You turned the knob of a 197cm by 99.2cm door (w/ mass 9.52kg) and pushed it with a perpendicular force of 8.75N to open it. What is the angular acceleration of the door (in rad/s2)?
To find the angular acceleration of the door, we need to use the equation:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
To calculate the torque, we need to know the applied force and the distance from the axis of rotation (door hinge) to the point where the force is applied. The distance can be obtained by using the given dimensions of the door.
In this case, the force is applied perpendicular to the door, and we'll assume it is applied at the edge of the door. Therefore, the distance from the axis of rotation to the point of force application is half the width of the door:
r = width / 2
Given the door width is 99.2 cm, which is 0.992 m, the distance is:
r = 0.992 m / 2 = 0.496 m
Now, we can calculate the torque:
τ = r * F
Given the force applied is 8.75 N, the torque is:
τ = 0.496 m * 8.75 N = 4.352 N·m
The moment of inertia of a rectangular door about its hinge is given by:
I = (1/3) * m * h^2
where m is the mass of the door and h is the height of the door.
Given the door height is 197 cm, which is 1.97 m, the moment of inertia is:
I = (1/3) * 9.52 kg * (1.97 m)^2 = 10.55 kg·m^2
Now we can use the torque equation to calculate the angular acceleration:
τ = I * α
4.352 N·m = 10.55 kg·m^2 * α
Rearranging the equation to solve for α:
α = 4.352 N·m / 10.55 kg·m^2 = 0.412 rad/s^2
Therefore, the angular acceleration of the door is 0.412 rad/s^2.