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.]
θ0=
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 find the answers to these questions, we need to apply the principles of rotational motion and equilibrium. Here's how we can approach each part:
(a) To find the angular speed of the ruler ω at an angle θ=30∘, we can use the conservation of mechanical energy. Since there is no friction and the ruler falls under the influence of gravity, the gravitational potential energy is converted to rotational kinetic energy. The equation for the conservation of mechanical energy is:
mgh = (1/2)Iω^2
where m is the mass of the ruler, g is the acceleration due to gravity, h is the height of the center of mass of the ruler, I is the moment of inertia of the ruler about its pivot point, and ω is the angular speed.
Given that the ruler has a uniform mass distribution, its center of mass is at the midpoint, so h = l/2 = 10 cm = 0.1 m. The moment of inertia of a thin rod rotating about one end is given by I = (1/3)mL^2, where L is the length of the rod. Plugging in the values, we have:
(0.2 kg)(10 m/s^2)(0.1 m) = (1/2)(1/3)(0.2 kg)(0.2 m)^2 ω^2
Simplifying the equation, we find:
ω^2 = (2/3)(10 m/s^2)(0.1 m) / (0.04 m)^2
ω^2 = (2/3)(250) / 0.0016
ω^2 = 312.5 rad^2/s^2 / 0.0016
ω^2 = 195312.5 rad^2/s^2
Taking the square root, we get:
ω = 441.36 rad/s
So, the angular speed of the ruler when it is at an angle θ=30∘ is 441.36 rad/s.
(b) To find the force exerted by the wall on the ruler at an angle θ=30∘, we need to consider the torque exerted by the gravitational force and set it equal to the torque exerted by the wall. Torque is given by the product of the force and the lever arm. The equation for torque is:
τ = Fl sin(θ)
where τ is the torque, F is the force, l is the lever arm (length of the ruler), and θ is the angle between the force and the lever arm.
The gravitational force acting on the ruler can be decomposed into its x and y components. The x component does not exert any torque since it acts along the line of action, perpendicular to the lever arm. The y component, however, exerts a torque. The magnitude of the y component of the force is given by Fy = mg sin(θ).
To find the force exerted by the wall, we can equate the torques:
τ_wall = τ_gravity
Fl_wall sin(θ) = Fl_gravity sin(θ)
Canceling out the lever arm and sin(θ), we find:
F_wall = F_gravity
Since the magnitude of the y component of the force is Fy = mg sin(θ), we have:
F_wall = Fy = mg sin(θ)
Plugging in the values, we have:
F_wall = (0.2 kg)(10 m/s^2) sin(30∘)
F_wall = 0.1 kg m/s^2
To find the x component of the force, we can use the fact that the net force in the x direction is zero, since there is no acceleration in that direction. Therefore, the x component of the force is equal in magnitude but opposite in direction to the x component of the gravitational force. The magnitude of the x component of the force is given by Fx = mg cos(θ).
Plugging in the values, we have:
Fx = (0.2 kg)(10 m/s^2) cos(30∘)
Fx = 0.1732 kg m/s^2
So, the force exerted by the wall on the ruler at an angle θ=30∘ is Fx = 0.1732 N (x component) and Fy = 0.1 N (y component).
(c) To find the angle θ0 at which the ruler loses contact with the wall, we need to determine when the force exerted by the wall on the ruler vanishes. When the ruler loses contact, the force exerted by the wall becomes zero.
From part (b), we found that F_wall = Fy = mg sin(θ). Setting F_wall = 0, we have:
0 = mg sin(θ0)
sin(θ0) = 0
θ0 = 0 degrees
Therefore, the falling ruler loses contact with the wall at θ0 = 0 degrees.