A 30.0 kg door is 1.00 m wide, as shown in the diagram. A force of 85.0 N is applied is applied 0.750 m from the hinges at a 60° angle from the door. What is the angular acceleration with which the door swings open? (Hint: The moment of inertia of the door can be calculated as a rod of uniform mass rotating from the end:  .) a. 5.52 rad/s2 b.1.84 rad/s2 c.6.38 rad/s2 d.2.60 rad/s2

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A 30.0 kg door is 1.00 m wide, as shown in the diagram. A force of 85.0 N is applied is applied 0.750 m from the hinges at a 60° angle from the door. What is the angular acceleration with which the door swings open? (Hint: The moment of inertia of the door can be calculated as a rod of uniform mass rotating from the end:  .)

a. 5.52 rad/s2

b.1.84 rad/s2

c.6.38 rad/s2

d.2.60 rad/s2

d = 0.750 m
Hinge
0 = 60%
F = 85.0 N
L = 1.00 m
Transcribed Image Text:d = 0.750 m Hinge 0 = 60% F = 85.0 N L = 1.00 m
Expert Solution
Step 1

from newtons second law of rotation we have 

τ=Iατ is torque about hingeI is moment of inertiaα angular accelerationI = ML23=10 kgm2τ=Fdsin60 =55.21 Nm

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