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)

Question

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.]

Answer 1

Hi, did you solve rocket problem. I gopt it wrong. How to find? Can you help please?

Answer 2

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 answer the other questions?

Answer 3

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

(a) To find the angular speed, we can use the conservation of angular momentum. Angular momentum is defined as L = Iω, where I is the moment of inertia and ω is the angular speed. In this case, the ruler starts from rest, so its initial angular speed is 0.

The moment of inertia for a uniform rod rotating about one end is given by I = (1/3)mL^2, where m is the mass of the ruler and L is its length. Plugging in the given values, we have:
I = (1/3) * 0.200 kg * (0.20 m)^2 = 0.00267 kg⋅m^2

Now, at angle θ = 30°, we want to find the angular speed ω. To do that, we can use the conservation of angular momentum. Since there are no external torques acting on the system, the angular momentum is conserved.

L(initial) = L(final)
0 = I * ω(initial) = I * ω(final)

Solving for ω(final), we have:
ω(final) = ω(initial) / (1 + (I / m * L^2))

Plugging in the given values, we have:
ω(final) = 0 / (1 + (0.00267 kg⋅m^2 / (0.200 kg * (0.20 m)^2)))
= 0 rad/s (the ruler will not rotate)

So, the angular speed of the ruler at θ = 30° is 0 rad/s.

(b) To find the force exerted by the wall on the ruler, we can consider the forces acting on the ruler at θ = 30°. There are two forces: the weight of the ruler and the force exerted by the wall.

The weight of the ruler acts downward and is given by W = mg, where m is the mass of the ruler and g is the acceleration due to gravity. Plugging in the given values, we have:
W = 0.200 kg * 10 m/s^2 = 2 N

The force exerted by the wall has two components, Fx and Fy. Fx is the horizontal component of the force and Fy is the vertical component. Since the ruler is not rotating, Fx = 0.

Using Newton's second law in the vertical direction:
Fy - W = 0
Fy = W

So, the force exerted by the wall on the ruler at θ = 30° is Fx = 0 N and 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 Fy = 0.

Using Newton's second law in the vertical direction:
Fy - W = 0
Fy = W

Setting Fy equal to zero, we have:
0 = W
W = mg

So, the ruler loses contact with the wall when the weight of the ruler equals zero, which occurs when m = 0. Therefore, the ruler will never lose contact with the wall.