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Rotational Mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsRotational Mechanics3 Marks
Question
A uniform square plate of mass 2.0kg and edge 10cm rotates about one of its diagonals under the action of a constant torque of 0.10N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
✓

Answer


I = 0.10N-m; a = 10cm = 0.1m; m = 2kg
Therefore $\Big(\frac{\text{ma}^2}{12}\Big)\times\alpha=0.10\text{N-m}$
$\Rightarrow\alpha=60\text{rad/s}$
Therefore $\omega=\omega_0+\alpha\text{t}$
$\Rightarrow\omega=60\times5=300\text{rad/s}$
Therefore angular momentum $=\text{l}\omega=\Big(\frac{0.10}{60}\Big)\times300=0.50\text{kg-m}^2/\text{s}$
And 0 kinetic energy $\frac{1}{2}\text{l}\omega^2=\frac{1}{2}\times\Big(\frac{0.10}{60}\Big)\times300^2=75\ \text{Joules}.$

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