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= 100 g and length l= 35 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=
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(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=
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Fy=
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(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=
To solve this problem, we need to apply the principles of rotational motion and equilibrium. We can start by finding the angular speed (omega, ω) of the ruler at an angle of θ = 30°.
(a) To find the angular speed, we can use the conservation of mechanical energy. The potential energy at the initial position (θ = 0°) is zero, and at any angle θ, the potential energy is given by mgh, where m is the mass of the ruler, g is the acceleration due to gravity, and h is the vertical distance between the ruler's center of mass and the initial position.
The potential energy can be converted into rotational kinetic energy when the ruler is at angle θ. The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia of the ruler and ω is the angular speed.
Since the ruler has a uniform mass distribution, its moment of inertia can be calculated using the formula I = (1/3)ml^2, where m is the mass of the ruler and l is its length.
The potential energy at angle θ is given by mgh = (1/2)Iω^2. Rearranging the equation, we can solve for ω:
ω^2 = (2mgh) / I
Substituting the given values:
- m = 100 g = 0.1 kg
- g = 10 m/s^2
- l = 35 cm = 0.35 m
ω^2 = (2 * 0.1 kg * 10 m/s^2 * 0.35 m * sin(30°)) / ((1/3) * 0.1 kg * (0.35 m)^2)
Simplifying the expression:
ω^2 = 2 m/s^2
Taking the square root of both sides:
ω = √2 rad/s
So, the angular speed of the ruler at θ = 30° is ω = √2 rad/s.
(b) To find the force exerted by the wall on the ruler at θ = 30°, we need to consider the forces acting on the ruler. There are two forces involved: the gravitational force acting on the center of mass of the ruler and the force exerted by the wall.
The gravitational force (mg) can be split into two components: Fgx (horizontal) and Fgy (vertical). The component Fgx does not contribute to the torque, so we can ignore it in this problem.
The force exerted by the wall has two components as well: Fwx (horizontal) and Fwy (vertical). The horizontal component Fwx balances the horizontal component Fgx, so Fwx = -Fgx = 0.
The only force left is the vertical component Fwy, which is the force exerted by the wall on the ruler in the y-direction. At equilibrium, Fwy balances the vertical component Fgy:
Fwy = Fgy
Fwy = mg * cos(θ)
Substituting the given values:
- m = 100 g = 0.1 kg
- g = 10 m/s^2
- θ = 30°
Fwy = 0.1 kg * 10 m/s^2 * cos(30°)
Simplifying the expression:
Fwy = 0.1 kg * 10 m/s^2 * √3 / 2
Fwy = √3 N
So, the y-component of the force exerted by the wall on the ruler at θ = 30° is Fy = √3 N.
Since there is no horizontal force, the x-component of the force exerted by the wall is Fx = 0 N.
(c) To find the angle θ0 at which the ruler loses contact with the wall, we must determine when the force exerted by the wall (Fwy) becomes zero.
Fwy = 0 N
0.1 kg * 10 m/s^2 * cos(θ0) = 0
cos(θ0) = 0
θ0 = 90°
Therefore, the falling ruler loses contact with the wall at θ0 = 90°.