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Haveagoodday
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1. Homework Statement
A door of width l = 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 (l=2 from the hinges) with a force of F = 100N, lasting a time
t = 0:.00 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 let's 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 (l 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 l=2 in part a) and b), and l in 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?
The Attempt at a Solution
Can someone check on my solutions, All answers appreciated![/B]
a)
τ= l/2*F=50
τ=M*(l/2)^2*α
α=τ/M*(l/2)^2= 13.33 rad/s^2
b)
ω=ωi+αt=2.7 rad/s
I=(1/12)*Ml^2=1.25 kg*m^2
L=I*ω=3.375 kg*m^2/s
K=(1/2)*I*ω^2=4.56 J
c)
θ=θ+ωt+(1/2)αt^2
π/2=2.7t+(1/2)13.33t^2
used the abc rule
t=0.323 s
d)
τ= l*F=100
τ=M*l^2*α
α=τ/M*l^2= 6.7 rad/s^2
ω=ωi+αt= 1.34 rad/s
I=(1/3)*Ml^2=5 kg*m^2
L=I*ω=6.7 kg*m^2/s
K=(1/2)*I*ω^2=4.489 J
θ=θ+ωt+(1/2)αt^2
π/2=1.34t+(1/2)6.7t^2
used the abc rule
t=0.512 s
e)
This solution i am sure is wrong
Fh(d)= τ(d)/I + F(d)/M
d)
d=l