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A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces. At the east end of the beam, a 200\ N200 N forces pushes downward. At the west end of the beam, a 200\ N200 N force pushed upward. What is the angular acceleration of the beam

Sagot :

Answer: [tex]240\ rad/s^2[/tex]

Explanation:

Given

Length of beam [tex]l=2\ m[/tex]

mass of beam [tex]m=5\ kg[/tex]

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

[tex]\tau =F\times l=200\times 2=400\ N.m[/tex]

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

[tex]I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2[/tex]

Torque is the product of moment of inertia and angular acceleration

[tex]\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2[/tex]

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