- #1
SEG9585
- 34
- 0
Hey all--
Can someone try to ifnd what I'm doing wrong with this question, I must have sort of brain lapse in the middle of a 4-day weekend and all:
Question: A (very large) door has a mass of 44,000kg, and a rotational inertia about a vertical axis through its hingesof 8.7*10^4 kg*m^2, and has a front face of 2.4m. NEglecting friction, what steady force perpendicular to the door can move it from rest through an angle of 90 degrees in 30s?
My incorrect solution: ("@" refers to the angle in radian measure), and "&" refers to angular acceleration)
@ = 2(pi)*r/4 @=2.4(pi)/2 radians
@ = .5(&)(t^2)
1.2(pi) = .5(&)(30^2)
& = .0084 rad/s^2
I(&) = r x F = rFsin90
(8.7*10^4)(.0084) = (2.4)(F)(1)
F= 304.5 N
However, the actual answer is apparently 130N. What did I do wrong?
Can someone try to ifnd what I'm doing wrong with this question, I must have sort of brain lapse in the middle of a 4-day weekend and all:
Question: A (very large) door has a mass of 44,000kg, and a rotational inertia about a vertical axis through its hingesof 8.7*10^4 kg*m^2, and has a front face of 2.4m. NEglecting friction, what steady force perpendicular to the door can move it from rest through an angle of 90 degrees in 30s?
My incorrect solution: ("@" refers to the angle in radian measure), and "&" refers to angular acceleration)
@ = 2(pi)*r/4 @=2.4(pi)/2 radians
@ = .5(&)(t^2)
1.2(pi) = .5(&)(30^2)
& = .0084 rad/s^2
I(&) = r x F = rFsin90
(8.7*10^4)(.0084) = (2.4)(F)(1)
F= 304.5 N
However, the actual answer is apparently 130N. What did I do wrong?
