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= 250 g and length l= 30 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)
(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)
(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?
please de others¡¡¡
a)
I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
Eini=Efin ->
w=sqrt(3*g (1-cos(theta))/L)
b)
alpha=3*mg/2*sin(theta)/L
ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
Fx=m*ax
Fy=m*g*cos(theta)-m*w^2*(L/2)
c)
cos(theta)=2/3
theta=48.19
To solve this problem, we can use the principle of conservation of angular momentum.
Angular momentum is defined as L = I * ω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for the ruler about its pivot point can be calculated using the parallel axis theorem: I = (1/3) * m * L^2, where m is the mass of the ruler and L is the length of the ruler.
(a) To find the angular speed of the ruler when it is at an angle θ = 30°, we need to find its angular momentum at that angle. We can use the conservation of angular momentum:
L_initial = L_final
For the initial position at θ = 0°, the angular momentum is zero since the initial angular velocity is zero.
0 = I_initial * ω_initial
For the final position at θ = 30°, we can calculate the moment of inertia as:
I_final = (1/3) * m * L^2
Substituting the values given:
I_final = (1/3) * 0.250 kg * (0.30 m)^2 = 0.075 kg•m^2
We can now solve for ω_final:
0 = 0.075 kg•m^2 * ω_final
Thus, ω_final = 0 rad/s. The angular speed of the ruler when it is at an angle θ = 30° is 0 radians/second.
(b) To find the force exerted by the wall on the ruler at θ = 30°, we need to consider the torque acting on the ruler at that angle. Torque is defined as τ = I * α, where α is the angular acceleration.
Since the ruler is under the influence of gravity, which creates a restoring torque, the net torque acting on the ruler is given by:
τ_net = τ_gravity
τ_net = m * g * L * sin(θ)
Substituting the values given:
τ_net = 0.250 kg * 10 m/s^2 * 0.30 m * sin(30°) = 0.375 N•m
Since τ = I * α, we can express α in terms of ω:
α = (dω/dt) = (dω/dθ) * (dθ/dt) = ω * (dω/dθ)
Using the chain rule of differentiation, we can rearrange the equation as:
(dω/dθ) * (dθ/dt) = α/ω
Now we can solve for α:
α = (τ_net / I) = (m * g * L * sin(θ)) / I
Substituting the values given:
α = (0.250 kg * 10 m/s^2 * 0.30 m * sin(30°)) / 0.075 kg•m^2 = 1.667 rad/s^2
Since the force exerted by the wall is perpendicular to the ruler, we can write:
F = m * g * sin(θ)
Substituting the values given:
F = 0.250 kg * 10 m/s^2 * sin(30°) = 1.250 N
The force exerted by the wall on the ruler when it is at an angle θ = 30° is 1.250 N.
To find the x and y components of the force, we can use trigonometry. The x component is given by:
Fx = F * cos(θ)
Substituting the values given:
Fx = 1.250 N * cos(30°) = 1.250 N * √3 / 2 ≈ 1.083 N
The y component is given by:
Fy = F * sin(θ)
Substituting the values given:
Fy = 1.250 N * sin(30°) = 1.250 N * 1/2 = 0.625 N
Thus, the x component of the force exerted by the wall on the ruler when it is at an angle θ = 30° is approximately 1.083 N and the y component is 0.625 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 = 0. Using the expression for F, we can solve for θ:
0 = m * g * sin(θ)
Substituting the values given:
0 = 0.250 kg * 10 m/s^2 * sin(θ)
sin(θ) = 0
This occurs when θ = 0° or θ = 180°. However, since the ruler starts falling from θ = 0°, it loses contact with the wall at θ = 180°. Therefore, the falling ruler loses contact with the wall at an angle θ0 = 180°.