Car door shutting at a certain acceleration

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SUMMARY

The discussion focuses on calculating the time it takes for a car door, modeled as a uniform square sheet of steel with a side length of 0.8 meters and a mass of 20 kg, to close when the car accelerates at 3 m/s². The torque is calculated using the formula Torque = Mass * radius * acceleration, resulting in a torque of 48 Nm. The angular acceleration is derived as 3.75 rad/s², leading to the conclusion that the door takes approximately 0.92 seconds to close. The calculations utilize fundamental physics principles, including torque and angular motion equations.

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Homework Statement


A car has its door open at 90 degrees. The door is considered a uniform square sheet of steel of side .8 m and mass 20kg. The hinges on the door are frictionless. At time t = 0 the car accelerates with constant acceleration a = 3 m/sec^2. How long does it take the door to close?

Homework Equations


Torque = Mass * radius * acceleration
acceleration = radius * alpha
Torque = Mass * radius^2 * alpha

The Attempt at a Solution


T = 20 (Mass) * (.8)(Radius) * 3 (Acceleration)
T = 48
alpha = 48(Torque) / (.8^2)(Radius) * (20)(Mass) = 3.75
Now that I have angular acceleration, I need only to integrate twice to find the time.
omega = 3.75t
theta = 3.75t^2 / 2 theta = 90 degrees or pi/2 in radians so,
pi/2 = 3.75t^2 / 2 pi/3.75 = t^2 t = sqr root (pi/3.75)

t = .92 seconds

Hopefully these are the correct formulas, but I could use someone to check if these are right. Thanks in advance.
 
Last edited:
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Hi Mastablade! :smile:

Sorry, I don't understand any of this. :redface:

What is the torque ?

Why are you not using moment of inertia ?

What physical principle are you using ?

Go back to your book, and read the chapter again. :smile:​
 

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