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प्रश्न
A flywheel of moment of inertia 5⋅0 kg-m2 is rotated at a speed of 60 rad/s. Because of the friction at the axle it comes to rest in 5⋅0 minutes. Find (a) the average torque of the friction (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.
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उत्तर
Let the angular deceleration produced due to frictional force be α.
Initial angular acceleration,
\[\omega_0 = 60 rad/s\]
Final angular velocity,
\[\omega = 0\]
t = 5 min =300 s
We know that
\[\omega = \omega_0 + \alpha t\]
\[\Rightarrow \alpha = - \frac{\omega_0}{t}\]
\[ \Rightarrow \alpha = - \left( \frac{60}{300} \right) = - \frac{1}{5} rad/ s^2\]
(a) Torque produced by the frictional force (R),
\[\tau = I\alpha = 5 \times $\left( \frac{- 1}{5}
right)\]
= 1 N - m opposite to the rotation of wheel
(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy
\[W = \frac{1}{2}I \omega^2 \]
\[ = \frac{1}{2} \times 5 \times \left( 60 \times 60 \right)\]
\[ = 9000 \text{ joule }= 9 kJ\]
(c) Angular velocity after 4 minutes,
\[\omega = \omega_0 + \alpha t\]
\[ = 60 - \frac{4 \times 60}{5}\]
\[ = \frac{60}{5} = 12 rad/s\]
So, angular momentum about the centre,
\[L = I\omega\]
\[ = 5 \times 12 = 60 kg - m^2 /s\]
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