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# A 5 kg uniform square plate is supported by two identical 1.5kg uniform slender rods AD and BE and is released from rest inthepostion theta 45 degrees. Knowing that the angular velocity ofAD as it passes through the vertical position is 5.2 rad/sdetermine the amount of energy dissipated infriction.

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A 5 kg uniform square plate is supported by two identical 1.5kg uniform slender rods AD and BE and is released from rest inthepostion theta 45 degrees. Knowing that the angular velocity ofAD as it passes through the vertical position is 5.2 rad/sdetermine the amount of energy dissipated infriction.

$$\displaystyle{P}{E}={m}{g}{h}={\left({1.5}\right)}\cdot{\left({9.81}\right)}\cdot{\left({200}-{\sin{{\left({45}\right)}}}\cdot{\left({\frac{{{200}}}{{{1000}}}}\right)}={0.86198}\ {J}\right.}$$
$$\displaystyle{K}{E}={0.5}\cdot{m}\cdot{v}^{{2}}={0.5}\cdot{\left({1.5}\right)}\cdot{\left({\frac{{{200}}}{{{1000}}}}\cdot{5.2}\right)}^{{2}}={0.8112}\ {J}$$