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)
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?
Can you please elaborate.. thanx
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 find the angular speed of the ruler when it is at an angle θ=30∘, we can use the principle of conservation of mechanical energy.
(a) Using the conservation of mechanical energy:
The initial mechanical energy is given by:
E_initial = mgl
The final mechanical energy is given by:
E_final = (1/2)Iω^2 + mglcosθ
where:
m = mass of the ruler = 250 g = 0.25 kg
l = length of the ruler = 30 cm = 0.3 m
g = gravitational acceleration = 10 m/s^2
θ = angle θ = 30∘
I = moment of inertia of the ruler
The moment of inertia of a uniform rod rotating about one end is given by:
I = (1/3)mL^2
where L is the length of the rod. In this case, L = 0.3 m.
Plugging in these values, we have:
E_initial = (0.25 kg)(10 m/s^2)(0.3 m)
E_final = (1/2)(1/3)(0.25 kg)(0.3 m)^2ω^2 + (0.25 kg)(10 m/s^2)(0.3 m)cos(30∘)
Since the ruler is released from rest (ω(θ=0∘) = 0), the angular speed ω when the ruler is at θ = 30∘ is such that:
E_initial = E_final
Simplifying the equation:
(0.25 kg)(10 m/s^2)(0.3 m) = (1/6)(0.25 kg)(0.3 m)^2ω^2 + (0.25 kg)(10 m/s^2)(0.3 m)(√3/2)
0.75 kg*m^2/s^2 = (1/6)(0.075 kg*m^2)ω^2 + 0.75 kg*m^2/s^2(√3/2)
0 = (1/6)(0.075 kg*m^2)ω^2 + (0.75 kg*m^2/s^2)(√3/2) - 0.75 kg*m^2/s^2
We can solve this quadratic equation to find the angular speed ω.
(b) To find the force exerted by the wall on the ruler when it is at an angle θ=30∘, we can use Newton's second law for rotation. The torque acting on the ruler is equal to the moment of inertia times the angular acceleration.
τ = Iα
The torque acting on the ruler can be calculated using the gravitational force and the distance from the pivot point to the center of mass of the ruler.
τ = mglsinθ
We know that α = dω/dt, and since dθ/dt = ω, we can rewrite the torque equation as:
τ = I(dω/dt)
Hence, mglsinθ = I(dω/dt)
From part (a), we have ω. So, we can differentiate that equation with respect to time to find dω/dt. You can solve for I, which is the moment of inertia of the ruler, using the formula I = (1/3)mL^2 where L is the length of the ruler.
Knowing ω, we can calculate dω/dt and substitute it back into the torque equation to find the force exerted by the wall on the ruler when it is at an angle θ=30∘, which can be calculated using F = τ/dL.
(c) To find the angle θ0 at which the falling ruler loses contact with the wall, the force exerted by the wall on the ruler must vanish. This occurs when the ruler starts to become fully vertical and the force becomes zero. At this point, the contact force is solely provided by gravity. Hence, we can equate the gravitational force acting at this angle to the force exerted by the wall to find the value of θ0. The gravitational force acting at this angle can be calculated using F_g = mgcosθ0. Equating this to zero and solving for θ0 will give us the angle at which the ruler loses contact with the wall.