What Is the Angular Speed of a Falling Ruler at 30 Degrees?

[Ruitker]

Homework Statement

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)

The angle is from the vertical.

Homework Equations

I'm not sure on which method to use. I was thinking either conservation of energy or using the centre of mass equations.

The Attempt at a Solution

Various attempts that have not yielded correct answers.

 

[HallsofIvy]

Personally, I wouldn't use either "conservation of energy" or "center of mass". I would apply the equations of motion to the end of the ruler. You know that the downward acceleration is always "-g" (-9.82 m/s^2). But the rule cannot fall straight because the ruler is "rigid". So separate the vector <0, -g> into one component along the length of the ruler and one at right angles to the ruler at each angle. Only the perpendicular component acts accelerates the end of the ruler.

 

[Ruitker]

Personally, I wouldn't use either "conservation of energy" or "center of mass". I would apply the equations of motion to the end of the ruler. You know that the downward acceleration is always "-g" (-9.82 m/s^2). But the rule cannot fall straight because the ruler is "rigid". So separate the vector <0, -g> into one component along the length of the ruler and one at right angles to the ruler at each angle. Only the perpendicular component acts accelerates the end of the ruler.

Thanks for your suggestions. I am not sure i understand what you mean. Shall i calculate the velocity and then convert to angular velocity or is there a trick i am missing.

 

[Tanero]

I would apply the equations of motion to the end of the ruler.

Isn't it better to write equation of motion for center of mass?
We need to find ω(θ) relationship. And to do this. Well, I am stuck too with this problem.

 

[HalfLostAndSearching]

I would approach this problem by first identifying the relevant physical principles and equations that can be applied to the situation. In this case, we are dealing with the motion of a falling ruler under the influence of gravity. This can be described by the equations of rotational motion, specifically the equation for angular velocity:

ω = ω0 + αt

Where ω is the final angular velocity, ω0 is the initial angular velocity (which is given as 0 in this problem), α is the angular acceleration, and t is the time.

To find the angular velocity at θ = 30∘, we need to determine the angular acceleration of the ruler. This can be done by using the equation for torque:

τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Since the ruler has a uniform mass distribution, we can use the equation for the moment of inertia of a rod:

I = (1/3)ml^2

Plugging this into the equation for torque, we get:

τ = (1/3)ml^2α

We know that the only force acting on the ruler is gravity, so we can write the torque as:

τ = mgl sinθ

Where m is the mass of the ruler, g is the acceleration due to gravity, l is the length of the ruler, and θ is the angle from the vertical.

Setting these two equations equal to each other, we get:

mgl sinθ = (1/3)ml^2α

Simplifying and solving for α, we get:

α = (3g/2l)sinθ

Now, we can plug this value for α into the equation for angular velocity and solve for ω:

ω = ω0 + (3g/2l)sinθ t

Since ω0 = 0, this simplifies to:

ω = (3g/2l)sinθ t

We can use the given information to solve for t:

θ = ω0t + (1/2)αt^2

Since ω0 = 0, this simplifies to:

θ = (1/2)αt^2

Plugging in the values for θ and α, we get:

30∘ = (1/2)(3g/2l)sinθ t^2