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Physics
The moment of inertia of a fly wheel is 0.2kgm2 which is initially stationary. A constant external torque 5Nm acts on the wheel. The work done by this torque during 10 seconds is
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Q. The moment of inertia of a fly wheel is $0.2kgm^2$ which is initially stationary. A constant external torque 5Nm acts on the wheel. The work done by this torque during 10 seconds is
A
1250 J
B
2500 J
C
5000 J
D
6250 J
Solution:
$I=0.2kg\,m^2$
$\tau=5\,Nm$
$\omega_i=0$
$t=10\,s$
we know $\alpha=\frac{\tau}{I}=25\,rad\,s^{-2}$
$\therefore \,\theta=\frac{1}{2}\alpha\,t^2=\frac{1}{2}(25)(100)=1250$
$\therefore W=\tau\,\theta=5(1250)=6250J$