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)

Question

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.]

fima · Answer

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?

socorro · Answer

please de others¡¡¡

Greco · Answer

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