A door of width 1 = 1.00 m and mass M 15.0 kg is attached to a door frame by two

Question

A door of width 1 = 1.00 m and mass M 15.0 kg is attached to a door frame by two hinges. For this problem, you may ignore gravity, as we are interested in rotational motion around the vertical axis. There is no friction of any kind. An angry mother-in-law slams the door shut, by pushing at the middle of the door (1/2 fromthe hinges) with a force of F 100N, lasting a time t = 0.200 s. The door is initially not rotating. The door can be taken to be a uniform rod, for the purpose of this exercise a) What is the angular acceleration of the door while it is being pushed? b) What is the resulting angular velocity, angular momentum and rotational kinetic energy from this push? c) Assuming that she lets go of the door (leaving it to slam shut) at the moment in the motion when the door is perpendicular to the wall, how long does it take for the door to close? d) What would be the result of a), b), c) and d) if she had pushed not in the middle of the door but at the edge ( from the hinges)? In addition to the force of the mother-in-law, the hinges also provide force both radial (centripetal) and tangential force while the mother-in-law pushes (they also compensate for gravity to keep the door upright, but ignore that for now) e) Combining angular acceleration and linear acceleration considerations, find the tangential force Fh supplied by the hinges as a function of d, the distance between the hinges and the point of application of the mother-in-law force (d was 1/2 in part a) and b), and n part c); now use a general d). Don't put in explicit numbers, just find the equation. f) For which d do the hinges not need to provide any tangential force?

Explanation / Answer

given door width, l = 1 m

mass, M = 15 kg

attached to door frame with two hinges, frictionless rotation

Force F = 100 N

distance of force from the hinges = l/2

time for which the force lasts, dt = 0.2 s

a. if the door is taken as a uniform rod

momen of inertia of the door wrt hinges = ml^2/3 = I

angular acceleration = alpha

hence I*alpha = F*l/2

ml^2*alpha/3 = F*l/2

alpha = 3F/2ml = 3*100/2*15*1 = 10 rad/s/s

b. initial angular velocity = 0

final angular velocity = w

w = 0 + alpha*dt = 10*0.2 = 2 rad/s

resulting angular momentum = ml^2*w/3 = 15*2/3 = 10 kg m^2 rad/s

final rotational kinetic energy = 0.5*ml^2*w^2/3 = 15*2^2/6 = 5*2 = 10 J

c. intial angular speed = 2 rad per sec

angular displacement = pi/2 rad

time taken = t

hence

pi/2 = 2*t

t = pi/4 s = 0.785 s

d. if she had pushed on the edge

alpha = 20 rad/s/s

w = 4 rad/s

angular momentum = 20 kg m^2 rad/s

final rotational KE = 40 J

t = pi/8 = 0.392 s

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