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?