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प्रश्न
A uniform square plate of mass 2⋅0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0⋅10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
उत्तर
Given
Torque acting on the plate = \[\tau = 0 . 10 N - m\]
Mass of the plate = \[m = 2 kg\]
On applying \[\tau = I\alpha,\] we get
\[\frac{m r^2}{12} \times \alpha = 0 . 10 N - m\]
\[ \Rightarrow \alpha = 60 rad/s\]
Let ω be the angular velocity after time t (t = 5 s).
Therefore, we have
\[\omega = \omega_0 + at\]
\[ \Rightarrow \omega = 60 \times 5 = 300\text{ rad/s}\]
Angular momentum,
\[I\omega = \left( \frac{0 . 10}{60} \right) \times 300\]
\[= 0 . 50 kg - m^2 /s\]
Kinetic energy,
\[\frac{1}{2}I \omega^2 = \frac{1}{2} \times \left( \frac{0 . 10}{60} \right) \times {300}^2 \]
\[ = 75\text{ joule}\]
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