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.]
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?
To solve this problem, we can use the principles of rotational motion and Newton's laws. Let's break down each part of the problem step by step:
(a) To find the angular speed of the ruler at θ = 30°, we can use the principle of conservation of angular momentum. Angular momentum is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia of the ruler, and ω is the angular speed.
In this case, we can assume the ruler rotates around its center of mass, so we can use the moment of inertia of a uniform rod rotating about its center, which is given by I = (1/12)ml^2, where m is the mass of the ruler, and l is its length.
Plugging in the values given: m = 200 g = 0.2 kg, l = 0.2 m, we have:
I = (1/12)(0.2 kg)(0.2 m)^2 = 0.001333 kg·m^2
Since the initial angular velocity is very small (ω(θ = 0°) = 0), conservation of angular momentum tells us that Iω = I(ω(θ = 30°)), where ω(θ = 30°) is the angular velocity we want to find.
Using this equation, we can solve for ω(θ = 30°):
I × 0 = I × ω(θ = 30°)
0 = 0.001333 kg·m^2 × ω(θ = 30°)
ω(θ = 30°) = 0 rad/s (since the ruler is not rotating at θ = 30° due to the absence of initial angular velocity)
Therefore, the angular speed of the ruler when it is at θ = 30° is 0 rad/s.
(b) To find the force exerted by the wall on the ruler at θ = 30°, we can use Newton's second law for rotational motion. The torque τ exerted on an object is equal to the moment of inertia I multiplied by the angular acceleration α, where α is given by α = τ/I.
In this case, the only torque acting on the ruler is the gravitational torque τ = mgl/2 sin(θ), where m is the mass of the ruler, l is its length, g is the acceleration due to gravity, and θ is the angle at which the ruler is inclined.
Since the ruler is not rotating (ω = 0) at θ = 30°, the angular acceleration α will also be 0. Therefore, the net torque acting on the ruler at this angle is 0.
The force exerted by the wall is a horizontal force due to the absence of any vertical net force. Since the ruler isn't rotating, the force exerted by the wall only has an x-component. So, Fx = 0 N.
As there is no vertical force, the y-component of the force exerted by the wall is Fy = Weight of the ruler = mg = (0.2 kg) × (10 m/s^2) = 2 N. Therefore, Fy = 2 N.
(c) The ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes, which happens when Fx = 0 N. Since we found in part (b) that Fx = 0 N when the ruler is at θ = 30°, we can conclude that the ruler loses contact with the wall at θ = 30°.