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.
W = 4.48
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)
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Please help
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?
Note
moment of inertia about cg rather than about end is (1/12)mL^2
following those equations I got a green mark for A.
But any idea about how b and c can be done?
To find the force exerted by the wall on the ruler when it is at an angle θ=30∘, we can use the equations of rotational dynamics.
First, let's calculate the moment of inertia (I) of the ruler. Since the ruler has a uniform mass distribution, the moment of inertia for a thin rod rotating about one end is given by:
I = (1/3) * m * l^2
Substituting the given values, we have:
I = (1/3) * 0.2 kg * (0.2 m)^2
I = 1/3 * 0.2 kg * 0.04 m^2
I = 0.0024 kg * m^2
Next, we can calculate the angular acceleration (α) of the ruler using Newton's second law for rotation:
τ = I * α
The torque (τ) is given by the gravitational force acting at the center of mass of the ruler and its perpendicular distance to the rotation axis. At an angle θ, the torque is given by:
τ = m * g * l/2 * sin(θ)
Substituting the given values, we have:
τ = 0.2 kg * 10 m/s^2 * 0.1 m * sin(30∘)
τ = 0.2 kg * 10 m/s^2 * 0.1 m * 0.5
τ = 0.01 N * m
Since the moment of inertia is constant, we can differentiate the equation τ = I * α to find the angular acceleration (α):
α = τ / I
α = 0.01 N * m / 0.0024 kg * m^2
α = 4.16 rad/s^2
Now, using the equation of rotational motion:
θ = ω * t + (1/2) * α * t^2
Since the initial angular velocity (ω) is given as zero and we want to find the time (t) taken to reach θ = 30∘, we can rearrange the equation to solve for t:
30∘ = (1/2) * 4.16 rad/s^2 * t^2
Simplifying:
t^2 = 30∘ / (2 * 4.16 rad/s^2)
t^2 = 7.21 s^2
Taking the square root:
t = 2.685 s
Now, we have the time it takes for the ruler to reach θ = 30∘. We can use this to find the angular velocity at that angle:
θ = ω * t
30∘ = ω * 2.685 s
ω = 30∘ / 2.685 s
ω = 11.17 rad/s
Finally, we can find the force exerted by the wall on the ruler using the equation:
F = I * α
Substituting the given values, we have:
F = 0.0024 kg * m^2 * 4.16 rad/s^2
F = 0.01 N
However, the question asks for the x and y components of the force.
Since the ruler is at an angle θ=30∘, the x component of the force (Fx) will be zero, as there is no horizontal force acting on the ruler.
The y component of the force (Fy) will be equal to the gravitational force acting on the ruler, which can be calculated as:
Fy = m * g * cos(θ)
Fy = 0.2 kg * 10 m/s^2 * cos(30∘)
Fy = 0.2 kg * 10 m/s^2 * 0.866
Fy = 1.732 N
Therefore, the force exerted by the wall on the ruler at θ=30∘ is Fy = 1.732 N in the y direction, and there is no force in the x direction.