Calculating Torque and Rotational Inertia for Opening a Heavy Door

[tymartin]

Homework Statement

You need to open a heavy door that's 30 inches wide. Instictively, you push near the edge that's farthest from the hinges. If, instead, you had pushed at a point only 10 inches in from the hinges, how much harder would you have had to push to open the door at the same speed? Explain your answer.

Homework Equations

Here is my attempt at the problem below. I don't know where I'm making my mistake. Any help would be appreciated. Thanks.

The Attempt at a Solution

30 inches= .76 m
Mass of Door-14 kg

I=1/3 (14kg)(.25)2 Rotational Inertia
1/3(14)(.063) I=1/3M/2
1/3(.88) I=1/3 (14kg) (.58)
.29 1/3 (8.12)
2.7
2.7/.29=9.3

 

[LowlyPion]

Is it that hard?

Isn't the same Torque required to open the door? Meaning same I and same a?

So isn't the question what force acting at 1/3 the distance is needed to accomplish the same result?

 

[nlightner]

Torque=Force x Distance

Torque at 30 inches= Force x .76m
Torque at 10 inches= Force x .25m

Since we want to open the door at the same speed, we can set the two torques equal to each other:

Force x .76m = Force x .25m

Solving for force, we get:

Force = (.25m/.76m) x Force = 0.33 x Force

Therefore, if we pushed at a point 10 inches from the hinges, we would have to apply 0.33 times the force compared to pushing at the edge farthest from the hinges. In other words, we would have to push with 3 times the force to achieve the same speed of opening the door. This is because the rotational inertia of the door is lower when pushing at a point closer to the hinges, so a smaller force is needed to achieve the same torque.