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= 25 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.

(a) What is the angular speed of the ruler ω when it is at an angle θ=30∘? (in radians/sec)

Apply conservation of energy as there is no friction.

Would i take it from the center of mass?

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 aswer the other questions?

To find the angular speed of the ruler when it is at an angle of 30 degrees, we can apply the principle of conservation of angular momentum. The angular momentum of an object is conserved when no external torques are applied to it.

The angular momentum, L, can be defined as the product of the moment of inertia, I, and the angular velocity, ω. For a uniform rod rotating about one end, the moment of inertia can be calculated using the formula:

I = (1/3) * m * L^2,

where m is the mass of the rod and L is its length.

Given that the mass of the ruler is 250 g (or 0.25 kg) and the length is 25 cm (or 0.25 m), we can calculate the moment of inertia:

I = (1/3) * (0.25 kg) * (0.25 m)^2
= 0.0052 kg.m^2.

At θ = 30°, the ruler has rotated through an angle of 30 degrees. The angular speed, ω, can be calculated by equating the initial and final angular momenta using the conservation principle:

I * ω(θ = 0°) = I * ω(θ = 30°).

Since the initial angular velocity is given as zero, we can solve for the final angular velocity, ω(θ = 30°):

0 = I * ω(θ = 30°).
ω(θ = 30°) = 0 / I
= 0 rad/s.

Hence, the angular speed of the ruler when it is at an angle of 30 degrees is 0 radians/sec.