Question

Sakai @ PC: Gateway: Welcome Google Slides Rotational Motion and Oscillations - EPS 101, section o -/4 POINTS OSUNIPHYS1 10.7.WA.057. MY NOTES ASK YOUR TEACHER A heavy swing door has a mass of m - 4,000 kg, a width w = 1.4 m, and a height H - 2.9 m. The door swings about a vertical axis passing through its center. The moment of inertia of this door about the vertical axis of rotation is given by I m w. (a) You stand on one side and push at the outer edge of the door and perpendicular to the face of the door with a force of F-16.0 N. Your friend pushes on the other outer edge of the door from the opposite side with the same force. What is the net torque applied to the door? Nm (b) What is the angular acceleration of the door? rad/s2 (c) Now consider a heavy door, with the same mass and dimensions as the one in part (a), that swings about a vertical axis passing through one long edge as shown in the diagram below. The moment of inertia of the door about this new axis of rotation is given by I mw?, where m is the mass and w is the width of the door

Answer 1

Mass of the doout is - 4000 kg w = 1.4 m. H = 2.9 m. of inertia of the door about moment vrotational anis is T I mw? 13 (a) F = 16 N. 1310 to the applied door is net touique I = F. W

16 x 1.4 II - 11.2 Nm of the doout I of acculeration a be the angular thim from condition of urotation T = Id 11.3 I mw? 11.2 I x4000 x (1.4) 12 rad / 2 0.0171 Lut I be the appeleied fortal.

met tourque about the anis of violation is T = F xw from part (0) T = 14.2 N.m. 11.2 = F x 1.4 IN 1 (d) Lut of the doout 2 be condition the angular acculeration of wrotation. from T = a' - I & I T - I'mwa I 2 - 11.2 1 x 4000x (1.4) - 4.285*10*) adla 4.385 X 103