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×[tex]k^{2}[/tex]
given that m=4.37kg
k=0.0710m
torque=1.2Nm
[tex]w_{o}=10000\times \frac{\pi}{30} rad /s[/tex]
[tex]\alpha =\frac{torque}{I}[/tex]
from the above equation we can calculate the angular acceleration of the hollow disc .
since [tex]w^{2}-w_o^{2} =2\alpha[/tex][tex]\theta[/tex]
from this above equation [tex]\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }[/tex]
no of revolutions = [tex]\frac{\theta}{2\pi}[/tex] = 1,601.1943.
Now to calculate time we know that time = [tex]\frac{2\theta}{w+w_{o}}[/tex]
so upon calculating we will be getting t=19.21 seconds
