Omega Shield

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Torque and Force?

I can't figure out what I am doing wrong....

The massive shield door at a neutron test facility at Lawrence Livermore Laboratory; this is the world's heaviest hinged door. The door has a mass of 44,000 kg, a rotational inertia about an axis through its hinges of 8.7 104 kg·m2, and a (front) face width of 2.4 m. Neglecting friction, what steady force, applied at its outer edge and perpendicular to the plane of the door, can move it from rest through an angle of 70° in 30 s? Assume no friction acts on the hinges.

So I found the angular velocity by taking
Omega= (70 degrees * (2pie/360))/30 s
= .0407 rads/s
Found angular acceleration
Alpha=.0407 rads/s /30 s
A=.0014 rads/s^2
Then torque
Rot. Inertia*Alpha=radius* Fsin of theta
8.7E4 kg/m^2 * .0014 rad/s^2=2.4m *Fsin70degrees
solved for F
F= 52.3650 N
Wrong answer
I know i never factored in the 44,000kg so....

Good news! You don't need the 44000 kg...
Bad: You assumed a constant ω; this problem is about accelerated motion.

Θ = ½αt² → α = 2Θ/t² = 2*(70*π/180)/30² = 2*1.222rad/900sec²

α = .002715 rad/sec²

T = I*α = 8.7E4 * .002715 = 236.2 N∙m

F = T/r = 236.2/2.4 = 98.417 N (about 10 kgf)