Question

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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asked 2021-04-04
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.

Answers (1)

2021-04-06
The first step is to find the total potential energy at 45 degrees:
To do this you must find the height using trigonometry
\(\displaystyle{P}{E}={m}{g}{h}={\left({1.5}\right)}\cdot{\left({9.81}\right)}\cdot{\left({200}-{\sin{{\left({45}\right)}}}\cdot{\left({\frac{{{200}}}{{{1000}}}}\right)}={0.86198}\ {J}\right.}\)
Since energy is conserved the kinetic energy at the bottom of thependulum must equal the potential at the top minus the energydissipated.
\(\displaystyle{K}{E}={0.5}\cdot{m}\cdot{v}^{{2}}={0.5}\cdot{\left({1.5}\right)}\cdot{\left({\frac{{{200}}}{{{1000}}}}\cdot{5.2}\right)}^{{2}}={0.8112}\ {J}\)
So energy dissipated =PE-KE=0.05078 J
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