The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg
The mass of the flywheel in the shape of a solid uniform disk of radius 1.2 m is 4 Kg.
Given :
N = 500 rpm
Work = 3.9 KJ = 3900 J
Radius = 1.2 m
Solution :
We know that the angular velocity is given by:
[tex]\rm \omega = \dfrac{2\pi N}{60}[/tex]
[tex]\rm \omega=\dfrac{2\pi\times 500}{60} = 52.36\;rad/sec[/tex]
We know that the Moment of Inertia for solid disk is given by:
[tex]\rm I = \dfrac{1}{2}mr^2[/tex] --- (1)
Now kinetic energy in terms of angular velocity is given by:
[tex]\rm KE = \dfrac{1}{2}I\omega^2[/tex] ---- (2)
From equation (1) and (2) we get,
[tex]\rm KE = \dfrac{1}{4}mr^2\omega^2[/tex] ---- (3)
Now put the values of KE, r and [tex]\omega[/tex] in equation (3) we get:
[tex]\rm 3900 = \dfrac{1}{4}\times m \times (1.2)^2\times (52.36)^2[/tex]
m = 3.9515 Kg
m [tex]\approx[/tex] 4 Kg
The mass of the flywheel in the shape of a solid uniform disk of radius 1.2 m is 4 Kg.
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