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2 years ago

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To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meter

s from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door

1 answer:

Zielflug [23.3K]2 years ago

6 0

Answer:

The moment of inertia is $I = 1.8 \ kg m^2$

Explanation:

From the question we are told that

The force applied is $F = 10 \ N$

The distance of the knob to the hinge is $d = 0.9 \ m$

The angular acceleration is $a = 5 \ rad/s$

The moment is mathematically represented as

$I = \frac{d Fsin(\theta)}{a}$

Here $\theta = 90^o$ This is because the force direction is perpendicular to the plane of the door

substituting values

$I = \frac{0.9 * 10 * sin (90)}{5}$

$I = 1.8 \ kg m^2$

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Answer:

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