Solve the following :

Question:

A uniform square plate of mass $2.0 \mathrm{~kg}$ and edge $10 \mathrm{~cm}$ rotates about one of its diagonals under the action of a constant torque of $0.10 \mathrm{~N}-\mathrm{m}$. Calculate the angular momentum and kinetic energy of the plate at the end of the fifth second after the start.

Solution:

$\mathrm{T}=\mathrm{l \alpha}$

$0.1=\frac{\mathrm{Ma}^{2}}{12}(\alpha)$

$0.1=\frac{(2)(0.1)^{2}}{12}(\alpha)$

$\alpha=60 \mathrm{rad} / \mathrm{sec}$

$\omega_{0}=0, t=5 \mathrm{sec}$

$\omega=\omega_{0}+\alpha t$

Angular momentum , L= la

$=\frac{(2)(0.1)^{2}}{12}(300)$

$=\left[\begin{array}{l}\frac{m^{2}}{}\end{array}\right]$

$=0.5 \mathrm{~kg}-\mathrm{sec}$

$\mathrm{KE}=2$

$\mathrm{I} \omega^{2}$

$=\frac{1}{2}\left[\frac{(2)(0.1)^{2}}{12}\right](300)^{2}$

$=\frac{1}{2}\left[\frac{(2)(0.1)^{2}}{12}\right](300)^{2}$

$K E=75 \mathrm{~J}$

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