A professor\'s office door is 0.91 m wide, 2.9 m high, and 4.0 cm thick; has a m

Question

A professor's office door is 0.91 m wide, 2.9 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots of frictionless hinges. A "door closer" is attached to the door and the top of the doorframe. When the door is open and at rest, the door closer exerts a torque of 5.21 N*m. The least amount of force to hold it open is F= 5.71 N.

a.) what is the moment of the inertia of the door about its axis through the hinges?

b.) what is the angular acceleration of the door once you let go of it?

Please show work! Thanks!

Explanation / Answer

mass m = 25 kg ,
l = 0.91 m ,
h = 2.9 m
t = 0.04 m

moment of inertia = m*l^2 /3 + m*t^2 /3

 = (25/3)*(0.91^2 + 0.04^2)

= 6.9 kg.m^2

 

b) torque = 5.21 = I*alpha

=> alpha = 5.21/6.9 rad/s^2

            =  0.75 rad/s^2

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