In a downtown office building you notice that each of the four sections of a rotating door has a mass of 90 kg. What is the width of each section of the door if a force of 58 N applied to the outer edge of a section produces an angular acceleration of 0.500 rad/s2?
To find the width of each section of the door, we need to apply Newton's second law for rotational motion. The equation for torque is:
τ = Iα
Where:
τ is the torque applied (in Nm),
I is the moment of inertia of the rotating object (in kgm²),
α is the angular acceleration (in rad/s²).
The moment of inertia for a rectangular section rotating around its edge can be calculated as:
I = (1/3) * m * r²
Where:
m is the mass of the section (in kg),
r is the distance from the rotation axis to the center of mass of the section (in m).
In this case, each section has a mass of 90 kg, so the moment of inertia for each section can be calculated as:
I = (1/3) * 90 * r²
Now, we can rearrange the torque equation to solve for the width of each section:
τ = Iα
τ = (1/3) * 90 * r² * α
The torque applied to each section is given as 58 Nm. The angular acceleration is given as 0.500 rad/s². Substitute these values into the equation:
58 = (1/3) * 90 * r² * 0.500
Now, solve for r²:
r² = (58 * 3) / (90 * 0.500)
r² = 116 / 45
r² ≈ 2.578
Take the square root of both sides to find r:
r ≈ √2.578
r ≈ 1.60 m
Finally, since each section of the door is bisected by the rotation axis, the width of each section is two times the value of r:
Width of each section ≈ 2 * r
Width of each section ≈ 2 * 1.60
Width of each section ≈ 3.20 m
Therefore, each section of the door has a width of approximately 3.20 meters.
moment of inertia: 4*1/3 90 (r)^2
torque applied*distance=I*ang accelearation
58*r=I*.5=.5*4*1/3 90 (r)^2
solve for r