Question

The moment of inertia of a flywheel is $0.2kg{m}^2$ which is initially stationary. A constant external torque $5Nm$ acts on the wheel. The work done by this torque during $10sec$ is:

• A. $1250J$
• B. $2500J$
• C. $5000J$
• D. $6250J$

Answer 1

Answer: (D)

$6250J$

Change in momentum $L=\int \tau .dt$ (basic property of torque)
$=5\times 10=50$
$Workdone=changeinkineticenergy$
$=\frac{L^2}{2I}=\frac{(50{)}^2}{2\times 0.2}=\frac{25000}{4}$
$=6250J$