The moment of inertia of a wheel is $1000 k g - m^{2}$ . At a given instant, its angular velocity is $10 r a d / s$ . After the wheel rotates through an angle of $100$ radians, its angular velocity increases to $100 r a d / s$ . Calculate the (a) torque applied on the wheel. (b) increase in the kinetic energy.

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Solution

Kinetic energy
given information.
initial angular speed of the wheel
$w_{0} = 10 r a d / s$
final angular velocity of the wheel $w_{f} = 100$ rad/s
angular displacement $θ = 100$ radian
$w_{f}^{2} = w_{0}^{2} + 2 α θ$
As we know
$(100)^{2} = 10^{2} + 2 \times 100 \times α$
$α = 49.5$
torque acting on the wheel $= I ≺$
$= 1000 \times 49.5$
$= 49500 N m^{2}$
change in rotational K.\epsilon $= \frac{1}{2} I [w_{f}^{2} - w_{0}^{2} = 49500005$

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