282.1 rev