A 1 meter wide door is initially open at an angle of 30o as shown (top view). You push with 20 N force in the middle of the door as shown and the door rotates around the hinge on the left. The door has a rotational inertia =3.0 kg m2. The angular acceleration of the door will be:

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DeniceSandidge DeniceSandidge · Answer 1 · 2020-05-28T05:06:50+02:00

Answer:

angular acceleration = 1.67 rad/s²

Explanation:

given data

door wide = 1 m

initially ope angle = 30°

push force = 20 N

rotational inertia = 3.0 kg m²

solution

we apply force at middle so length will be here r1 = [tex]\frac{1}{2}[/tex] = 0.5 m

and

now we get here torque that is express as

torque τ = Force × r1 × sin30 ......................1

put her value and we get

torque τ = 20 × 0.5 × sin30

torque τ = 5 Nm

and we know

torque = rotational inertia × angular acceleration .......................2

put her value and we get angular acceleration

angular acceleration = [tex]\frac{5}{3}[/tex]

angular acceleration = 1.67 rad/s²