How Long Does It Take to Rotate a Door to 60 Degrees?

The question of how long it takes to rotate a door to 60 degrees from a closed position is a classic example of a physics problem that can be solved using principles of mechanics. This article delves into the process of calculating the time required, step-by-step, and highlights the importance of the underlying physics concepts.

Understanding the Physical Scenario

Imagine you are pushing a heavy door that is partially open. The door has a mass of 35 kg, a width of 0.60 meters, and you apply a force of 23 N at a distance of 0.40 meters from the axis of rotation. This scenario involves multiple fundamental physics concepts such as torque, moment of inertia, and angular acceleration. We'll walk through the calculations needed to find the time required to rotate the door to 60 degrees.

Calculated Torque

Torque ((tau)) is a measure of the force applied perpendicularly to the axis of rotation. It is given by the product of the force and the perpendicular distance from the axis. In this case:

[tau 0.40 , text{m} times 23 , text{N} 9.2 , text{Nm}]

Moment of Inertia of the Door

The moment of inertia ((I)) is a measure of an object’s resistance to changes in rotational motion. For a door of width 0.60 meters and mass 35 kg rotating about one edge, the moment of inertia is calculated as follows:

[text{Assuming uniform density, } I frac{1}{3} times 35 , text{kg} times (0.60 , text{m})^2]

This results in:

[I 0.6667 times 35 times 0.36 7.9986 approx 8.0 , text{kg m}^2]

Near the axis, the moment of inertia is:

[I frac{1}{3} times 35 times (0.4)^2 1.87 , text{kg m}^2]

Angular Acceleration

Angular acceleration ((alpha)) is related to torque by the formula (tau I alpha). Substituting the values, we find:

[alpha frac{tau}{I} frac{9.2 , text{Nm}}{1.87 , text{kg m}^2} 4.93 , text{rad/s}^2]

Time to Rotate to 60 Degrees

The angular displacement ((theta)) for 60 degrees is (theta frac{pi}{3}) radians. The kinematic equation for rotational motion is:

[theta frac{1}{2} alpha t^2]

Solving for (t), we get:

[theta frac{1}{2} (4.93 , text{rad/s}^2) t^2] [frac{pi}{3} frac{1}{2} (4.93) t^2] [frac{1.0472}{2.465} t^2] [t sqrt{0.425} approx 0.65 , text{sec}]

Imagining the Process

Let's imagine the door being pushed. Each component—torque, moment of inertia, and angular acceleration—plays a crucial role in determining how long it takes to rotate the door to 60 degrees. This process simplifies a complex real-world scenario into manageable calculations.

Real-world Considerations

It is important to note that real-world scenarios can vary. Factors such as the uniformity of the door's density, friction of the hinges, and air resistance based on current atmospheric conditions can affect the calculated time. Nonetheless, the theoretical calculation provides a solid framework for understanding the physics behind door rotation.

Conclusion

In conclusion, the time required to rotate a 35 kg door 60 degrees from a closed position, when applying a force of 23 N at a distance of 0.40 meters from the axis of rotation, is approximately 0.65 seconds. This calculation is based on simplified assumptions and highlights the principles of torque, moment of inertia, and angular acceleration. Understanding these principles is crucial in various applications, from daily tasks to engineering designs.