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=

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?

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.

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

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.

Can you guys help me with questions 2 and 3 ? I just have one more try but i am really confused in this two ;(

To solve this problem, we can use the principles of rotational motion and Newton's laws. Let's go step by step:

(a) To determine the angular speed of the ruler when it is at an angle θ=30∘, we can use the principle of conservation of angular momentum. The initial angular momentum is zero (ω=0), and the final angular momentum can be calculated using the moment of inertia of the ruler.

The moment of inertia of a uniform rod rotating about one end is given by I = (1/3) * m * l², where m is the mass and l is the length of the rod.

In this case, m=200g and l=20cm. Let's convert them to SI units:
m = 200g = 0.2kg
l = 20cm = 0.2m

Plugging these values into the equation for moment of inertia, we have:
I = (1/3) * 0.2 * 0.2²
I = 0.004 kg·m²

Now, we can use the conservation of angular momentum:
I₁ * ω₁ = I₂ * ω₂

Since ω₁=0, we can solve for ω₂:
0 * 0 = 0.004 * ω₂
0 = 0.004 * ω₂
ω₂ = 0

Therefore, the angular speed of the ruler at an angle θ=30∘ is 0 radians/sec.

(b) To find the force exerted by the wall on the ruler when it is at an angle θ=30∘, we need to analyze the forces acting on the ruler at that position. There are two forces: the gravitational force and the normal force exerted by the wall.

The gravitational force can be calculated using:
F_grav = m * g
F_grav = 0.2 * 10
F_grav = 2 N

The normal force exerted by the wall can be decomposed into its x and y components. The y component, F_y, will be equal to the gravitational force:
F_y = F_grav
F_y = 2 N

The x component, F_x, will be 0 since there is no force acting horizontally on the ruler.

Therefore, the force exerted by the wall on the ruler when it is at an angle θ=30∘ is:
Fx = 0 N
Fy = 2 N

(c) The ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes. In other words, when F_x and F_y both become zero.

From part (b), we know that F_x = 0. Therefore, we need to find the angle θ for which F_y becomes zero.

When the ruler is at an angle θ, the gravitational force can be decomposed into its x and y components. The y component is given by F_y = m * g * cosθ. Setting this to zero, we have:
F_y = 0
m * g * cosθ = 0

Since m * g is nonzero and constant, cosine of θ must be zero. This occurs when θ = 90∘.

Therefore, the falling ruler loses contact with the wall at an angle θ0 = 90∘.