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A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influen...Asked by Anonymous
                A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m= 200  g and length l= 20  cm. Use g=10 m/s2 for the gravitational acceleration, and the ruler has a uniform mass distribution. Note that there is no friction whatsoever in this problem. (See figure)
(a) What is the angular speed of the ruler ω when it is at an angle θ=30∘? (in radians/sec)
ω=
(b) What is the force exerted by the wall on the ruler when it is at an angle θ=30∘? Express your answer as the x component Fx and the y component Fy (in Newton)
Fx=
Fy=
(c) At what angle θ0 will the falling ruler lose contact with the wall? (0≤θ0≤90∘; in degrees) [hint: the ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes.]
θ0=
            
        (a) What is the angular speed of the ruler ω when it is at an angle θ=30∘? (in radians/sec)
ω=
(b) What is the force exerted by the wall on the ruler when it is at an angle θ=30∘? Express your answer as the x component Fx and the y component Fy (in Newton)
Fx=
Fy=
(c) At what angle θ0 will the falling ruler lose contact with the wall? (0≤θ0≤90∘; in degrees) [hint: the ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes.]
θ0=
Answers
                    Answered by
            fima
            
    apply conservation of energy:
U=m*g*hcm
EK= 1/2*I*w^2, I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
solve for Eini=Efin -> w=
 
any one knows how to aswer the other questions?
    
U=m*g*hcm
EK= 1/2*I*w^2, I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
solve for Eini=Efin -> w=
any one knows how to aswer the other questions?
                    Answered by
            C_P
            
    did this work for you? are you sure that we shouldn't use parallel axis theorem to find I? Because rod is rotating about end. And you assume that it's a point mass in the rotational motion.
    
                    Answered by
            an0nym801
            
    I don't think so.
omega comes out too low
I think part a) has to be done with energy conservation
U at the top - U at 30° = K of rod
    
omega comes out too low
I think part a) has to be done with energy conservation
U at the top - U at 30° = K of rod
                    Answered by
            fima
            
    the equation is Ok, and the w value got green checked. So YES, I'm sure about A. 
And I did for conservation of energy.
But anyone knows how to do B and C. And please, try to post the procedure, because we all have different data.
    
And I did for conservation of energy.
But anyone knows how to do B and C. And please, try to post the procedure, because we all have different data.
                    Answered by
            koala
            
    Can you guys help me with questions 2 and 3 ? I just have one more try but i am really confused in this two ;(
    
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